Mathematics-in-modern-world.pptx

Our Lady of Fatima University Valenzuela • Quezon City • Antipolo • Pampanga • Cabatuan

Mathematics in the Modern World Authors :

Clarenz Dalisay, LPT Engr. Elise Erandio Renato Nieva, LPT, MA Rey-ann Rivera, LPT, MSMath(c) Engr. Ernesto Villarica

Grace Ann Delos Reyes, LPT, MATMath Cheryl Lumiguen, LPT, MAT Melanio Nipas Jr, LPT, MA Engr. Angelica Sicat

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PREFACE This is worktext is a collaborative efforts of the different faculty members from Mathematics and Physics Department of all campuses of this university. This will help the College students to appreciate Mathematics as an important tool in their lives. This work will also give them a new perception of Mathematics, not as just pure calculation but instead a body of knowledge that can be seen everywhere and add colors in their lives. This worktext is divided into two parts that covers the mandated and elective topics that are identified by the Commission of Higher Education (CHED) in the GE Course.

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Copyright, 2018 by: Clarenz Dalisay, LPT Engr. Elise Erandio Renato Nieva, LPT, MA Rey-ann Rivera, LPT, MSMath(c) Engr. Ernesto Villarica

Grace Ann Delos Reyes, LPT, MATMath Cheryl Lumiguen, LPT, MAT Melanio Nipas Jr, LPT, MA Engr. Angelica Sicat

Disclaimer: All literary works that appear on the book are copyrighted by their respective owners. We claim no credit for them unless otherwise noted. If you own the rights to any of the works and do not wish them to appear on the book, please contact us and they will be promptly removed. 4

TABLE OF CONTENTS Title Page Disclaimer Preface Table of Contents UNIT 1

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Page Nature of Mathematics: Mathematics in our 4 World 1.1 Patterns and Numbers in Nature and the 4 World 2. Fibonacci Sequence 8 3. Golden Ratio

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Nature of Mathematics: Mathematical

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Language and Symbols

3

2.1 Language of Mathematics

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2. Sets

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3. Elementary Logic

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Nature of Mathematics: Problem

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Solving and Reasoning

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4

5

6

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1. Inductive Reasoning

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2. Deductive Reasoning

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3. Polya’s Strategy

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Mathematics as a Tool: Data Management

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1. Statistics, Origin and Terms

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2. Measure of Central Tendency

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3. Measure of Variation

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4. Quartile, Decile and Percentile

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5. Normal Distribution

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6. Correlation and Linear Regression

86

Mathematics as a Tool: Codes and Cryptography 5.1 Coding Theory

95

5.2 Cryptography

115

Mathematics as a Tool: Apportionment and Voting 6.1 Apportionment

127

6.2 Voting Methods

141

Mathematics as a Tool: Graph Theory

158

7.1 Basic Terms

158

7.2 Applications of Graph Theory

181

95

127

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CHAPTER

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NATURE OF MATHEMATICS: MATHEMATICS IN OUR WORLD

LEARNING OUTCOMES: At the end of the chapter the students are expected to : 1. Argue about the nature of mathematics, what it is, how it is expressed, presented and used. 2. Express appreciation for mathematics as human endeavor.

LESSON 1 : PATTERNS AND NUMBERS IN NATURE AND THE WORLD Humans developed a system for recognizing thoughts, classifying and organizing patterns which are vital clues to the rules governing natural processes. It is called Mathematics. Early Greek Philosophers studied different patterns in attempt to explain the order of nature. The humanity became explicitly aware of the two types of patterns known as fractals and chaos. Fractals are geometric shape that repeat their structure on an even finer scale, while chaos shows apparent randomness whose origins are entirely deterministic. Pythagoras explained the harmonies in music as arising from a number. Plato considered that there is a universal pattern or ideal forms to which objects in nature are imperfect copies. For example, a flower can be circular but not a mathematical circle. World, by nature, is a combination of patterns and they can be seen by careful observations in spite of regularities and irregularities. Any form of these observations can be modelled mathematically. These natural patterns include symmetry, spirals, tessellation, stripes, and so on.

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Waves and dunes are clues to the flow of water, sand and air. Regular nightly motion of stars also confirms that the Earth rotates, and that rotation was used by ancients to predict time. Rainbows tells us the scattering of light, indirectly confirming that raindrops are spheres. The tiger’s stripes and the hyena’s spots attest to mathematical regularities in biological growth and form. Animals have a mirror symmetry, so do leaves and some flowers. Echinoderms like starfish, sea urchins and sea lilies have five-fold symmetry. Snowflakes have six-fold symmetry, and a water splash approximates a radial symmetry. Considering numbers as simplest mathematical objects, a year is roughly three hundred sixty-five days, people have two legs, cats have four while spiders have eight. Have you gone for a trip such as mountain climbing and noticed the world around you? The differences in leaves, in hues of the sky, the transformations and movements of clouds, how high a rabbit can jump nor how fast a turtle run? Have you ever wondered how much water one must take? Are you watching your weight and your food caloric intake? Or how much time you allot in preparing things for such a trip? Have you not wondered how most vendors do not make a mistake in giving your change after buying even without calculators? Consciously or unconsciously, all of these activities engage some form of mathematics. Mathematics aside from being related to numbers, is also considering symbols, notations, operations and functions thus providing new questions to think about. Pythagoreans obsession with number was not baseless. Too many ‘coincidences’, too many connections—between number and number, number and shape, number and music—led them to re-examine the accepted world view of their day. Numbers, as originally conceived, were the pragmatic progeny of accounting and commerce; but slowly, as they ‘grew’, were seen by the Pythagoreans (and others) as having a meaning that stretched far beyond the mere representation of ‘quantities’ and the collections of material objects that had spawned their introduction. The Pythagorean vision marked a fundamental turning point in the attitude not only to numbers, but indeed to nature itself. Nature has its laws. Natural laws are not like laws in our society. Societal laws concerns human actions and are determined by governing bodies. Natural laws on the other hand are determined by fundamental forces in the nature. These arises from careful application of systematic studies of the natural world trough experimentation and observation called as scientific methods. This method provides scientist a rigorous framework to objectively

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study the natural world. Natural laws can be verified through experiments conducted by independent observers. The law can be a simple statement in words, as well as a mathematical equation. Examples of these natural laws are (1) Laws of Planetary Motion by Johannes Kepler, (2) Law of Universal Expansion by Edwin Hubble, (3) Laws of Motion and Universal Gravitation by Sir Isaac Newton, (4) Fluid Mechanics by Sir George Stokes, and even the Theory of Relativity of Albert Einstein would not have gone farther without mathematics. Many breakthroughs in the field of sciences and engineering are already evident. Theoretical investigations for treatment procedures by modelling and simulation of biological processes were made possible through creative employment of mathematics. Improvements in Information and Communication Technology also contributed to the fast pacing on human endeavours in terms of communications, commerce and lifestyles. However, though “millennials” are hesitant in considering courses that involve mathematics, they are unknowingly applying mathematics as they show their interests into “being connected” as they use gadgets and other technologies appealing to their senses. One can never deny the fact that mathematics is everywhere not only because it can be seen in nature but also because of its practical applications in our daily lives.

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FIBONACCI SEQUENCE Fibonacci sequence, named after Leonardo Pisano or Leonardo of Pisa, is a series of numbers where a number is found by adding the two numbers before it. Starting with 0 and 1, the succession goes on 0,1,1,2,3,5,8,13,21,34 and so on. As a trivia, Fibonacci is a shortened word for the Latin term “filiusBonnaci” which means “son of Bonaccio”. His father’s name was Guglielmo Bonnacio who was a wealthy Italian merchant, representing the Republic of Pisa trading in Bugia, Algeria in North Africa. Fibonacci was an extensive traveller around Mediterranean and met numerous merchants. He studied different numerical systems and methods, including the Hindu- Arabic numeral system which he learned from North Africa and eventually making it as popular to Europe through his Liber Abaci (Book of Calculation). Today, the system is the most common representation of numbers. The original problem Fibonnacci investigated was about how fast rabbits can breed in ideal circumstances. Beginning with a newborn male and female rabbit, we will assume the following conditions: a. Rabbits reach sexual maturity after one month. b. The gestation period for a rabbit is one month. c. Female rabbit always gives birth to one male and one female rabbit every month. d. Rabbits do not die. This problem can be illustrated as:

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Considering the conditions given, the newborn pair of rabbit are not yet at sexual maturity after one month, therefore, they can’t mate. At two months, they have mated but not yet given birth resulting to having the same number of pairs. After three months, the female rabbit will give birth to a pair, resulting in two pairs. At the fourth month, the original pair gives birth again, the second pair mates but not give birth, resulting to three pairs in all. Continuing this process, this scenario answered Fibonacci’s question as to how many pair of rabbits could be born in a year. Fibonacci in Computer Science The Fibonacci numbers also occur as the sums of “shallow” diagonals in Pascal’s Triangle. Fibonacci numbers is also a complete sequence where every positive integer can be written as a sum of Fibonacci numbers, where one number is used once at most. In computer science, a search technique was developed for searching a sorted array with aid from the sequence.

Fibonacci and several Biological Settings Fibonacci sequence can be found in other places in nature like branching in trees, arrangement of leaves on a stem, the fruitlets of a pineapple, the flowering of artichoke, an uncurling fern and the arrangement of a pine cone. Also on many plants, the number of petals is a Fibonacci number. Many plants including buttercups have 5 petals; lilies and iris have 3 petals; some delphiniums have 8; corn marigolds have 13 petals; some asters have 21 whereas daisies can be found with 34, 55 or even 89 petals.

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The Fibonacci numbers are also well represented in honeybees. Honeybee lives in a colony called hive and they have an unusual Family tree. In a colony, there is one special female called queen. There are male bees who does not work and are produced by the queen’s unfertilized eggs called drones. They only have a mother but no father. There are also other female bees who are called worker but they do not produce eggs. Female bees are produced when the queen mated with a male bee so they have two parents. They usually end up being a worker bees but some are being fed with a substance called royal jelly which makes them grow into queens ready to go off when the bees are ready to swarm and leave their home in search for a new place to build their hive. If you take any bee hive and follow the pattern, it would look like:

Number of

Male Bees Female Bees

Parents

1 2

Grand Parents

2 3

Great

Gt-Gt

Gt-Gt-Gt

Gran

Gran

Grand

d

d

Parent

Parent

Parent

s

s 3 5

5 8

s 8 13

Since a drone has only one parent (only a mother and no father), it also has an interesting genealogical tree. Generation 1 has 1 member (the male). One generation back there is also 1 member (the mother). Two generations back there are 2 members (the mother and father of the mother). The number of members in each generation follow the Fibonacci sequence.

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Fibonacci Rectangles Start with two small squares of size 1 next to each other, draw on top of both of these a square of size 2 (=1+1). Touching both a unit square and the latest square of side 2, having sides 3 units long; and then another touching both the 2-square and the 3-square (which has sides of 5 units). Adding squares whose side is as long as the sum of the latest two square's sides produces a new rectangular image that follows Fibonacci sequence. This set of rectangles whose sides are two successive Fibonacci numbers in length and which are composed of squares with sides which are Fibonacci numbers, we will call the Fibonacci Rectangles. In each square of Fibonacci Rectangles, a spiral can be drawn by putting together the quarters of a circle in each square. The spiral is not a true mathematical spiral (since it is made up of fragments which are parts of circles and does not go on getting smaller and smaller) but is a good approximation to a kind of spiral that does appear often in nature. Such spirals are seen in the shape of shells of snails and sea shells and in the arrangement of seeds on flowering plants too. The distance of the line from the center of the spiral increase by a factor of the golden number in each square. So points on the spiral are 1.618 times as far from the center after a quarter-turn, that is, equivalent to 6.854 times further out than when the curve last crossed the same radial line.

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FIBONACCI SEQUENCE AND THE GOLDEN RATIO Two consecutive Fibonacci numbers approaches Golden Ratio. It is easier to see what is happening if we plot the ratios on a graph where Fib(i) can be any Fibonacci number and Fib(i-1) is any number preceding it:

The golden ratio often represented by a Greek letter phi with a value of 1.618034 is also called the golden section or the golden mean or Divine Proportion. The proportion is also said to be aesthetically pleasing due to which several artists and architects, including Salvador Dali and Le Corbusier, have proportioned their work close to the golden ratio. The Fibonacci sequence and the golden ratio are intimately interconnected.

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CHAPTER

2

NATURE OF MATHEMATICS: MATHEMATICAL LANGUAGE ANd SYMBOLS

LEARNING OUTCOMES: At the end of the chapter the students are expected to : 1. Discuss the language, symbols and conventions of mathematics. 2. Explain the nature of Mathematics as a language. 3. Discuss the language, symbols and conventions of mathematics. 4. Perform operations on Mathematical expressions completely. 5. Acknowledge that mathematics is a useful language.

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LESSON 1 : LANGUAGE OF MATHEMATICS Language is a fundamental tool that bridges the gap among people from varying origins and cultures without prejudice to their background and upbringing. It can be communicated in either spoken or written manner as long as words are arranged in a structured and systematic way. As a general concept, “language” may refer to the ability to learn and use systems of complex communications or to describe the set of rules that makes up these systems. Considering the Mandarin Language, a different sets of characters are available for the sun, moon, stars, house, chair, table, trees, plants, flowers and relationships like father, mother, brother and sister. These unfamiliar characters may make learning Mandarin more difficult that the Greek Language. Even the Greek letters are different from the English alphabet. Mathematics is also a language. It has its own symbol system, the same way as English or Greek languages have their own alphabet. For example, try solving the given problem: Ana est deux fois plus âgée que son frère et la somme de leurs âges est de 36 ans. Quel âge ont-ils? Since the language used may seem unfamiliar to you, understanding the entire problem creates another problem: failure to understand what the given is all about. To understand the ideas or concepts of Mathematics and figure out clearly the logic behind problems is one of the many reasons why we need to know the language of mathematics. The language itself is precise, it can make very fine distinctions among a set of symbols. It is concise because it can briefly express long sentences. It is also powerful because of the relative ease it gives upon expressing complex thoughts.

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Mathematics is a symbolic language. Some of the symbols you may encounter as you read this book are the following:

∑ Ǝ Ɐ ∈ ∉ ⊆ ⇒ ⟺ ℜ N Z Q ∞

The sum of There exists For every (for any) Element of (or member of) Not an element of (or not a member of) Subset of If … , then If and only if Set of real numbers Set of natural number Set of integers Set of rational numbers Infinity

The subsets of the real world can be described using this mathematical language symbols above. Problems in Physics, like the free-falling bodies, speed and acceleration, quantities like the chemical content of vegetables, the use of mathematical modelling in biological disease modelling and the formula employed in the social sciences can all be expresses using mathematical sentences. Even abstract structures can also be described by mathematics. Examples of such are abstract algebra, linear algebra, topology, real analysis and complex analysis, all of which have no known physical counterparts. Mathematics, therefore, is the language of the sciences, business, economics, music, architecture, the arts, and even politics. There is an intimate connection between the language of mathematics and the English language. The brain hemisphere responsible for controlling language is also the same hemisphere responsible for the tasks involving mathematics. This part of the brain is the left

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brain hemisphere that coordinates logical or analytical thinking while the right brain hemisphere is responsible for creative thinking.

English Language VS Mathematics Language

SYMBOLS

ENGLISH MATHEMATICS English Alphabet English Alphabet, Numerals, and punctuation Greek Letters, Grouping Symbols,

Name Complete thought Action

Noun Sentence

What is in a sentence

Verbs

Attribute of a sentence Synonyms

Fact or fiction

Verbs

Different words but the same meaning

Special Symbols Expressions Sentence Operations and other actions (e.g. simplify, rationalize) Equality, inequality, membership in a set True or false The same object but different names

The table below shows the similarities or comparison of English language and Mathematics.

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Comparison: English Language VS Mathematics Language Name In English, a word that functions as the name of a thing or a set of things that serve as the main word in the subject is called Noun. For simplicity, we will simply use it in this bookas the name of things we want to talk about.

I.

Example:

a. Carol loves Mathematics.

However, in Mathematics, instead of using Noun, we call it Expression which simply refers to the object of interest. Other types of expressions are Numbers, set, functions, ordered pairs, matrices, vectors, groups. Example:

a. 5

b. 1.2 + 6, c.3x – 3 d.ordered pair (a, b) e.a function f(x) II.

Complete Thought Both in English and Mathematics that a Sentence must show complete thought (noun and verb) and express true, false or sometimes true or sometimes false idea. Mathematical sentences however may neither be true or false but not both. Example:

a. He loves Mathematics.

b. 1 + 3 =4

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III.

Synonyms In English, Synonyms are different words with the same meaning. Example: Group - association In Mathematics, Synonyms are the same object but with different names. Example:

1 + 2 + 5 and 8

½ + ½, 2 - 1, 5/5

Some Difficulties in the Math Language 1.Different meaning/use of words in Math and English “and” is equivalent to plus “is” may have different meaning 2.The different uses of numbers: cardinal, ordinal or nominal Cardinal numbers are the ones used for counting, ordinal numbers are those that tells the position of a thing while nominal numbers are numbers used only as a name to identify something, not as an actual value or position. 3.Mathematical objects may be expressed in many ways such as sets and functions.

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Translation: English Language to Math Language and vice versa I.

English to Math English 1. The sum of a number and 10 2. The product of two numbers 3. The product of -1 and a number 4. One-half times the sum of two numbers 5. Twice a number

Mathematics X + 10 x •y -1 • x ½ (X + Y)

2X

Practice: 1. 1. Five less than a number

2. A number, less 8 3. Six more than a number 4. A number, plus 6 5. The square of a number 6. Four times the square of a number 7. One –half a number 8. Three less than twice a number 9. Five more than three a number 10.The square of the sum of 5 and a number

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II. Mathematics to English Choose a quantity to be represented by a variable, then write the mathematical expression for each number. 1. A three-digit numbers whose hundreds digit is half the tens digit and the tens digit is 2 more than the units digit. Let x = be the unit’s value ½

(x + 2)

x+2

x

2. The total interest earned after one year when P 100 000 is invested part at 6 % annual interest rate and the remaining part at 7.5 % annual interest rate. Let

x = be the part to be invested to at 7.5% y = be the total earned interest (P 100 000 – x)0.06 + 0.075x = y

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LESSON 2 : SETS A set is a collection of numbers, peoples, letters of the alphabet, any other sets and other distinct objects. They are conventionally represented by capital letters. These objects are called elements. Sets can be described or specified either by using a description or by listing each member of the set. That is, A is the set of the first four positive integers A = {1, 2, 3, 4} Note that in listing, the order of the elements is irrelevant and in case that the set has many elements, the listing can be abbreviated. B is the set of the first thousand positive integers B = {1, 2, 3, ... , 1000} Considering the given sets A and B, every member of set A is also a member of set B, therefore, it can also be stated as “A is a subset of B” or “A is contained in B”. The relation between sets A and B is established by ⊆ is called inclusion or containment. If, however, there is another number of sets available and they are all contained in B, set B will be considered as Universal set. Also, a set with no element is called null set.

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Lesson 3 : Elementary logic Logic allows us to determine the validity of arguments in and out of mathematics. Proposition is a declarative statements that may expressed an idea which can be true or false but not both. They can be expressed by the symbols P, Q, R or p, q, r. Types of Propositions 1.Simple – means single idea statement 2.Compounds – conveys two or more ideas. Example: Each of the following statement is a proposition. Some are true and some are false. Can you tell which are true and which are false? If it is false, state why. a.9 is a prime number b. 5 + 3 = 8 c. X2 + y2≥ 0 d. 10 < -3

Answers: a. b. c. d.

False, Prime numbers have no other factors than 1 and itself True True False, A negative number is always less than a positive number

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Logical Connectives Statements

Connective Symbolic Form

Type of Statements

Not P

Not

~P

P and Q

And

P ∧Q

Conjunction

Or

P ∨Q

Disjunction

If…. Then

P→ Q

Conditional

P ↔Q

Bi-conditional

P or Q If P, then q

P if and only if If and only if q

Negation

Quantifier 1. Universal Quantifier “ for all” or “for every” , denoted by ∀ 2. Existential Quantifier “ there exists” denoted by ∃ Converse, Inverse and Contrapositive of a Conditional

The Converse of

p→q

is

q →q

The Inverse of

p→q

is

~p →~q

The Contrapositive of

p →q

is

~q →~p

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Examples of Logic A. Write the following in symbolic form using P, Q, R statements and the syml ~,→,↔,∨,∧ where P: The sun is shining Q: It is raining R: The ground is wet 1.If it is raining then the sun is not shining Answer: Q →~P 2.It is raining and the ground is wet. Answer: Q ∧ R 3. The ground is wet if and only if it is raining and the sun is shining. Answer: R ↔ (Q ∧ P)

B. Write the statementusingⱯand ∃ as needed. 1. Everyone in the room is a registered student . 2. Not all men are mortal. Answer: 1. ⱯPerson x in the room, x is registered nurse. 2. ∃ men x such that x is immortal.

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CHAPTER

3

NATURE OF MATHEMATICS: PROBLEM SOLVING AND REASONING

LEARNING OUTCOMES: At the end of the chapter the students are expected to : 1. Apply inductive and deductive reasoning to solve problems. 2. Solve problems involving patterns and recreational problems following Polya’s Strategy. 3. Organize one’s method and approaches for proving and solving problems.

SOLVING PROBLEMS BY INDUCTIVE REASONING Egyptian’s and Babylonian’s approach in solving problems was an example of “do thus and so” method, that is, to solve a problem or perform an operation, a cookbook-like recipe was given, and it was performed over and over to solve similar problems. They concluded the same method would work for any similar type of problem by observing that a specific method worked for a certain type of problem. Such conclusion is called conjecture. A conjecture is an educated guess based on repeated observations of a particular process or pattern. The method of reasoning we have described is called inductive reasoning, where we use specific examples to reach a general conclusion of something. However, it should be carefully noted that a conjecture is an idea that may or may not be correct.

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Use inductive reasoning to predict the next number. 1.

Consider the counting numbers:

1,2,3,4, 5,…

Solution: The three dots indicate that the numbers continue indefinitely in the pattern established. By observing, we can see that the pattern adds 1 to the previous number to get the next number. So by applying inductive reasoning, we can conclude that the next number would be 6. 2. 1, 3, 6, 10, 15, ? Solution: The first two numbers have a difference of 2. The second and third has a difference of 3. The third and fourth, has 4. From this we can see that the difference between any two numbers is 1 more than the preceding difference. Since 10 and 15 differs by 5, we can predict that the next number will be 6 larger than 15, that is, 21. 3. Now, let’s consider the following list of natural numbers, 2, 9, 16, 23, 30,… By applying the same reasoning we used on the first example, we might conclude that the next number would be 37 by observing that any number in the list is 7 more than the preceding number. As we have said, the conjecture may or may not be correct.

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For this example, you were tricked into coming up with a wrong conclusion. Not that your logic was faulty, but the person who makes the list thinks of another scenario. The numbers given are actually dates of Mondays of June where the 1st day falls on a Sunday. From this, the answer should have been 7, since the next Monday that follows the 30th of June is 7th of July. The process we used in predicting the next number for examples 1 and 2 may reveal a flaw for our third example. We can never be sure that what is true to a specific case will be true in general. Inductive reasoning does not guarantee a true result, but it provide a means of arriving into a conclusion. A statement that disproves the conjecture is called counterexample. 4. Verify that the statement x2> x if false by finding a counterexample. Solution: For x =1, we have 12 = 1. Since 1 is not greater than 1, we can say that this is a counterexample for the given problem. Thus, “for all numbers x, x2> x” is a false statement. Practice: Verify that each of the following statements is a false statement by finding a counterexample for each. 𝑥

a. = 𝑥 1 𝑥+3 b =𝑥+ 3 . 1 c. √𝑥2 + 16 = 𝑥 + 4

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SOLVING PROBLEMS BY DEDUCTIVE REASONING Deductive reasoning is the process of solving problems by applying premises, syllogisms, and conclusions. It goes from general case to a specific case. Key Terms: Argument - is the reason or reasons offered for or against something Premises - minor or major propositions or assertions that serve as the bases for an argument. It can be an assumption, law, rule, widely held idea or observation. Syllogism - an argument composed of two statements or premises followed by a conclusion. Conclusion - the last step in a reasoning process 1. “Today is Thursday. Tomorrow is Friday.” Solution: There is only one premise in this statement, “Today is Thursday”. The other statement “Tomorrow is Friday” is called conclusion. Following the days in a week, the fact that Friday is the day after Thursday is not explicitly stated. Since the conclusion comes from general facts that apply to this case, deductive reasoning was used. 2. All men are mortal. Socrates is a man. Therefore, Socrates is mortal. Solution: This example is reasoning that uses syllogism. The first two statements are the premises and the third one is the conclusion. 3. All doctors are men. My mother is a doctor Therefore, my mother is a man.

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Solution: Saying that an argument is valid does not mean the conclusion is true. Thus, if the premises are wrong, the argument may be valid but the conclusion may not be true. 4. General case : “In any right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse”- Pythagorean Theorem Specific case: Given: a = 3, b = 4 Problem: Find the value of the hypotenuse, c Solution: Applying the Pythagorean Theorem to find the value of the hypotenuse, we will use the mathematical translation of the theorem, that is, a2 + b2 = c2. Working out this formula to get c, we have, 𝑐 = √𝑎2 + 𝑏2. Substituting the values of a and b as indicated from the given, we can conclude that c = 5.

SOLVING PROBLEMS BY POLYA’S STRATEGY POLYA’S STRATEGY named after George Polya (1887 – 1985), is a four-step problem solving strategy that are deceptively simple. Polya's four-step approach to problem solving 1. Preparation: Understand the problem ∙

Learn the necessary underlying mathematical concepts

∙

Consider the terminology and notation used in the problem: 1. 2. 3. 4. 5. 6.

What sort of a problem is it? What is being asked? What do the terms mean? Are the information given enough? What is known or unknown? Rephrase the problem in words. Note specific examples of the conditions given in the problem.

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2. Thinking Time: Devise a plan ∙

You must start somewhere so try something. How are you going to attack the problem?

∙

Possible strategies: (i. e. reach into your bag of tricks.) 1. Make a list of the known and unknown information. 2. Use a variable for unknowns. 3. Draw pictures, diagrams or even tables. 4. Be systematic. 5. Solve a simpler version of the problem. 6. Guess and check. Trial and error. Guess and test. (Guessing is OK.) 7. Look for a pattern or patterns. 8. Work backwards. Guess at a solution and check your result.

Tip! ☺ ✔ Once you understand what the problem is, if you are stumped or stuck, set the problem aside for a while. Your subconscious mind may keep working on it. 3. Insight: Carry out the plan Work out an idea or a new approach carefully and see if it leads to a solution. If the plan does not seem to be working, then start over and try another approach. Often the first approach does not work. Do not worry, just because an approach does not work, it does not mean you did it wrong. You actually accomplished something, knowing a way does not work is part of the process of elimination. Remember to keep an accurate and neat record of all your attempts. The key is to keep trying until something works.

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4. Verification: Look back and review your solution Check to see if your potential solution it works.

∙

1.Did you answer the question? 2.Is your result reasonable? 3.Double check to make sure that all of the conditions related to the problem are satisfied. 4.Review your computations in finding the solution. ∙

∙ ∙

If your solution does not work, there may only be a simple mistake. Try to modify your current attempt before scrapping it. Though you have to remember likely that at least part of it will end up being useful. Is there a simpler way to solve the problem? (You need to become flexible in your thinking. There could be another way.) Can the problem or method be generalized so as to be useful for future problems?

“A great discovery solves a great problem but there is a grain of discovery in the solution of any problem. Your problem may be modest; but if it challenges your curiosity and brings into play your inventive faculties, and if you solve it by your own means, you may experience the tension and enjoy the triumph of discovery.” – George Polya Examples: 1. The sum of two numbers is 30. The first number is twice as large as the second number. What are the numbers?

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Solution: Understand the Problem: We need to determine two distinct numbers which when added will give a sum of 30, and when compared, one is twice the other. Devise a Plan: Let x be the value of the second number and since the first is twice the second, we will represent it as 2x. Carry out the Plan: 2𝑥 + 𝑥 = 30 3𝑥 = 30 So, the second number is 𝑥 = 10 and the first number is 2𝑥 = 20. Review the Solution: Since the problem requires two numbers whose sum is 30, adding the two numbers 20 and 10 gives 30. Also, 20 is twice of 10. Therefore, our answer satisfies the conditions given in our problem. 2. In tossing two coins at a time, what is the probability of having 2 heads in a single throw? Solution: Understand the Problem: A coin has two faces, a head and a tail respectively. Two coins when tossed can show several combination. As per required, the two coins should be thrown at once and they should both show heads up. Devise a Plan: Let H represent heads and T for tails. We are going to list down all possible combinations without duplication.

40

Carry out the Plan: We will start to list the possibilities having the First coin show heads and then tail. That is,

First Coin

Second Coin

H H T T

H T H T

The probability that will show both heads will be computed as the number of possibilities of having two heads divided by the number of trials, that is 1 𝑥 100% = 25%. 4

Review the Solution: The list is organized and has no duplication, so out of a single toss of two coins, there is only a 25% probability or chance of having two heads.

41

42

Score: Date:

43

44

45

46

47

CHAPTER

4

MATHEMATICS AS A TOOL : DATA MANAGEMENT

LEARNING OUTCOMES: At the end of the chapter the students are expected to : ∙ ∙ ∙

∙ ∙ ∙ ∙

Use a variety of statistical tools to process and manage numerical data. Compute the z-scores of a given data. Locate and calculate the different values of measures of relative position and interpret these values. Rule out relationship between two different variables and Use the correlation coefficient in making predictions Show knowledge on normal distribution. Interpret in standard deviation units any particular score in normal distribution

LESSON 1 : STATISTICS, ORIGIN AND TERMS Statistics itself came from the Latin word “status” which means state. From the ancient times, statistics was used by state leaders to know how much tax to levy their subjects and how many soldiers are needed in an expected war. In capitalism, no also the leaders of the state but also capitalists, are interested in statistical surveys resulting to increased demand for data processing for their increasing benefits such as insurance.

48

STATISTICAL DATA Data in statistics is always a result of experiment, observation, investigation and other means and often appears as a numerical figure and then evaluated to make it into useful knowledge. For most people, “statistics” is a scary thing that must be avoided as much as possible

because they think that it is a collection of numbers and vague

formulas. People, without

noticing, applies statistics in their everyday life,

such as to the amount of food they eat, how much money was in their purse, how far is their work area to respective homes, color of their hair, number of rebounds and assists of a player in a basketball game, their height, gender, and so on. Statisticians develop and apply appropriate methods in collecting and analyzing data. They guide the design of the research study and then analyze the results. The interpretation of the result is the basis of the statistician in making inferences about the population. BASIC CONCEPTS 1. Descriptive Statistics – deals with the collection and presentation of data and collection of summarizing values to describe its group characteristics. The most common summarizing values are the measure of central tendency and variation. 2. Inferential Statistics – deals with the predictions and inferences based on the analysis and interpretation of the results of the information gathered by the statistician. Some of the common statistical tools of inferential statistics are the t-test, z-test, analysis of variance, chi-square, and Pearson r. 3. Variable – a numerical characteristic or attribute associated with the population being studied. They are further classified as categorical or qualitative and numerical or quantitative. 4. Discrete variables – values obtained by counting. 5. Continuous variables – values obtained by measuring, all of which cannot be put into a list because they can have any value in some interval of real numbers.

49

6. Scales of Measurement – subdivided into four categories and upon drawing inferences on a random sample, the type of measurement scale must be carefully chosen. a. Nominal – classifies elements into two or more categories or classes, the numbers indicating that the elements are different but not according to order or magnitude b. Ordinal –a scale that ranks individual in terms of the degree to which they possess a characteristic of interest c. Interval – in addition to ordering scores from high to low, it also establishes a uniform unit in the scale so that any equal distance between two scores is of equal magnitude. There is no absolute zero in this scale. d. Ratio – in addition to being an interval scale, it also has an absolute zero. 7. Population – defined as groups of people, animals, places, things or ideas to which any conclusions based on characteristics of a sample will be applied. 8. Sample – a subgroup of the population. 9. Parameter – a numerical measure that describes a characteristic of the population. 10. Statistic – numerical measure that is used to describe a characteristic of a sample.

50

51

52

LESSON 2: MEASURES OF CENTRAL TENDENCY To describe a whole set of data with a single value that represents the middle or centre of its distribution is the purpose of measure of central tendency (measures of centre or central location). To put in other words, it is a way to describe the center of a data set. Why Is Central Tendency Important? ∙ ∙ ∙

It lets us know what is normal or 'average' for a set of data. It also condenses the data set down to one representative value, which is useful when you are working with large amounts of data. Central tendency also allows the comparison of one data set to another, as well as one piece of data to the entire data set. For example, you could easily draw comparisons between the girls’ and boys’ heights by calculating the average height for each sample group.

Three main measures of central tendency ∙ ∙ ∙

Mean Median Mode

1. Mean (balance point) ∙ The mean by definition is the sum of all the values in the observation or a dataset divided by the total number of observations. This is also known as the arithmetic average. ∙ The mean can be used for both continuous and discrete numeric data as well as for categorical data, as the values cannot be summed. ∙ As the mean includes every value in the distribution the mean is influenced by outliers (which are numbers that are much higher or much lower than the rest of the data set) and skewed (asymmetric) distributions. ∙ This measurement is applicable to use for ratio and interval data.

53

Examples: 1. Find the mean of the following numbers:

47, 53, 67

Solution: From the given problem, we can see that there are only 3 given values. This is considered as ungroup. Therefore, Mean = (47+53+67)/3 = 55.67 2. Find the mean of the following math test scores: 33,44,55,65,55,33,67,78,54,55,55,44 Solution: There are 12 observations or values in this given problem. Usually, data are grouped if they are above 30 observations. We can work this case either by using the grouped or ungrouped formula: Using Ungrouped data

Mean =

sum of observations 638 = total number of observations 12

x̄ = 53.17

54

Using Grouped Data Test scores 33 44 55 65 67 78 54

x̄ = x=

Frequency 2 2 4 1 1 1 1 N = 12

∑𝑓𝑥 𝑁

(33∗2)+(44∗2)+(55∗4)+(65∗1)+(67∗1)+(78∗1)+(5 4∗1) 12

x̄= 53.17 2. Median (physical middle point) ∙

∙

∙

∙ ∙

The median is considered as the physical middle point in a distribution because it is located at the center position when the values are arranged in ascending or descending order, which in turn divides the distribution in half (there are 50% of observations on either side of the median value). If a distribution has an odd number of observations, the median value is the middle value. If it is an even number, the median value is the mean or average of the two middle values. The median is usually the preferred measure of central tendency when the distribution is not symmetrical because it is less affected by outliers and skewed data than the mean. The median cannot be identified for categorical nominal data, as it cannot be logically ordered. This is widely used for ordinal type of information.

55

Examples: 1. (a) Ungrouped data: Arrange to ascending order:

3,1,5,2,2,7,5,6,8 1,2,2,3,5,5,6,7,8

There are 9 values in the distribution, therefore the median is the one on the fifth position based on the arranged values: Median is 5 (b) Ungrouped data:

3,1,5,2,2,7,5,6,8,4

Arrange to ascending order:

1,2,2,3,4,5,5,6,7,8

Since there are 10 values in the given, we will take the average of the fifth and sixth numbers in the arrangement, which is 4 and 5 respectively, therefore, Median is 4.5.

56

2. Find the median of the following math test scores. Test scores 25-30 31-36 37-42 43-48 49-54 55-60 61-66

Frequency 4 7 3 5 8 12 6 N = 45

Solution: n/2 = 45/2 = 22.5 Test scores 25-30 31-36 37-42 43-48 4954 55-60 61-66

frequency 4 7 3 5 8


12 6 N = 45

39 45

57

3. Mode ∙ The mode can be found for both numerical and categorical (nonnumerical) data. It is the most commonly occurring value in a distribution. There can be more than one mode for the same distribution of data, (bi-modal, or multi-modal), thus limiting the ability of the mode in describing the center of the distribution. ∙ In some particular cases, the distribution may have no mode at all (i.e. if all values are different).In such case, it may be better to consider using the median or mean, or group the data in to appropriate intervals, and find the modal class.

Examples: 1. Ungrouped Data a. 33,44,55,65,55,33,67,78,54,55,55,44 b. 33,44,65,55,33,67,78,54,55,55,44 c. 1,2,3,4,5,6,7,8,9 mode

Mode is 55 Mode is 33, 44 Mode is 1,2,3,4,5,6,7,8,9 or no

58

2. Determine the mode of the following math test scores. Test scores 25-30 31-36 37-42 43-48 49-54 55-60 61-66

Frequency 4 7 3 5 8 12 6 N = 45

Modal class is: 56-60 Mode = 55.5 + ((

12−8

)(6)

12−8)+(12−6)

Mode = 57.9

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LESSON 3 : MEASURES OF VARIATION/DISPERSION Measures of variation is used to describe the distribution of the data. How is the data distributed? Is it cluster in one area or is it really spread out? Different measurements of variation ∙ ∙ ∙ ∙

Range Mean absolute deviation Variance Standard deviation

RANGE - The range is the simplest measure of variation to find. It is simply the highest value minus the lowest value. RANGE = MAXIMUM - MINIMUM Since the range only uses the largest and smallest values, it is greatly affected by extreme values, that is - it is not resistant to change. Example 1. Find the range of the following ages of mountaineers: 23,17,15,45,55,37,44,60 Range = 60-23 = 37 Interpretation: the ages of mountaineers are slightly spread out. Their ages have big variation.

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MEAN ABSOLUTE DEVIATION ∙ ∙ ∙

Mean absolute deviation is the average of how much the data values differ from the mean. A small MAD value indicates a clustered data values. A big MAD value indicates a spread out data values

Example 1. Find and interpret the mean absolute deviation of the test scores obtained by these two students. Anna’s test scores 95 93 95 96 95 95+93+95+96+ 95 5

Anna Mean = Anna MAD = Joy Mean =

Joy’s test scores 98 95 90 95 95

= 94.8

(

94.8−95)+(94.8−93)+(94.8−95)+(94.8−96)+(94 5 .8−95) 98+95+90+95+ 95 5

= 0.72

= 94.6 61

Joy MAD =

(

94.6−98)+(94.6−95)+(94.6−90)+(94.6−95)+(94 5 .6−95)

= 1.84

Interpretation: Anna has a smaller MAD value than Joy which means Anna is more consistent in her test results. VARIANCE AND STANDARD DEVIATION ∙

∙

Standard deviation is used to quantify the amount of variation or dispersion of a set of data values and represented by the Greek letter sigma σ or the Latin letter s. A low standard deviation is an indicator that the data points tend to be close to the mean while a higher value indicates a widerspread from the mean. The variance (represented by the Greek letter sigma that is squared or the letter s2 ) is the square of the deviation of data sets from its mean.It is use to determine how far or clustered a random data points from their average value.

62

Example 1. Determine and interpret the variance and standard deviation of the following number of votes obtained by each party list. Partylist A

Votes 26 33 29 24 30 35

Partylist B

Votes 35 26 24 24 30 35

Miguel Dom Ann Gab Hazel Cherry

Arman Daniel Elise Dona Mary John

Solution: Partylist A Mean

= 35.4

Partylist A variance

= 58.872

Partylist A Standard Deviation

= 7.67

Partylist B Mean

= 34.8

Partylist B variance

= 66.768

Partylist B Standard Deviation

= 8.17

Interpretation: Partylist A has a smaller standard deviation, therefore, they have a higher chance to win or lose together than Party List B. 63

LESSON 4. QUARTILE, DECILE AND PERCENTILE A measure of position determines the position of a single value in relation to other values in a sample or a population data set. We commonly refer to these measure of position as quantiles or fractiles. Quantiles It is a score distribution where the scores are divided into different equal parts. There are three kinds of quantiles: Quartile, Decile and Percentile. Quartile- A measure of position that divides the ordered observations or score distribution into 4 equal parts. Decile- A measure of position that divides the ordered observations or score distribution into 10 equal parts. Percentiles - A measure of position that divides the ordered observations or score distribution into 100 equal parts. Formula for Quantile of Ungrouped Data

64

Example 1. A classroom teacher gave a quiz to 9 students. The scores obtained are as follows:

10 5 9 4 2 6 3 4 8 Find the following: a. 2rd Quartile b. 7th Decile c. 94th Percentile Solution: a. 2nd Quartile

2

3

4

4

Q

5

6

8

Q

1

9

10

Q

2

3

ANSWER: 5 b. 7th Decile 2

3

4

4

5

6

8

9

10

D

D

D

D

D

D

D

D

D

1

2

3

4

5

6

7

8

9

ANSWER:8 c. 2 P

10

94th Percentile 3 P

20

4 P

30

4 P

40

5 P

50

6 P

60

8 P

70

9 P

80

10 P

90

ANSWER:10

65

Interpretation: a.Fifty percent of the data falls below and above the value of 5. b.70 % of data falls below the value 8 and while 30 % of students scored higher than 8 Example 2. Calculate the 3rd quartile. (refer to the formula of grouped data)

Math Score(x) Frequency(f)

4

6

10

13

17

3

5

6

4

2


3

8

14

18

20

66

67

68

69

70

71

72

73

74

LESSON 5. THE NORMAL DISTRIBUTION The Normal Distribution Standard deviation is a statistic that characterizes a distribution of score. It increases indirect proportion as the scores spread out more widely, the larger the standard deviation, wider the spread of scores. The meaning of standard deviation is best defined by normal distribution of scores. The normal distribution is illustrated by the normal curve. Normal curve is a symmetrical curve having a bell-like shape. The total area under the normal curve is equal to 1, and represent all of the scores in a normal distribution. In such a curve, the mean, the median, the mode are identical or equal, so the mean falls at the exact center of the curve. The curve has no boundaries in either direction, for the curve never touches the baseline no matter how far it is extended. The curve is a curve of probability, not certainty.

Figure 2.1 Percentage under the Normal Curve

75

The formula for the standard normal distribution is

𝑦=

−𝑧2

𝑒

2

√

2𝜋

where z = standard score, 𝑒 ≈ 2.1718.. and 𝜋 ≈ 3.14159..

Since the curve is symmetrical, this holds true for both side of the mean. As presented in Figure 1, approximately 68.26% of the scores lie between +𝜇 + 1𝜎 and 𝜇 + 1𝜎. Furthermore, about 13.6% of the scores fall between 𝜇 + 1𝜎 and 𝜇 + 2𝜎. All of the scores in a normal distribution lie between the mean plus or minus standard deviations. If a set of scores is normally distributed, one can interpret any particular score if he knows how far, in standard deviation units, it is from the mean. Example 1.The mean of the normal distribution is 50 and standard deviation is 10. a. How does an individual’s score of 60 compare with all the other scores? b. How about the score of 30? Illustration 1.

8

19

30

41

52

63

Figure 2.2 A Normal Distribution of Score 76

a. If a person’s score is 60, then slightly more than 84. 135% of all the other scores in a distribution lie below his scores. b. If a person’s score is 30, then slightly more than 97. 725% of all the scores in a distribution fall above his scores. Practical Application of Normal Curve In the field of educational research, there are a number of practical applications of the normal curve, among which are: 1. To calculate the percentile rank of scores in a distribution, 2. To normalize a frequency distribution; which is an important process in standardizing a psychological test or inventory 3. To test the significance of observe measure in experiments, relating them to the chance fluctuations or errors that an inherent in the process and sampling and generating about population form which the samples are drawn.

Standard Scores use a common scale to indicate how an individual compare to other individuals in a group. These scores are particularly helpful in comparing an individual’s relative position on different instruments. Two most frequently used Standard Scores Z-SCORES. These standard scores tell how far a raw score is from the mean in standard deviation units. The formula is 𝑧=

𝑥−𝑥̅

or

𝑧 =

𝑠

Where

𝑥−𝜇

𝜎

x- raw score 𝑥̅- mean 𝑠− standard deviation

T-SCORES. These are z-scores that are expressed in another way. The formula is

𝑡 = 50 + 10 (

𝑥−𝑥̅ 𝑠

)

or 𝑡 = 50 + 10𝑧 77

Example 2. 1. Jerry got a grade of 85 on the final examination in Linguistic for which the mean grade was 75 and the standard deviation was 10. On the final examination in Trigonometry for which the mean grade was 80 and the standard deviation was 24, he received a grade of 92. In which subject was his relative standing higher? Given. Linguistic: 𝑥̅ = 75, 𝑠 = 10 and 𝑥 = 89 and Trigonometry: 𝑥̅ = 80, 𝑠 = 17 and 𝑥 = 93. z-scores

t-scores

Linguistics 𝑧 = 𝑥−𝑥̅ = 89−75 = 1.4 𝑠

10

𝑡 = 50 + 10𝑧 = 50 + 10(1.4) = 64

Trigonometry 𝑧 = 𝑥−𝑥̅ = 93−80=0.77 𝑠

17

𝑡 = 50 + 10𝑧 = 50 + 10(0.77) = 57.7

Interpretation: The score of Jerry in Linguistic is 1-unit standard deviation above the mean. His score in Trigonometry is 0.5-unit standard deviation above the mean. Thus, his relative standing was higher in Linguistic.

78

2. Professor Elise wanted to get a student’s equally weighted mean achievement in Algebra test and Geometry test. The data are shown below. Course

Test Score

Mean

Standard Deviation

50

Highest possible score 70

Algebra

40

Geometry

95

125

200

20

5

Solution

Algebra test z score ̅

𝑧=

Geometry test z score ̅ 𝑧= Mean of z score

𝑥 − 𝑥 40 − 50 = = −2.0 𝑠 5

𝑥 − 𝑥 95 − 125 = = −1.5 𝑠 20

{(- 2.0) + (- 1.5)} / 2 = - 1.75

Interpretation: The mean standard score of – 1.75 indicates that on an equally weighted mean, the performance of the student was fairly consistent that is - 2.0 standard scores below the mean in Algebra and - 1.50 standard scores below the mean in Geometry.

79

The Normal Curve Areas ∙ ∙

Computing Normal Probabilities There are several different situations that can arise when asked to find normal probabilities.

80

This can be shortened into two rules. 1. If there is only one z score given, use 0.5 for the second area, otherwise look up both z scores in the table. 2. If the two numbers are the same sign, then subtract; if they are different signs, then add. If there is only one z score, then use the inequality to determine the

second sign (< is negative, and > is

positive). Example. If a random variable has the standard normal distribution, what are the probabilities that it will take on a value: 3. Less than 1.25 4. Between 0.49 and 1. 25 3. 3. Between -0.75 and 1.45 4. Greater than or right of 2.67 5. Less than or left of 0.75

81

82

Application of the Normal Curve You are now ready to solve normal distribution problems. In doing so, you must always bear in mind the difference between continuous random variable and discrete random variables. Each category has its own way of calculating the areas under their standard normal distribution. Continuous random Variable (Measurable) Example 1. The scores of the grade 6 pupils have a mean of 5.35 and standard deviation of 0.15. a. What percentage of all these scores are lower than 4? b. What percentage of these scores are between 5 and 6? Solutions c. Lower than 4. Given 𝜇 = 5.35, 𝜎 = 0.15 and 𝑥 = 4

The area between 0 and is . Hence, the percentage of all these scores are lower than 6 is

%.

b. Area between 5 and 6

83

Area between 5 and 6 = Area between 5 and 5.35 + Area between 5. 23 and 6. =___________+ _________ = _________or _____ % Thus, percentage of these scores are between 5 and 6 is %. Discrete Random Variable(Countable) Example 2. Princess Joy is a Home Economics teacher. She knows from experience that the number of budget meals she sells each day is a random variable having approximately a normal distribution with the mean equal to 43.5 and standard deviation equal to 5.6. What are the probabilities that in any value day she will sell? a. Exactly 30 budget meals; b. At most 30 budget meals; c. At least 30 budget meals. Solution: Because the data are discrete, the values of this discrete random variable shall be spread over a continuous scale by representing each whole number k by interval form 𝑘−

1

2

to

𝑘+

1

2

84

In this example, 𝑘 = 30,

𝑥 = 𝑘 − 21= 30 − 1

1

2

= 29.5

𝑥2= 𝑘 − 2 = 30 + 12= 30.5 1

a. Exactly 30 budget meals Area between 29.5 to 43.5 𝑥−𝜇 𝑧1 = 𝜎

The area is .

Area between 43.5 to 30.5 𝑥−𝜇 𝑧2 = 𝜎

The area is .

Illustration:

The area between 29.5 and 30.5 (exactly 30) = 𝑧1−𝑧1=

or

%.

Hence, the probability that she will sell exactly 30 budget meals is__ b. At most 30 budget meals. 𝑥−𝜇 𝑧= 𝜎

Illustration:

85

%.

The area is .

Area at most = 0.5 - _________ = ____ or __ %. Thus, the probability that she will sell at most 30 budget meals is ______ %.

86

87

88

89

90

LESSON 6 : CORRELATION AND LINEAR REGRESSION Correlation measures the association or the strength of the relationship between two variables say x and y. Definitions ⮚ Two variables are positively correlated if the values of the two variables both increase. ⮚ Two variables are negatively correlated if the values of one variable increase while the values of the other decrease. ⮚ Two variables are not correlated or they have zero correlation if one variable neither increases nor decreases while the other increases.

Verbal Interpretation The degree of correlation can be determined by correlation coefficient. Its value represents an interpretation as shown in the table below.

r

Verbal Interpretation

0.00

No correlation

±0.01 to ±0.20

Slight Correlation

±0.21 to ±0.40

Low Correlation

±0.41 to ±0.70

Moderate Correlation

±0.71 to ±0.80

High Correlation

±0.81 to ±0.99

Very High Correlation

±1.0

Perfect Correlation

91

Pearson Product- Moment Correlation(r) ⮚ The most familiar sort of statistical tool in quantifying the linear relationship between two random variables, x and y. ⮚ Data are parametric( numerical measurement describing a characteristic of a sample). Formula

Steps in Solving Correlation 1. 2. 3. 4.

State the Null Hypothesis (Ho) and the Alternative Hypothesis ( Ha). Determine the tabular value (TV), degree of freedom(df) = N-2. Determined computed value (CV). State the conclusion. a. Decision: i. rc>rt (means Reject Ho) and ii. rc
Example: 1. Calculate and analyze the correlation coefficient between the number of study hours and the number of sleeping hours of different students at 0.05 level of significance.

Number of Study Hours (x) Number of

2

4

6

8

10

10

9

8

7

6

Sleeping Hours (y) 92

Solution: 1. Ho: There is no significant relationship between the number of study hours and the number of sleeping hours of different students. Ha: There is significant relationship between the number of study hours and the number of sleeping hours of different students. 2. Tabular Value: 𝛼 = 0.05 and df= N-2= 5-2=3 (3, 0.05)=0.878 3. Computed Value: 𝑦2

Student

𝑥

𝑦

𝑥∙𝑦

𝑥2

1

2

10

20

4

100

2

4

9

36

16

81

3

6

8

48

36

64

4

8

7

56

48

49

5

10

6

60

100

36

∑𝑥

∑𝑦

∑ 𝑥𝑦

∑ 𝑥2

∑ 𝑦2

= 30

= 40

= 220

= 220

= 330

𝑁=5

4. Conclusion Based on the result of 𝑟 = −1 which is less than the tabulated value of 0.878, Do not reject Ho. This implies that there is no significant relationship between the number

93

of study hours and the number of sleeping hours of different students. The result of r also implies perfect correlation. Linear Regression Regression Analysis is very powerful tool in the field of statistical analysis specifically in predicting the value of one variable to the given value of another variable, and those variables that are related to each other. Therefore, it is used when predicting the behavior of a variable. The regression equation explains the amount of variations visible in the independent variable x. It is actually an equation of a straight line. The purpose of regression is to determine the trend of the two variables as related to each other whether the trend is rising or falling. Formula:

𝒚 = 𝒂 + 𝒃𝒙 where: y= criterion measure, x= predictor, a= ordinate or the point where the regression line crosses the yaxis, and b= beta weight or the slope of the line

94

To get the regression equation, the value of a and b are computed using the formula below.

(∑ 𝒚)( ∑ 𝒙𝟐) − (∑ 𝒙)(∑ 𝒙𝒚) 𝒂= 𝒏(∑ 𝒙𝟐) − (∑ 𝒙)𝟐 and

𝒏(∑ 𝒙𝒚) − ∑ 𝒙 ∑ 𝒚 𝒃= 𝒏(∑ 𝒙𝟐) − (∑ 𝒙)𝟐 where: n= number of pairs

Example1. The data in the table represent the membership at a university mathematics club during the past 5 years.

Number of Years (x)

Membership (y)

1

25

2

30

3

32

4

45

5

50 95

Form a curve of the form𝒚 = 𝒂 + 𝒃𝒙to predict the membership 5 years from now. Solution: x

y

𝒙𝟐

𝒙𝒚

1

25

1

25

2

30

4

60

3

32

9

96

4

45

16

180

5

50

25

250

∑ 𝒙 = 𝟏𝟓

∑ 𝒚 = 𝟏𝟖𝟐

∑ 𝒙𝟐 = 𝟓𝟓

∑ 𝒙𝒚 = 𝟔𝟏𝟏

Since you need to predict the membership five years from now, or at year 10, substitute 10 for x in the equation. Thus, 5 years from now, 𝐲 = 𝟏𝟔. 𝟗 + 𝟔. 𝟓(𝟏𝟎) = 𝟖𝟏. 𝟗 ≈ 𝟖𝟐. Therefore, five years from now, the club would have 82 members.

96

97

98

99

CHAPTER

5

MATHEMATICS AS A TOOL : CODES AND CRYPTOGRAPHY

LEARNING OUTCOMES: At the end of the chapter the students are expected to : ∙ ∙

Perform the coding of a word text using the different method Decrypt the different message.

LESSON 1 : CODING THEORY Introduction Communication systems and storage devices nowadays are not reliable as it seems. The communication system is usually composed of sender (or message source), communication channel, and the receiver. Communication channel is the physical medium through which information is transmitted. Some examples of such channels are telephone lines, internet cables, fiberoptic lines, and air. Storage of data can be also be considered as channels, like hard drives, CD-ROMs, DVDs, and the likes. As the message travels from the sender to the receiver through the channel, it encounters several “noises” that alters the message in the channel that will cause disruption and errors in the messages. Examples of noises are caused by sunspots, lightning, meteor showers, or even human errors like poor typing and poor hearing. One of the tasks of coding theory is to detect, or even correct, those errors. In transmitting messages, coding is defined as source coding and channel coding. There are

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two processes in coding; encoding and decoding. Encoding is transforming messages into bits of message that is suitable to the communication channel while decoding is the opposite process of encoding.

Source Coding

Sender

Receiver

Noise

Communication Channel

Source Encoder

Source Decoder

Source coding is defined as converting the messages from the sender into bits suitable to the communication channel. An example of this is the ASCII code that converts each character in the message to a byte of 8 bits. The commonly used alphabet is the binary system where it uses the numbers 0 and 1.

Example 1.a. Consider the source encoding of four directions north, south, east, and west as follows: north 00 , south

01 , east

10 ,

west

11 .

Suppose the message “north”, which is encoded as 00, is transmitted over a noisy channel. The message may encounter errors and may be received as 01. The receiver will get the message bit 01 and decode it as “south” without realizing that the message is corrupted.

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With this problem at hand, additional process of encoding is required. Channel Coding Receiver

Sender

Source Encoder

Channel Encoder

Noise

SourceDecoder

Communication Channel

Channel Decoder

Channel coding is defined as adding some form of redundancy to the source encoded message so that the errors can be detected or even corrected. Parity Check Example 1.b. Consider the Example 1.a, we perform the channel encoding by adding a redundancy bit of size 1 as follows: 00

000 , 01

011 , 10

101 , 11

110 .

Suppose again that the message “north”, which is source encoded as 00 and again channel encoded as 000, is transmitted over a noisy channel. Suppose that the message encounter only one error in the communication channel. Then the received word can be any of the following messages; 001, 010, or 100. In this manner, we can detect that there is an error in the message since none of the three possible messages is among the source encoded messages. This encoding scheme is called parity check, where in a single bit is added to the message as redundancy bit. A bit string is said to have an odd parity if there is an odd

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number of 1s, and even parity if there is an even number of 1s. We add redundancy bit to message such that it will become an even parity. In this encoding scheme, we can only detect if the messages received encountered a disruption because of errors or noises, however it does not allows us to correct them. For example, if we receive 100, we do not know if it is from 000, 101, or 110. However, adding more redundancy will allow us to correct errors. Example 1.c. Consider the Example 1.a, we perform the channel encoding by adding a redundancy bit of size 3 as follows: 00 00000 , 11001 .

01

01111 ,

10

10110 ,

11

Same with Example 1.b, the message “north” will be encoded up to the message 00000. Because of the noises in the channel, we say that only one error is introduced to the message that will give us five different possible messages 10000, 01000, 00100, 00010, or 00001. Assume that 10000 is received. Then we can be sure that 10000 is from the correct message 00000 since if we are going to compare 10000 to the other encoded messages such as 01111, 10110, and 11001, there will be at least two errors in the message. Note that in this encoding scheme that can correct errors, information transmission speed will be at cost since we have to transmit 5-bit message for a message of a size of 2 bits. Repetition Code Suppose that the source encoding is already done and that the encoded message is of fix length k. Then the channel encoding by repetition is performed by taking the k bits, then repeat it 2r + 1times, where r ≥ 1 is a fixed integer. Example 2.a. Suppose that the source encoded message is 101 where k = 3. Then choose = 2 , so we must repeat 101, 5 times. This will give us 101101101101101. Assume that the message is transmitted through a noisy channel and distorted. The received message is 111101100011100. The decoding process will be done in this manner. Consider the positions 1, 4, 7, 10, 13 of the received message. We will take the most frequent bit as the first decoded

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bit. Then, consider 2, 5, 8, 11, 14 positions and do the same process to get the second bit. For the third and last bit, consider the positions 3, 6, 9, 12, and 15. With this, the decoded message will be 101. In this encoding scheme, we can detect and correct more than one error for the cost of transmission speed of the messages since like the example above, you will transmit 15-bit message for a message of size of 3 bits.

Modular Arithmetic Congruence Let a, and b are integers and m is a natural number. We say “a is congruent to b modulo m” in symbols a ≡ b (mod m) , if m divides a b or if m divides b - a. In other words, the difference of a and b is divisible by m. In addition, if m > 0, and r is the remainder when b is divided by m then r ≡ b (mod m) . The integer r is called the least residue of b (mod m) . Examples I. Verify if the following congruence are true. 1.) 3 ≡ 9 (mod 2)

2.) 4 ≡ 15 (mod 3) 3.) 7 ≡ 15 (mod 4) 4.) 5 ≡ 20 (mod 3) 5.) 4 ≡ 20 (mod 3) Answers: 1.) It is true since 9 - 3 = 6, which is divisible by 2. 2.) It is false since 15 - 4 = 11, which is not divisible by 3. 3.) It is true since 15 - 7 = 8, which is divisible by 4. 4.) It is true since 20 - 5 = 15, which is divisible by 3. 5.) It is false since 20 - 4 = 16, which is not divisible by 3.

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II.Solve the least residue of the following. 1.) 29 (mod 3)

2.) 35 (mod 4) 3.) 50 (mod 7) 4.) 30 (mod 4) 5.) 25 (mod 5) Answers: 1.) The answer is divided by 3. 2.) The answer is divided by 4. 3.) The answer is divided by 7. 4.) The answer is divided by 4. 5.) The answer is is divided by 5.

2, since 2 is the remainder when 29 is 3, since 3 is the remainder when 35 is 1, since 1 is the remainder when 50 is 2, since 2 is the remainder when 30 is 0, since there is no remainder when 25

III.Solve the following modular arithmetic. 1.) 10 − 4 (mod 5)

2.) 11 + 7 (mod 3) 3.) 8 + 7 (mod 7) 4.) 20 − 7 (mod 5) 5.) 31 − 6 (mod 4) Answer s: 1. 1 2. 0 3. 1 4. 3 5. 3

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Check Digits and Check Codes There are several methods in producing identification numbers which are unique. In the following methods, modular arithmetic is used to produce and verify identification numbers. The examples are; the Universal Product Code (UPC), United States Postal Services (USPS), the Credit Card, and the International Standard Book Number. Each examples uses their last digit as the check digits to verify the identification number. Universal Product Code (UPC), is mainly used in products sold in department stores and groceries. The UPC consists of barcodes with 12 digits where the last one is the check digit. The International Standard Book Number (ISBN) is used on books where usually found at the last page of the book. ISBN can be ISBN-10 or ISBN-13 where they used 10 digits or 13 digits string of number respectively with the last digit as the check digit. Lastly, the United States Postal Services (USPS) is consist of 11 digits, while the Credit Card uses 16 digits where both of them uses their last digits as the check digits. The table below summarize the formulas for the check digits.

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d =9− d +d +d +d +d +d +d +d +d +d

Check Digit d

(

11

1

2

3

4

5

6

7

8

9

10

)(mod 9)

11

Credit Card d

Check Digit d

= 10 − 2d + d + 2d + d + 2d + d + 2d + d + 2d + d + 2d + d + 2d + d + 2d

(

16

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

)

(mod10)

Note: Add all the digits, treating the two-digit numbers as two single digits.

16

Examples: 1. Determine the check digit for a new product given that the first 11 digits are 8-21345- 67132- ? Solution: 1

Let d

2

= 8, d

3

= 2, d

4

= 1, d

5

= 3, d

6

= 4, d

7

= 5, d

8

= 6, d

9

= 7, d

10

= 1, d

11

= 3, and d

= 2.

Then, d12 = 10 − (3(8) + 2 + 3(1) + 3 + 3(4) + 5 + 3(6) + 7 + 3(1) + 3 + 3(2))(mod10) d12 = 10 − 86 mod10 d12 = 10 − 6 d12 = 4 Hence, the check digit is 4. 2. Determine the check digit of book with an ISBN-10 first 9 digits of 017-316444- ? Solution: 1

Let d

2

= 0, d

3

= 1, d

4

= 7, d

5

= 3, d

6

= 1, d

7

= 6, d

8

= 4, d

9

= 4,and d

= 4.

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Then, d10 = 11 − (10 (0) + 9 (1) + 8(7) + 7 (3) + 6 (1) + 5(6) + 4 (4) + 3(4) + 2 (4))(mod11) (

)

d = 11 −158 mod11 10

d10 = 11 − 4 d10 = 7 Hence, the check digit is 7. 3. Verify if the given Credit Card identification number 4301-1234-8888-3751 is valid. Solution: 1 = 4, d2 = 3, d3 = 0, d4 14 = 7,and d15 = 5.

Let d

5

= 1, d

6

= 1, d

7

= 2, d

8

= 3, d

9

= 4, d

10

= 8, d

11

= 8, d

12

= 8, d

13

= 8, d

3, d

It is given that the check digit is 1. We will verify it with the formula. ( ) + 3 + 2 (0) + 1 + 2 (1) + 2 + 2 (3) + 4

d = 10 − (2 4 16

()

( ) + 8 + 2 (3) + 7 + 2 (5)) (mod10)

+2 8 +8+2 8

d16 = 10 − (8 + 3 + 0 + 1 + 2 + 2 + 6 + 4 (

)

+ 16 + 8 + 16 + 8 + 6 + 7 + 10) mod10

d16 = 10 − (8 + 3 + 0 + 1 + 2 + 2 + 6 + 4 (

)

+ 1 + 6 + 8 + 1 + 6 + 8 + 6 + 7 + 1 + 0) mod10

( )(mod10)

d = 10 − 70 16

d16 = 10 − 0 d16 = 1 + 0 d16 = 1 Hence, the check digit is correct and the identification number is valid.

108

=

109

110

111

112

113

114

115

116

117

118

119

LESSON 2 : CRYPTOGRAPHY ∙ ∙ ∙

∙

Originated from two Greek words such as KRYPTO which means hidden and GRAPHENE means writing. Is a method of making and breaking of secret codes. It the method that is commonly used in military, some government agency’s transaction, business firms such as bank, insurance, etc. to secure secrecy of the information. It uses two processes such as encryption and decryption.

Example:

(Plain text) Maths is fun

(Coded Form) PDWKV LV IXQ

ENCRYPTION ∙

Is the process of transforming plain text into code form using a certain algorithm

DECRYPTION ∙

Is the process of returning/converting back the coded message into plain text.

** PLAIN TEXT = refers to the original message ** CIPHERTEXT = refers to the coded message

KEY = refers to the strings of information that is used to reveal the encrypted message into readable form. Example:

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Note : Every letter of the English Alphabet should be paired to at least one character of the Coded Alphabet. Simple Methods of Cryptography SHIFT CIPHER (CEASAR CIPHER) = it is the simple type of substitution cipher. It uses shift in forming the key of cryptography. The cipher text is obtained from taking an equivalent of a single letter of the alphabet to another letter by doing a uniform number of shifts either left or right. Each letter of the English alphabet should be matched exactly to one letter of the cipher alphabet. Illustrations: Using a shift of 3 to the left (the commonly used number of shifts) A B C D E F G H I J K L M N O P Q R S T U V W X YZ D E F G H I J K L MNO P Q R S T U V W X Y Z A BC ( http://en.wikipedia.org/w/index.php?title=File:Caesar3.svg&page=1)

Using a shift of 5 to the left, A B C D E F G H I J K L M N O P Q R S T U V W X YZ V W X Y Z A B C D E F G H I J K L M N O P Q R S TU

Take note : 1st line of letters are the English Alphabet while the letters on the 2nd row are the equivalent cipher character/alphabet

Examples: 1. Using the key of shift of 3 to the right, encrypt the Word “ CRYPTOGRAPHY”

CRYPTOGRAPHY

FUBSWRJUDKCT

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2. Decrypt the cipher text, “BJY DN BJJY”, using the key of 5-shifts to the left.

“BJY DN BJJY”

God is good.

Using a Modulo Operator The sender of the message uses the key K to encrypt and to decrypt the secret message. The key K may have any integer value from 0 – 25 and this will be shared to the person/s who will be receiving the secret message.

To encrypt the message: 1.Express the letters of the alphabet into an integer from 0 to 25, that matches its order, for example A = 0, B = 1, C = 3 ….., then label them as C 2.Calculate Y = (C+ K) mod 26 , for every letter of the message. 3.Convert the number Y into a letter following the order of the letter of the alphabet. Illustration:

Encrypt the message “MMW is fun to learn”

Let K = 5 Step 1.

M = 12 , W = 22, I = 8, S = 18, F = 5, U = 20, N = 13, T = 19, O = 14, L = 11, E = 4, A = 0, R = 17 122

Step 2. M

M

12 + 5 17

W

F

U

N

T

O

L

12 22

8 18 5

20

13

19

14

11

5

5

5

5

5

5

5

18

24

19

16

1 9

16

9

5

17 27

I

S

5 5

13 23 10 25

Y = (17

17 27

13 23 10 25

Y = 17

17 1

13 23 10 25

18

18

24

2 4

19

16

E

A

R

N

0

17

13

5

5

5

5

9

5

22

18

4

5

22

18 mod26 )

9

5

22

18

Step 3. RRB NX KZS YT QJFWS

ENCRYPTED MESSAGE

To Decrypt the Message: 1.Express the letters of the alphabet into an integer from 0 to 25, that matches its order, for example A = 0, B = 1, C = 3 ….., then label them as Y 2.Calculate C = (Y - K) mod 26 , for every letter of the decrypted message. 3.Convert the number C into a letter following the order of the letter of the alphabet.

123

Illustration: Decrypt the cipher text RRB NX KZS YT QJFWS Let K = 5 Step 1

R = 17 , B = 1, N = 13, X = 23, K = 10, Z = 25, S = 18, Y = 24, T = 19, Q = 16, J = 9 , F = 5, W = 22, Step 2 R

R

B

N

X

K

17 17 1

13

23

-5

5

5

5

5

12 12 - 4 8 18 C = ( 12 12 - 4 8 18 C = 12

12

22

Z

S

Y

T

Q

J S

F

W

10

25

18

24

19

16

9 18

5

22

5

5

5

5

5

5

5

5

5

5

5

20

13

19

14

11

4

0

17

13

5

8

20 13 19 13) mod 26

18 17

5 13

14 11

20

4

13

0

17

19 14 11 4 0

Step 3 MMW IS FUN TO LEARN

DECRYPTED MESSAGE

3. Encrypt the word text “FATIMA” using the Modulo Operator, given k = 20

124

Step 1

F = 5,

A = 0,

T = 19,

I = 8,

M = 12

Step 2 F

A

T

I

M

A

5

0

19

8

12

0

20

20

+

20 25

Y = (25 Y = 25

20

20

20

20

39

28

32

20

20

39

28

32

20) mod 26

20

13

2

6

20

Step 3 ZUNCGU

ENCRYPTED MESSAGE

Review: Arithmetic Modulo (Modular Arithmetic) Modular Arithmetic = also known Clock Arithmetic = it is an operator (mod), which seeks for a remainder when two numbers are divided. = it is denoted by the symbol k mod M = r, (which read as k modulo M equals r)

125

How to Calculate the Arithmetic Modulo? A. Let k and M are any positive Integers ∙ ∙

Divide k by the value of M to obtain the quotient (q) and the remainder (r) Such that k = Mq + r , 0 ≤ r < M

Examples: 1. Find 15 mod 6. Solution :

15 ÷ 6 = 2 remainder 3 ,

hence , 15 mod 6 = 3

Checking : k = Mq + 3, 3

where k = 15, M = 6, q = 2, and r =

15 = 6 * 2 + 3 15 = 12 + 3 15 = 15 2. Solve 8 mod 10. Solution: 8 ÷ 10 = 0 remainder 8, Checking : k = Mq + 3,

hence, 8 mode 10 = 8 where k = 8, M = 10, q = 0, and r = 8

8 = 10* 0 + 8 8=8 B. Let k be any negative integer and M is positive integer. ∙ ∙

Divide /k/ by the value of M to obtained the r’, (r’ ≠ 0) Therefore, r = M – r’.

126

Example: Evaluate – 6 mod 4. Solution: /- 6/ ÷ 4 = 1 remainder 2, r’ = 2 r=4–2=2

, hence - 6 mod 4 = 2

127

128

129

130

131

CHAPTER

6

MATHEMATICS AS A TOOL : APPORTIONMENT AND VOTING

LEARNING OUTCOMES: In this lesson, the students are expected to: ∙ Solve apportionment problems involving the different method. ∙ Differentiate : Hamilton apportionment method, Jefferson apportionment method and Huntington – hill method. ∙ Conduct voting using different methods. LESSON 1 : APPORTIONMENT ⮚ Is a method of dividing a whole into various parts. ⮚ The process originated in 1790 in the U.S. congress. ⮚ They want to established or select a fairly number of representatives of each state based on state population in the U.S. congressional seat. ⮚ It means that the numbers of representative ( the seat) is proportion to the population size being represented. ⮚ Sometimes called “ the equal proportion “. ⮚ Different plans were introduce to select right numbers of representative . ⮚ has expanded into different applications in the modern world in economics, accounting , business, law , etc.

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The Hamilton Plan ⮚ Early apportionment method used in US congress was introduced by Alexander Hamilton ⮚ Given the number of seat in the US congress will be apportioned between states proportionally to their population. ⮚ Standard divisor (D) –the number of voters represented by each representative. ⮚ Standard quota (Q) – the whole part of the quotient when the population of the sub – group is divided by the standard divisor. Formulas: Standard Divisor (D)=

𝑡𝑜𝑡𝑎𝑙 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑡𝑖𝑣𝑒 𝑁

D

𝑅

= Standard quota (Q)= Q=

𝑠𝑢𝑏−𝑔𝑟𝑜𝑢𝑝 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑖𝑣𝑖𝑠𝑜𝑟 𝑛 𝐷

Note : 1. Standard quota , Q must be an integer. In case of decimals, just drop the decimal values . 2. When the total standard quota is not equal to given total apportioned or the number of representative , place an additional representative to the next the sub – group representative with the highest decimal value until the representatives are complete.

133

Example: A new school offering the complete six grades in high school has the following enrollment in the different grades below. The administration are to apportioned the 20 teachers for each grade. Calculate a. The standard divisor b. The standard quota

Grades

students

Grade 12

40

Grade 11

35

Grade 10

22

Grade 9

38

Grade 8

25

Grade 7

39

Total

199

Solution : a.

b.

Standard Divisor , D

=

𝑁 𝑅

19

=9

20

grade 12 Standard quota , Q = 12 grade 11 Standard quota , Q = 11 grade 10 Standard quota ,Q

10

=

= 9.95

𝑛

𝑛 𝐷

=

𝑛

=

𝐷

=

𝐷

40 9.95

= 4.02

35 9.95

= 3.51

22

= 2.21

9.95

134

Grade

student

Grade 12

40

Grade 11

Q=

𝒏

Q

Q

40 9.95

4.02

4

Corrected no. of teacher s 4

35

35 9.95

3.51

3

4

Grade 10

22

22 9.95

2.21

2

2

Grade 9

38

38 9.95

3.81

3

4

Grade 8

25

25 9.95

2.51

2

2

Grade 7

39

39 9.95

3.91

3

4

Total :

199

17

20

𝑫

Analysis : The above example is already an application of apportionment. If the given problem represents the states , voters and representatives related to the origin of apportionment , Answer the following questions: 1. Which represent the different state in the problem ? 2. Which represent the different voters in the problem ? 3. Which represent the number of representatives in the problem ? 4. Define apportionment in your own words ?

135

The Jefferson Plan ⮚ Another apportionment method used in US congress was introduced by Thomas Jefferson ⮚ This method uses a modified standard divisor the arrives at the correct or exact numbers of representative using trial and error . ⮚ The modified uses an assume value always smaller than the standard divisor. Example : A new school offering the complete six grades in high school has the following enrollment in the different grades below. The administration are to apportioned he 20 teachers for each grade. Calculate the number of teachers for each grade using the Jefferson apportionment method. Grades

students

Grade 12

40

Grade 11

35

Grade 10

22

Grade 9

38

Grade 8

25

Grade 7

39

Total

199

136

Solution : a .

assume values for Modified Standard divisor ( D ) m

say , D = 9 m

b. by trial and error, solve the number of teachers 𝑛

=

40

𝑛 𝐷

=

35 9

= 3.88

𝑛

=

22

= 2.44

grade 12 Standard quota , Q = 12

𝐷

grade 11 Standard quota ,Q

=

grade 10 Standard quota , Q = 10

𝐷

11

9

9

= 4.44

note: grade 12 supposedly must have 4 teachers.

c.

Another assumption and trial.

say ,D

m

= 8.7

grade 11

Standard quota , Q = 11

𝑛 𝐷

= 8.7 = 4.02

grade 10

Standard quota , Q

𝑛 𝐷

= 8.7 = 2.52

10

=

𝐷

=

40

Standard quota , Q

12

=

𝑛

grade 12

8.7

= 4.59

35

22

137

Q=

𝒏

Grade

student

Grade 12

40

40 8.7

4.59

correct no. of teacher s 4

Grade 11

35

35 8.7

4.02

4

Grade 10

22

22 8.7

2.52

2

Grade 9

38

38 8.7

4.36

4

Grade 8

25

25 8.7

2.87

2

Grade 7

39

39 8.7

4.48

4

Total :

199

𝑫

Q

20

Analysis : The above problem , the number of apportionment is exact using the modified standard divisor.

Apportionment principle

⮚ A new representative is added to a sub – group due to an increase in population. ⮚ The representative is assigned to the group in such a way it gives the smallest relative unfairness of apportionment.

138

Formula : Where :

R

=

𝑨 𝑪

R = relative unfairness of apportionment 1 - C2 / C = average population of the sub – group receiving the new Representative. A = absolute unfairness of apportionment = / C

𝑠𝑢𝑏−𝑔𝑟𝑜𝑢𝑝

C=

𝑛𝑜.𝑜𝑓 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑡𝑖𝑣𝑒

Example:

RBSN company wants to add a new call center agent in one of its office. Report indicate an increase in the daily calls of the offices in the past month. Determine which office should get the additional agent. Use the apportionment Principle to justify your answer.

Office branch

Number of agents

Ave. no. of call / day

Makati

62

882

Ortigas

48

996

Solution

C

Office branch Makati

882

Ortigas

882

63

62

C

1

= 14.00 = 14.22

2

996 48 996 49

A=/C –C / `1

2

R=

𝑨 𝑪

= 20.75

6.75

0.48

= 20.32

6.10

0.30

139

Answer. R = 0.30 (the lowest ) means the new agent or representative will goes to Ortigas office. Analysis. 1 . 2 .

The additional one in C and C indicates that we assume that office or state

1

2

receives the additional agent or representative.

C and C with no additional indicates that office or state does not 1

receives

2

the additional agent or representative.

3. Repeat same procedure when adding another new representative is apportioned. Huntington – Hill apportioned method

⮚ The present method of apportionment being use by the US congress ⮚ The method that make use of equal proportion. ⮚ The new additional representative to a sub – group must have the highest Huntington number.

Formula :

H=

(𝑃𝐴 )2 𝑎 ( 𝑎+1 )

Where : PA = population of the sub – group a = the current number of representative of sub – group A H = Huntington – Hill number

140

Example

RBSN company wants to add a new call center agent in one of its office. Report indicate an increase in the daily calls of the offices in the past month. Determine which office should get the additional agent. Use the Huntington – Hill apportionment to justify your answer.

Office branch

Number of agents

Ave. no. of call / day

Makati

62

882

Ortigas

48

996

Solution H=

H=

∙

(𝑃𝐴 )2 882 = 𝑎 ( 𝑎+1 ) 62 2( 62+1 )

(𝑃𝐴 )2

=

𝑎 ( 𝑎+1 )

= 199.16

996 2 48 ( 48+1 )

= 421.77

Since H = 421.77 has the highest Huntington – Hill number, So that the Ortigas office will get the new agent.

141

142

143

144

145

LESSON 2 : VOTING METHODS 1. Plurality Method ∙ ∙

Each voter selects one candidate or choice. The winner is the candidate or choice with the most votes.

Example 1: If we vote on a coffee shop for the kind of drinks and 14 people vote for black coffee, 9 for mocha frappe, and 10 people for tea, then the winner of the plurality election is black coffee. Note:Black coffee did not win by a majority, because it has fewer than 50% of the votes. Black coffee won by a plurality. Example 2: (Minimum Votes Needed to Win) There are 100 voters in plurality elections between Duterte, Roxas, and Poe. After 70 votes have been counted, Duterte has 38 votes, Roxas has 18 votes, and Poe has 14 votes. How many remaining votes does Duterte need to guarantee he wins? How to solve this: First, pick your candidate’s biggest competition, in this case Roxas. Pretend all 30 votes go to Roxas and Duterte. Let x be the number of votes Duterte needs to tie Roxas in this scenario. Then Roxas gets 30 – xof the remaining votes. Since it’s a tie, 38 + x = 18 + (30 – x). Solve for x to get x = 5. If Duterte gets more votes than this, he is guaranteed to win, and so the answer is the smallest number bigger than x, in this case 5 votes. Strategic Voting: When a person votes in a way that does not reflect his or her true preferences in an attempt to improve the outcome of the election form that person’s point of view, it is called strategic voting.

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Runoff Election: · First a plurality vote is taken. · If one candidate has more than 50% of the vote, that candidate wins. · If no candidate has a majority of the votes, a second plurality election is held with a designated number of the top candidates. · This process repeats until one candidate has more than 50% of the votes.

Example 3: In May 2016 Presidential elections, Duterte with 16, 601, 997 votes, Roxas with 9, 978, 175 votes, and Poe with 9, 100, 991 votes. If there had been a runoff between Duterte and Roxas, what percentage of Poe’s supporters would have needed to vote for Duterte for him to have a majority of the vote? How to solve this: First, adding up the votes from all three candidates, we see that there are 35, 681, 163 votes total. A majority is 1 more than half of this, so 1+ (35, 681, 163) = 17, 840, 582.5 or 17, 840, 583. Duterte already has 16, 601, 997 votes so he needs 17, 840, 583 – 16, 601, 997 = 1, 238, 586. This is

2. Borda Count Method ∙ Award points to candidates based on preference schedule, then declare the winner to be the candidate with the most points. Ballo t 1st 2BndD

Point s B gets 4 points D gets 3 points

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In general, if N is the number of candidates… ∙ Each first-place vote is worth N points. ∙ Each second-place vote is worth N – 1points. ∙ Each third-place vote is worth N – 2 points. ∙ … ∙ Each Nth-place (i.e., last place) vote is worth 1 point.

Whichever candidate receives the most points wins the election. Example 1: A corporation would like to invite a new investor. The possibilities are Ayala (A), Bonifacio (B), Calixto (C), and Dancel (D). All investors of the said corporation are polled. The results: Number of Voters 1st choice

35 30 20 15 B

D

C

D

2nd choice

C

A

A

C

3rd choice

A

B

B

A

4th choice

D

C

D

B

Who is the winner under the Borda Count?

148

How to solve this: A Borda Count Election Number of Voters 1st choice (4 points) 2nd choice (3 points) 3rd choice (2 points) 4th choice (1 point)

35 30 20 15 B: 140 D: 120 C: 80 D: 60 C: 105 A: 90 A: 60 C: 45 A: 70 B: 60 B: 40 A: 30 D: 35 C: 30 D: 20 B: 15

Ayala (A): 70 + 90 + 60 + 30 = 250 points Bonifacio (B): 140 + 60 + 40 + 15 = 255 points Calixto (C): 70 + 30 + 80 + 45 = 225 points Dancel (D): 35 + 120 + 20 + 60 = 235 points Bonifacio wins.

149

Example 2: A Borda Count Election is held between Theo, Nicole, and Rafael. Number of Voters 1st choice 2nd choice 3rd choice

10

8

7

5

Theo Nicole Nicole Rafae l Nicole Rafael Theo Nicole Rafael Theo Rafae Theo l

How to solve this: Number of Voters 10 8 7 5 st 1 choice (3 Theo: 30 Nicole: Nicole: Rafael: points) 24 21 15 nd 2 choice (2 Nicole: Rafael: Theo: 14 Nicole: points) 20 16 10 rd 3 choice (1 Rafael: Theo: 8 Rafael: 7 Theo: 5 point) 10 Theo: 30 + 8 + 14 + 5 = 57 points Nicole: 20 + 24 + 21 + 10 = 75 points Rafael: 10 + 16 + 7 + 15 = 48 points Nicole is the winner.

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Example 3: Contestant

Rankings

A

5

4

1

1

3

B

4

1

5

2

2

C

3

5

4

3

1

D

2

2

2

4

5

E

1

3

3

5

4

Number of Voters 120 90 56 123 31

How to solve this: Contestant A

Rankings 5 (1 point)

4 (2 points)

1 (5 points)

1 (5 points)

3 (3 points)

2 (4 points)

2 (4 points)

4 (2 points)

3 (3 points)

1 (5 points)

4 (2 points)

B

4 (2 points)

C

3 (3 points)

D

2 (4 points)

2 (4 points)

2 (4 points)

E

1 (5 points)

3 (3 points)

3 (3 points)

1 (5 points) 5 (1 point)

5 (1 point)

5 (1 point)

5 (1 point) 4 (2 points)

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Number of Voters

120

90

56

123

31

Contestant A: 120(1) + 90(2) + 56(5) + 123(5) + 31(3) = 1288 points Contestant B: 120(2) + 90(5) + 56(1) + 123(4) + 31(4) = 1362 points Contestant C: 120(3) + 90(1) + 56(2) + 123(3) + 31(5) = 1086 points Contestant D: 120(4) + 90(4) + 56(4) + 123(2) + 31(1) = 1341 points Contestant E: 120(5) + 90(3) + 56(3) + 123(1) + 31(2) = 953 points

Thus, using the Borda Count method of voting, contestant B wins.

3. Plurality by Elimination ∙ ∙

The plurality with elimination voting method is also known as an instant run-off voting and sequential run-off voting. It is a preferential voting method and candidates that have the least first place votes get eliminated until one candidate has majority of first place votes.

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Example 1:Let’s say we have a town of 20,000 people electing a mayor using the Plurality with Elimination Voting Method. There are 4 candidates, candidate A, candidate B, candidate C, and candidate D. The results of the election are shown below in a preference schedule:

Number of Voters 4000 7000 3000 6000 1st

D

C

A

B

2nd

A

B

B

D

3rd

B

D

C

C

4th

C

A

D

A

Because no candidate received a majority of first place votes, the candidate with fewest first place votes is eliminated, which is candidate C. The adjusted preference schedule is shown below: Number of Voters 4000 7000 3000 6000 1st

D

B

A

B

2nd

A

D

B

D

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B

3rd

A

D

A

Again, because no candidate has the majority of first place votes, the candidate with the fewest first place vote is eliminated, which is candidate D. The adjusted preference schedule is shown below: Number of Voters 4000 7000 3000 6000 1st

A

B

A

B

2nd

B

A

B

A

Finally, there is candidate who receives the majority of first place votes, which is candidate B with 13,000 votes. So candidate B wins the election using the Plurality with Elimination Voting Method. Preference Ranking: A preference ranking is a voter’s order of preference of the candidates. Example 2: Three candidates are running for chairman of a institutional organization. The preference ranking of the voters are as follows.

Number of Voters 16 5 10 8 10 7 Perona 1

1

2

3

2

3

Reyes

2

3

1

1

3

2

Santos 3

2

3

2

1

1

a)Who would win in a plurality election with a runoff between the top 2 finishers? b)In a plurality election, could the seven voters who ranked Santos first and Reyes second achieve a preferable outcome by voting strategically?

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c) In a plurality election with a runoff between the top candidates, could the seven voters who ranked Santos first and Reyes second achieve a preferable outcome by voting strategically? How to solve this: a)We saw in the plurality election, the top two candidates are Perona and Reyes. We have to see how Santos’ voters will change their votes. According to the preference ranking, 10 will switch to Perona so Perona would have his 21 original votes and 10 more, giving him 31. Reyes would have his original 18 and another 7, giving him 25 votes. Perona would thus still win. b)Yes, if the 7 voters had voters had voted for Reyes in the plurality election instead of Santos, Reyes would have 25 votes, and so Reyes would have won. These 7 votes prefer Reyes to Perona, and so this is a preferable outcome. c)First, if the 7 voters switch their votes in the first round of voting, they can’t change who ends up in the runoff, since Reyes is already in the runoff. Second, in the runoff, they prefer Reyes but voting him doesn’t affect the outcome, since Perona still wins. Thus, the answer is no. 4. Pairwise Comparison Method ∙ Compare each two candidates head-to-head. ∙ Award each candidate one point for each head-to-head victory. ∙ The candidate with the most points wins. Example 1:

Number of Voters 18 11 9

∙

5

2

1st choice

A

C

D B C

2nd choice

B

D

C D B

3rd choice

C

B

B C D

4th choice

D

A

A A A

Compare A to B. 155

⮚ 18 voters prefer A. ⮚ 11 + 9 + 5 + 2 = 27 voters prefer to B. ⮚ B wins the pairwise comparison and gets 1 point. ∙

Compare A to C. ⮚ 18 voters prefer A. ⮚ 11 + 9 + 5 + 2 = 27 voters prefer to C. ⮚ C wins the pairwise comparison and gets 1 point.

∙

Compare A to D. ⮚ 18 voters prefer A. ⮚ 11 + 9 + 5 + 2 = 27 voters prefer to D. ⮚ D wins the pairwise comparison and gets 1 point. Compare B to C. ⮚ 18 + 5 = 23 voters prefer B. ⮚ 11 + 9 + 2 = 22 voters prefer to C. ⮚ C wins the pairwise comparison and gets 1 point.

∙

∙

Compare B to D. ⮚ 18 + 5 + 2 = 25 voters prefer B. ⮚ 11 + 9 = 20 voters prefer to D. ⮚ B wins the pairwise comparison and gets 1 point.

∙

Compare C to D. ⮚ 18 + 11 + 3 = 32 voters prefer C. ⮚ 9 + 5 = 14 voters prefer D. ⮚ C wins the pairwise comparison and gets 1 point.

Contestant A

Total Number of Points 0

156

B

2

C

3

D

1

Therefore, C wins using pairwise comparison method. Example 2:Let’s say we have a town of 20,000 people electing a mayor using the Plurality with Elimination Voting Method. There are 4 candidates, candidate A, candidate B, candidate C, and candidate D.

Number of Voters 4000 7500 3000 5500 1st

D

C

A

B

2nd

A

B

B

D

3rd

B

D

C

C

4th

C

A

D

A

∙

Compare A to B. ⮚ 4000 + 3000 = 7000 voters prefer A. ⮚ 7500 + 5500 = 13000 voters prefer to B. ⮚ B wins the pairwise comparison and gets 1 point.

∙

Compare A to C. ⮚ 4000 + 3000 = 7000 voters prefer A. ⮚ 7500 + 5500 = 13000 voters prefer to C. ⮚ C wins the pairwise comparison and gets 1 point.

∙

Compare A to D. ⮚ 3000 voters prefer A. ⮚ 4000 + 7500 + 5500 = 17000 voters prefer to D. ⮚ D wins the pairwise comparison and gets 1 point.

157

∙

∙

∙

Compare B to C. ⮚ 4000 + 3000 + 5500 = 12500 voters prefer B. ⮚ 7500 voters prefer to C. ⮚ B wins the pairwise comparison and gets 1 point. Compare B to D. ⮚ 7500 + 3000 + 5500 = 16000 voters prefer B. ⮚ 4000 voters prefer to D. ⮚ B wins the pairwise comparison and gets 1 point. Compare C to D. ⮚ 7500 + 3000 = 10500 voters prefer C. ⮚ 4000 + 5500 = 9500 voters prefer D. ⮚ C wins the pairwise comparison and gets 1 point.

Contestant A

Total Number of Points 0

B

3

C

2

D

1

Therefore, B wins using pairwise comparison method.

158

159

160

161

162

CHAPTER

7

MATHEMATICS AS A TOOL : GRAPH THEORY

LEARNING OUTCOMES: ∙ ∙ ∙

At the end of the chapter, the students are expected to : Define Graph theory, nodes, edges, paths and different kind of paths. Determine the Euler and Hamiltonian paths and circuits. Solve real life problems using concepts, algorithms and theorems of Graph Theory

LESSON 1 : Basic Terms Graph ∙

A diagram that contains information and depicts connection and relationship between the various parts of the diagram.

Examples:

163

1.) Road Map 2.) Circuit Diagram 3.) Flow Chart 4.) Transportation Route 5.) Tree Diagram

Essential Features of a Graph 1.) The Objects – referred to as the nodes or vertices 2.) Edges – the connecting lines Example. Consider the following graphs:

1.) The graph has 4 vertices and 4 edges

2.) The graph has 4 vertices and 3 edges. 3.) The graph has 5 vertices and 8 edges

164

Note: From example 3; 1. Edge e4 is known as Loop. Loop is an edge that connects a vertex to its self. 2. Edges e7 and e8 are called multiple edges or parallel edges. Parallel Edges – are edges that connect the same vertices. Path - A path in graph theory is a sequence of edges.

Example: Consider the following graphs below. G1:

Paths 1.) ade 2.) adc 3.) bce 4.) abced 5.) adecb Note:

Length 3 3 3 5 5

1. A path can repeat edges. 2. The length of a path refers to the number of edges connected.

165

3. If the direction is not indicated in the graph by an arrow, the movement is can be in any direction to find a path (you can move backward and forward). However if there is an arrow indicating direction the movement in finding a path is in accordance with the indicated arrow . G2 :

Paths 1. e4 e3 2. e3 e2 3. e1 e5 4. e1 e5 5. e1 e5

e4 e4 e4 e3 e2 e5 e4

Length 2 3 3 5 4

Vertex Sequence of a Path - a path is written in terms of edges. A path determines a sequence of vertices. G3:

Path

Length

Vertex Seq.

# of Edges

# of Vertex 166

1. adeecbd 7 ) 2. cbdec 5 ) 3. eecbdecc 8 )

XYZZZWYZ

7

8

ZWYZZW

5

6

ZZZWYZZWZ

8

9

Note: 1.) The number of vertices in vertex sequence is always one larger than the number of edges in the path. 2. ) If a path passes through a loop, the vertex of the loop is repeated in the vertex sequence. Brief History of Graph Theory -From TeroHarju Lecture Note Graph theory may be said to have its beginning in 1736 when EULER considered the (general case of the) Königsberg bridge problem: Does there exist a walk crossing each of the seven bridges of Königsberg exactly once?(Solution Problema tisadgeometriamsituspertinentis,CommentariiAcademiaeScientiarumImpe rialisPetropolitanae 8 (1736), pp.128-140.) It took 200 years before the first book on graph theory was written. This was “TheoriederendlichenundunendlichenGraphen” (Teubner,Leipzig,1936) by KÖNIG in 1936. Since then graph theory has developed into an extensive and popular branch of mathematics, which has been applied to many problems in mathematics, computer science, and other scientific and not- so- scientific areas. For the history of early graph theory, see N.L. BIGGS, R.J. LLOYD AND R.J. WILSON, “Graph Theory1736 – 1936”, Clarendon Press, 1986.-

167

The Konigsberg Bridge Problem

Closed Path - A closed path is said to be closed path if the first and the last vertices of its vertex sequence are the same. Example: G4:

Path 1.) acef 2.) acb 3.) fed 4.) ecabd

Vertex Sequence LMNOP LMNL PONO ONMLNO

(not closed) (closed) (not closed) (closed)

Cycle - a path is called a cycle if the following conditions are satisfied: a.) the path is closed, b.) the path repeat no edges, c.) the vertices of the vertex sequence of the path are all distinct except for the 1st and last vertices which are the same vertices.

168

Example: G5:

Path

Vertex Sequence

a.) e1 e2 e5 e3

ABDCB

b.) e4 e5 e2 e3

CCDBC

c.) e2 e3 e5

DBCD

(not closed, not a cycle) (closed, not a cycle) (closed, cycle)

Connected Graph - a graph is called connected graph if for any two given vertices there is a path connecting them. Example: G6: Consider the following graphs;

not connected

connected

169

connected

Complete Graph -is a connected graph where every pair of vertices is joined by an edge. Example: G7: a.)

b.)

c.)

Simple Graph -a graph is simple if it has neither loop nor parallel edges. Example: G8: a.)

Simple graph

Not a simple graph b.)

170

c.)

Not a simple graph

Degree of a Vertex -

The degree of a vertex is defined as the number of edges connected to the vertex. If a graph contains a loop, the loop contributes 2 to the degree of the vertex.

Example: G9:

Vertex A B C D

a.)

Degree 3 6 4 5

Acyclic Graph - A graph is called acyclic graph if it has no cycle. Example: G10: a.)

No cycle; acyclic

171

b.) Cycle: CDHG - wsrv Not Acyclic

Weighted Graph - A graph whose edges are assigned with weights. Weight may represent mileage, time, cost, or some other quantities. Example: Consider the transportation route of a salesman; G11:

Note: The vertices correspond to the different cities while edges represent the distances in kilometers. Tree - An acyclic connected graph. - Properties: a.) acyclic graph b.) no cycle path c.) connected graph -Forest -refers to several trees

172

Example: G12:

Acyclic, connected -tree

Spanning Tree -a sub-graph tree of a graph that contains all the vertices of the graph. Example: G13:

G13.1 G13.2 G13.1 and G13.2 are spanning trees of G13.

Note: If a graph has an n-edges then there are n! Spanning tree in the graph. 173

Example: G13.3

5 edges 5! = 120 spanning tree

Minimal Spanning Tree - the minimal spanning tree of a graph is a spanning tree of the graph with a minimum total weights - a connected graph has always at least one minimal spanning tree. Example: G14:

G14.1

G14.1 is the minimal spanning tree of G14,i.e., 5+8+10 (23 total weights) 174

Two Special Circuits A. Euler Circuit: - A closed path in a graph which uses each of the edges exactly once, - Named after Leonard Euler(April 15, 1707 –September 18, 1783), a Swiss Mathematician and Physicist. He started working on graphs from year 1736. Euler Theorem: -The graph has an Euler Circuit if the graph is connected and the degree of the vertices must be even. Euler Path - a path that uses each edge of a graph exactly once. Euler Path Theorem -a connected graph has an Euler path if and only if the graph has two vertices of odd degree with all other vertices of even degree. Example: G15: a.)

b.)

The graph is connected, Contains Euler path, Two of the vertices have odd degree, By Euler Theorem, the graph has no Euler Circuit

Not connected, Does not contain Euler path and Euler circuit.

175

c.)

Connected, contains Euler path, All vertices has an even degree, Contains Euler Circuit.

Solution in Konigsberg Bridge Problem

Konigsberg bridge problem: Does there exist a walk crossing each of the seven bridges of Königsberg exactly once? Graph Representation: G16:

By Euler Theorem, the graph contains vertices having an odd degree; the graph has no Euler Circuit. There is no closed path that will bring us to each island by crossing each bridge exactly once. Therefore, the Konigsberg Bridge problem has no solution. 176

Fleury’s Algorithm -Is used to find an Euler Circuit in a graph if the graph has one. Steps: 1. Select any of the vertices in the graph as the starting point. 2. Select any edge connected to the vertex selected in step 1. Remove the edge. (The removal of the edge must not disconnect the starting vertex or the starting point). After the removal of the edge a new vertex is reach. 3. Select an edge connected to the new vertex and repeat step 2. 4. Repeat step 3 until the starting vertex is reach. Example: G17: Let Vertex X be the starting point. Remove edge f

Remove edge d

Remove edge e

Remove edge c

177

Remove edge b

Remove edge a Therefore, the Euler Circuit is fdecba (with a vertex sequence: XYZZWYX). B.

Hamilton Circuit - A closed path which uses each vertex of the graph exactly once, except for the last vertex which duplicate the 1st vertex. - Also known as Hamiltonian Circuit. - Named after Sir William Rowan Hamilton (1805-1865) He marketed a puzzle in the mid 1800 in the form of a dodecahedron that contains the name of a city in each corner. The problem is to visit each city exactly once by travelling along the edges and be able to return to the starting city. (D.S. Malik and M.K. Sen)

The Puzzle Example: G18: 178

a.)

b.)

The graph has Hamilton circuit: V1V2V3V4V5

The graph has no Hamilton circuit, since it will make use of the middle vertex (V) twice.

Dirac’s Theorem - Consider a connected graph with at least 3 vertices and no multiple edges. Let n be the number of vertices in the graph. If every vertex has degree of at least n/2, then the graph must be HAMILTONIAN.

Example: G19:

Algorithm used to find a Hamiltonian Circuit Greedy Algorithm -

Also known as shortest path algorithm, 179

-

Developed by Dijkstra.

Steps: 1. Choose a vertex to start at, then travel along the connected edge that has the smallest weight.( if two or more vertices have the same weight, pick any one) 2. After arriving at the next vertex, travel along the edge of smallest weight that connects to a vertex not yet visited. Continue this process until you have visited all vertices. 3. Return to the starting vertex.

Example: Consider Graph 11. G11:

Let C1 be the starting point, choose edge 80,

From C2, choose edge 90,

180

From C4, choose edge 160 (in this case edge 80 is not advisable since we will cross C3 twice),

From C7, choose edge 110 instead of edge 70 (since we will cross C5 twice if we choose edge 70),

Frrom C8, choose edge 105,

181

From C5 choose edge 60,

From C6, choose edge 200

From C3, choose edge 90,

182

The Hamilton Circuit is now completed, with a vertex sequence: C1C2C4C7C8C5C6C3C1 total weight : 80+90+160+110+105+60+200+90=835. The Edge-picking Algorithm Steps: 1. Mark the edge of the smallest weight in the graph.(if two or more edges have the same weight, pick only one) 2. Mark the edge of next smallest weight in the graph, as long as it does not complete a circuit does not add a third marked edge to a single vertex. 3. Continue this process until you can no longer mark any edges. Then mark the final edge that completes the Hamiltonian circuit. Planar Graph -is a graph that can be drawn so that no edge intersects each other (except at vertices) Example: G20:

G20.1

a.)

G20 is a planar graph since it can be re-drawn (G20.1) without any intersecting edges. b.) G20.2

G20.3

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G20.2 is NOT a planar graph since it cannot be re-drawn (G20.3) without any intersecting edges. Matrix Representation of a Graph Rule: If a graph has N vertices its matrix representation has NxN shape, denoted by M. Example: Write the matrix representation of the graph below. G21:

M=

W X Y Z

W 0 2 1 1

X 2 0 0 0

Y 1 0 0 1

Z 1 0 1 2

Note:

1.) The entries indicate the edges between the vertices. M(W,X) = 2 means that there are 2 edges that connects W & X. M(X,Y) = 0 means that there is no edge that connects X & Y. 2.The diagonal indicates the loop/s in the graphs. M(Z,Z) =2 means that there is a loop that is connected to vertex Z 3.)The degree of the vertex is equal to the sum of the entries in its row or column. Deg(W) = 4 (sum of its column entry or of its row entries)

184

4.) The matrix is symmetric with respect to its diagonal.

Adjacency Matrix of Simple Graph Theorem: If M is the adjacency of a simple graph, then the row* column entry of Mn is equal to the number of length n from vertex from row to the vertex in column, n=1,2,3…

Example. Consider the Graph G22:

How many paths of length 2 can be used to reach a vertex from another? Solution: Path of Length 2 = M2 = MxM The Matrix (M) of the is; W X W 0 1 M X 1 0 Y 1 1 = Z 1 1

Y 1 1 0 1

Z 1 1 1 0 185

Determine M 2 = M x M

MxM=

W X Y Z

M 2=

W 0 1 1 1 W X Y Z

X 1 0 1 1

Y 1 1 0 1 W 3 2 2 2

Z 1 1 1 0 X 2 3 2 2

W X Y Z

x

Y 2 2 3 2

W 0 1 1 1

X 1 0 1 1

Y 1 1 0 1

Z 1 1 1 0

Z 2 2 2 3

M2(W,W)= 3, means there 3 paths of length 2 that connects vertex W to itself. That is; e1e1, e6e6, e4e4. 2 M (X,Z)=2 means there are 2 paths of length 2 that connects vertex x to vertex Z, that is; e1e4 and e2e3. Since the Matrix is symmetric, we can consider the upper diagonal or the lower diagonal. Therefore the total number of paths of length is 3+3+3+3+2+2+2+2+2+2 =24 paths. Note: Paths of length 3 is M3 of length 4 is M4. Lesson 2: Applications of Graph Teory The Pipeline Problem The MAYNILAD is considering 8 cities to be connected by a pipeline. The distances (in km) between cities are given in the graph below:

186

Determine the minimum length of pipe that MAYNILAD is needed to connect the 8 cities. Use the minimal spanning tree. Solution:

Total weight of the minimal spanning tree = 80 + 90+ 80+150 +60+70+105 = 635 Therefore, Maynilad needs 635km length of pipeline to connect the 8 cities. The Map Coloring Problem The Four Color Theorem - Every map in the plane consisting of connected region without a hole can be colored with four different colors without coloring two adjacent regions with the same color. Example: Consider the Map that shows some of the states of US.

187

How the map should be colored with 4 colors if no two adjacent states should have with the same color? Solution: Draw the graph representation of the map. a.) Vertices are the regions or states b.) Edges –to connect two vertices if the states or regions corresponding to these vertices are adjacent. Graph representation,

Use 4 colors to be assigned to its vertices. Assigned 2 different colors to any 2 adjacent vertices.

188

Travelling Salesman Problem (Hamiltonian Problem) Example: In 1859, Sir William Rowan Hamilton marketed a game called Around the World. The game consisted of a regular dodecahedron made of wood. Each corner bore the name of a famous city of the world. The game was to find a path starting at any city, travelling along the edges of the dodecahedron, visiting each city exactly once and returning to the starting city. The diagram below represents the game. (Source: D.S. Malik and M.K. Sen)

Around The World Solution: Using the Hamiltonian Circuit: 189

Therefore, one of the Hamiltonian Circuit that can be an answer to the puzzle is the path with vertex sequence of V1V2V3V4V5V14V13V12V11V10V9V8V7V17V18V19V20V16V15V6V1.

190

191

192

193

194

195

196

References:

Altares, Priscilla S., Copo, A. R., et al, (2013)Elementary Statistics with Computer Applications, REX Book Store Baltazar, E. C., Ragasa, Carmelita, Evangelista, J. (2018). MATHEMATICS in the Modern World, C& E Publishing, Inc. F.J. MacWilliams, and N.J.A. SLoane, (1981), "The Theory of ErrorCorrecting Codes", Bell Laboratories, Murray Hill, J.A. Buchmann,(2002), "Introduction to Cryptography", SpringerVerlag, New York, Lipmann, David, (2007), “Math in Society, Edition 2.5”, Piece College Ft Steilacom Development. Malik and Sen, (2010), “ Discrete Mathematics”, C& E Publishing, Inc. Nocon, R. & Nocon E. (2018), “Essential Mathematics for the Modern World” , C& E Publishing, Inc. Punsalan, Twila G., Uriarte, Gabriel G. (2013) Simplified Approach, REX Book Store.

STATISTICS: A

Parreno, Elizabeth B., Jimenez, Ronel O., ( 2006) BASIC STATISTICS: A Worktext, C &E Publishing, Inc. W.C. Huffmann, and V. Pless, Cambridge University Press,

(2003) "Fundamentals of Error Correcting Codes",

Website : www. Wikipedia

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GLOSSARY A Acyclic graph - if graph has no cycle. Apportionment - Is a method of dividing a whole into various parts. Apportionment principle – when a new representative is added to a sub – group due to an increase in population. Argument - reasons offered or against something. B Borda Count Method - Award points to candidates based on preference Schedule, then declare the winner to the candidate with most points. C Ciphertext - refers to the coded message. Closed path - if the first and the last vertices of its vertex sequence are the same. Communication channel - is the physical medium through which information is transmitted. Communication system - is usually composed of sender (or message source), communication channel, and the receiver. Compound proposition - conveys two or more idea. Conclusion – last step in the reasoning process. Conjecture - is an educated guess based on repeated observation of a particular Process or pattern. Connected graph - if for any two given vertices there is a path connecting Connected graph - if for any two given vertices there is a path connecting them. Continuous variables - can assume an infinity of many possible values Corresponding to the point on a line interval. Correlation - measures the association or the strength of the relationship between two variables say x and y. Counterexample - a statement that disproves the conjecture. Cryptography - Is a method of making and breaking of secret codes.

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cycle - if the path is closed and the path repeat no edges. D Data - a result of experiment, observation, investigation and often appears as numerical figure. Decile- is a measure of position that divides the ordered observations or score into ten equal parts. Decoding - is the opposite process of encoding. Decryption - is the process of returning/converting back the coded message Into plain text. Deductive reasoning - process of solving problems by applying premises , syllogism and conclusion. Degree of a vertex - is defined as the number of edges connected to the vertex. Descriptive statistic - division of statistics that summarizes or describes the important characteristics of a given data set. Discrete variables - can assume a finite or countable number of values. Dispersion - measure the degree of clustering of data about a central point. E Edges – the connecting lines Encoding - is transforming messages into bits of message that is suitable to the communication channel. Encryption - Is the process of transforming plain text into code form using a certain algorithm. Euler Circuit: - a closed path in a graph which uses each of the edges exactly once, Existential quantifier - means “ there exists “ F Fibonacci sequence - is a series of numbers where a number us found by adding the two numbers before it. G Golden ratio - also called the golden section, golden mean or divine proportion. Graph - a diagram that contains information and depicts connection and relationship between the various parts of the diagram. Graphene – means writing.

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H Hamilton Circuit - a closed path which uses each vertex of the graph exactly once, except for the last vertex which duplicate the 1st vertex. Hamilton plan method - given the number of seat in the US congress will be apportioned between states proportionally to their population. Huntington – Hill apportioned method - the method that make use of equal proportion. I Inductive reasoning - same method would work for any similar type of problem to arrived at a conclusion. Inferential statistics - aims to give information about the population by studying the characteristics of the sample drawn from it. Interval - it gives meaningful amount of differences between data but does not have a true – zero starting point. J Jefferson plan method - is a method that uses a modified standard divisor the arrives at the correct or exact numbers of representative using trial and error . K Key - refers to the strings of information that is used to reveal the encrypted message into readable form. Krypto - means hidden.

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L Language - Is the fundamental tool that bridges the gap among people from varying origins and culture. Linear Regression - is very powerful tool in predicting the value of one variable to the given value of another variable . Loop - is an edge that connects a vertex to its self M Mean - sum of all the values in the observation divided by the total number of observation. Median -is considered as the physical middle point of the distribution. Minimal spanning tree - is a spanning tree of the graph with a minimum total weights. Mode - is the most occurring value in the distribution. Modular Arithmetic - is an operator (mod), which seeks for a remainder when two numbers are divided. N Nominal data - data that consist names, labels or categories only. Normal curve - is a symmetrical curve having a bell-like shape. O Objects – referred to as the nodes or vertices Ordinal data - measurements which deals order or rank but difference between rank does not exist. P Pairwise Comparison Method - compare each two candidates head-tohead and award each candidate one point for each headto head victory. Parallel Edges – are edges that connect the same vertices. Parameter - numerical measures that describe the population of interest. Parity check - encoding scheme where in a single bit is added to the message as

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redundancy bit. Path - in graph theory is a sequence of edges. Pearson Product- Moment Correlation - Is the most familiar sort of statistical tool in quantifying the linear relationship between two random variables, x and y. Percentiles – is a measure of position that divides the ordered observations or score distribution into 100 equal parts. Plain text - refers to the original message. Plurality by Elimination - a voting method which is also known as an instant run-off voting and sequential run-off voting Plurality Method - the winner is the candidate or choice with the most votes. Polya’s strategy - a four step problem solving strategy. Population - set of complete collection of all possible values of the variable. Premises – assertions that serve as the basis for an argument. Proposition - is a declarative statements that may expressed an idea which can be true or false but not both. Q Quantiles - is a measure of position determines the position of a single value in relation to other values in a sample or a population data set. Quartile- is a measure of position that divides the ordered observations or score distribution into four equal parts. R Range - the highest value minus the lowest value. Ratio - a modified interval level to include the starting point zero. S Sample - portion of the population. Set - is a collection of, numbers, people , letters of the alphabet. Shift cipher - uses shift in forming the key of cryptography. Simple graph - a graph is simple if it has neither loop nor parallel edges. Simple proposition - means single idea. Source coding - is defined as converting the messages from the sender into bits suitable to the communication channel. Spanning tree - a sub-graph tree of a graph that contains all the vertices of the

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graph. Standard divisor - the number of voters represented by each representative. Standard quota - the whole part of the quotient when the population of the sub – group is divided by the standard divisor. Statistics – is a collection, presentation ,analysis and interpretation of data . Strategic Voting - when a person votes in a way that does not reflect his or her true preferences in an attempt to improve the outcome of the election. Syllogism - argument composed of two statements followed by a conclusion. Synonyms - different words with same meaning. T Tree - an acyclic connected graph. U Universal quantifier - means “ for all “ or “ for every “ V Variable - attribute of interest observable of each entity in the universe. Variance - is the square of the deviation of data sets from its mean. Vertex Sequence - a path is written in terms of edges. W Weighted graph - a graph whose edges are assigned with weights. Weight may represent mileage, time, cost, or some other quantities.

Z Z-score - these standard scores tell how far a raw score is from the mean in standard deviation units.

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