# 18.1: Complex numbers - Mathematics

Informally, a complex number is a number that can be put in the form

[z=x+icdot y, ]

where (x) and (y) are real numbers and (i^2=-1).

The set of complex numbers will be further denoted by (mathbb{C}). If (x), (y), and (z) are as in 18.1.1, then (x) is called the real part and (y) the imaginary part of the complex number (z). Briefly it is written as

[x= ext{Re} z ext{and} y= ext{Im} z.]

On the more formal level, a complex number is a pair of real numbers ((x,y)) with the addition and multiplication described below; the expression (x + icdot y) is only a convenient way to write the pair ((x,y)).

[egin{aligned} (x_1+icdot y_1) + (x_2+icdot y_2) &:= (x_1+x_2) + icdot(y_1+y_2); (x_1+icdot y_1)cdot(x_2+icdot y_2) &:= (x_1cdot x_2-y_1cdot y_2) + icdot(x_1cdot y_2+y_1cdot x_2). end{aligned}]

This text book is part of the reform of the school curriculum in Rwanda, that is changes in what is taught in schools and how it is taught. It is hoped that this will make what you learn in school useful to you when you leave school. In the past, the main thing in schooling has been to learn knowledge, that is, facts and ideas about each subject. Now, the main idea is that you should be able to use the knowledge you learn by developing skills or competencies. These skills or competencies include the ability to think for yourself, to be able to communicate with others and explain what you have learnt, and to be creative, that is developing your own ideas not just following those of the teacher and the text book. You should also be able to find out information and ideas for yourself rather than just relying on what the teacher or text book tells you.

Activity-based learning

This book has a variety of activities for you to do as well as information for you to read. These activities present you with materials or things to do which will help you to learn things and find out things for yourself. You already have a lot of knowledge and ideas based on the experiences you have had and your life within your own community. Some of the activities, therefore, ask you to think about the knowledge and ideas you already have.

In using this book, therefore, it is essential that you do all the activities. You will not learn properly unless you do these activities. They are the most important part of the book. In some ways, this makes learning more of a challenge. It is more difficult to think for yourself than to copy what the teacher tells you but if you take up this challenge, you will become a better person and become more successful in your life.

You can learn a lot from other people in your class. If you have a problem, it can often be solved by discussing it with others. Many of the activities in this book, therefore, involve discussion in groups or pairs. Your teacher will help organise these groups and may arrange the classroom so that you are always sitting in groups facing each other. You cannot discuss properly unless you are facing each other.

One of the objectives of the competence based curriculum is to help you find things out for yourself. Some activities, therefore, ask you to do research using books in the library, the internet if your school has this, or other sources such as newspapers and magazines. This means that you will develop the skills of learning for yourself when you leave school. Your teacher will help you if your school does not have a good library or internet.

To guide you, each activity in the book is marked by a symbol or icon to show you what kind of activity it is. The icons are as follows:

## Complex Numbers

When we combine a Real Number and an Imaginary Number we get a Complex Number:

### Can a Number be a Combination of Two Numbers?

Can we make up a number from two other numbers? Sure we can!

We do it with fractions all the time. The fraction 3 /8 is a number made up of a 3 and an 8. We know it means "3 of 8 equal parts".

Well, a Complex Number is just two numbers added together (a Real and an Imaginary Number).

Balbharati solutions for Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board chapter 1 (Complex Numbers) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the Maharashtra State Board Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board solutions in a manner that help students grasp basic concepts better and faster.

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Concepts covered in Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board chapter 1 Complex Numbers are Introduction of Complex Number, Concept of Complex Numbers, Algebra of Complex Numbers, Square Root of a Complex Number, Fundamental Theorem of Algebra, Argand Diagram Or Complex Plane, De Moivres Theorem, Cube Root of Unity, Set of Points in Complex Plane.

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## 18.1: Complex numbers - Mathematics

Final Exam Review Session. A review session for the final exam will be held on Monday, December 13, 3:00pm - 4:30pm in Room 242, Science and Technology 1.

Exercises for Chapter 18. The exercises for Chapter 18 are as follows:

Section 18.1 1, 3, 5, 7, Collected exercises: 2, 6

Section 18.2 1, 5, Collected exercise: 2

Section 18.3 3, 5, Collected exercise: 2

The last homework assignment is due on Tuesday December 14 and is optional. Your grade on this assignment will replace your lowest current homework grade (or not if your grade on this assignment is your lowest homework grade). If you choose not to turn it in it will not effect your homework average.

The Final Exam. The final exam for this course will be given on Tuesday December 14, 10:30am - 1:15pm in the same room where we have class. The final exam will be cumulative with approximately half of the exam covering Chapters 16 and 17, and the rest of the exam being approximately equally distributed among the other Chapters covered in the course. You are permitted to bring two 8.5" x 11" sheets to class with formulas. Calculators are permitted but you must show all work in order to receive credit for a problem. I will shortly be updating this space with information on which sections will not appear on the final exam.

The following sections will be covered on the final. Understand that you will be expected to know material from sections that are not covered on the final to the extent that that material appears in sections that are covered.

18.1, 18.3 (one problem only at most).

Grades for Midterm Exam 2. Your score for Exam 2 will be counted toward your final average as though the exam were out of 54 points. So for example if you got 44/60 = 73% on the exam, the score used for calculating your semester grade will be 44/54 = 81%.

Midterm Exam 2. The second midterm exam will be given on Thursday November 18. It will cover all sections in Chapters 13-15. You will have the full class period to do this exam. The problems on the exam will be similar to the odd numbered assigned exercises. You are permitted to bring with you one 8.5 x 11 inch sheet of paper with formulas on it (both sides). A calculator is also permitted but you must show all work in order to receive credit for a problem.

Homework #10. Homework #10 is now due on Tuesday, November 23 so that I might have time to cover the material adequately and so that you might have time to study for Thursday's exam. This change is reflected in the due dates below.

New Due Dates for Homeworks. The due dates and coverage of homework sets to be handed in have been changed below. The new due dates start with the assignment due this Thursday.

Midterm Exam 1. The first midterm exam will be given on Thursday October 7. It will cover all sections in Chapters 9 and 10 except for Section 10.9. You will have the full class period to do this exam. The problems on the exam will be similar to the odd numbered assigned exercises. You are permitted to bring with you one 8.5 x 11 inch sheet of paper with formulas on it (both sides). A calculator is also permitted but you must show all work in order to receive credit for a problem.

Additional Clarification on Expectations Regarding Collected Homework Sets: All homework sets must be stapled if they exceed one page. Any other method of joining pages together, including paper clips, fold-overs, or chewing gum is not acceptable.

Clarification on Expectations Regarding Collected Homework Sets: Please understand that homework sets are not to be turned in by email. Allowing homework sets to be emailed is a privilege I am extending to prevent someone losing credit on a homework assignment because he or she is late to class for reasons beyond his or her control. I do expect you to turn in a printout of your homework assignment as soon as you arrive to class. It is not acceptable to email the assignment in and then not show up for class. If you know you will be absent because of an unavoidable conflict, we can make a suitable arrangement, but under normal circumstances a printout must be turned in before the end of class on the day the assignment is due.

New due date for Homework #3: Homework #3 is due Thursday September 23, not Tuesday, September 21. This update has been changed below as well.

VT Math and Putnam Competition. The Math Department is once again assembling a GMU math study group/team for participating in the Virginia Tech Math Competition and the Putnam Math Competition.

Interested in mathematics (you do not need to be a math major)

Weekly (informal) meetings to work on Putnam/VT problems from previous years (date/time TBD).

VT Exam - Saturday Oct. 30, 2010.

Putnam Exam - Saturday Dec. 4, 2010.

Who: Contact Dan Anderson [email protected].

Uncollected Homeworks: When a homework set is graded, I will return them to you during the next class. All uncollected homework will be put in an envelope outside my office door. You can come by and pick it up anytime. I will not bring uncollected homework to class more than once.

Expectations Regarding Collected Homework Sets: I want to be absolutely clear on my expectations regarding submitted homework sets. All of these must be met or else the homework set will not be accepted and you will get a zero for it.

All homework sets must be submitted on time. "On time" means at the beginning of class (that is 10:30am) on the day the assignment is due. You should arrange to arrive to class early on those days in order to make sure your homework set is turned in on time. If there are issues in your life that are unpredictable (e.g. traffic, children, etc.) and undermine your confidence that you can get to class early, then you may submit the assignment to me by email before class and then bring the printout to class. Even if you turn in the printout late, I will consider that the homework has been submitted on time. Also if you want to turn in the assignment early that is fine.

All homework sets must be typed using some kind of software that can reproduce mathematical symbols. If there are some symbols that your software is lacking you may leave space on your homework and write those symbols in by hand. If you want to include a hand-drawn diagram or some other illustration on your homework, that is fine, but any calculations you do must be typed. You should make sure to leave enough time before class to get a decent printout of your assignment. "The printer did not work" is not a valid excuse.

All homework assignments that exceed one page must be stapled together. Paper clips, binder clips, brass fasteners, scotch tape, chewing gum, fold-overs, or anything else you can think of that is not a staple, is not acceptable.

All work must be your own. You may collaborate all you wish (and in fact it is an excellent idea to do so) on the odd-numbered exercises assigned as practice, but the work on the collected homework sets must be your own.

Section 9.1 Vectors in 2-space and 3-space. pdf

Section 9.2 Inner Product (Dot Product). pdf

Section 9.3 Vector Product (Cross Product). pdf

Section 9.4 Vector and Scalar Functions and Fields. Derivatives. pdf

Section 9.5 Curves. Arc Length. pdf

Section 9.7 Gradient of a Scalar Field. Directional Derivatives. pdf

Section 9.8 Divergence of a Vector Field. pdf

Section 9.9 Curl of a Vector Field. pdf

Section 10.1 Line Integrals. pdf

Section 10.2 Path Independence of Line Integrals. pdf

Section 10.4 Green's Theorem in the Plane. pdf

Section 10.5 Surfaces for Surface Integrals. pdf

Section 10.6 Surface Integrals. pdf

Section 10.7 Triple Integrals. Divergence Theorem of Gauss. pdf

Section 10.8 Further Applications of the Divergence Theorem. pdf

Section 10.9 Stokes's Theorem. pdf

Section 13.1 Complex Numbers. Complex Plane. pdf

Section 13.2 Polar Form, Powers and Roots. pdf

Section 13.3 Derivative. Analytic Functions. pdf

Section 13.4 Cauchy-Riemann Equations. pdf

Section 13.5 Exponential Function. pdf

Section 14.1 Complex Line Integrals. pdf

Section 14.2 Cauchy's Integral Theorem. pdf

Section 14.4 Derivatives of Analytic Functions. pdf

Section 15.1 Sequences, Series, Convergence Tests. pdf

Section 15.3 Functions given by Power Series. pdf

Section 15.4 Taylor and Maclaurin Series. pdf

Section 16.1 Laurent Series. pdf

Section 16.2 Singularities and Zeros. Infinity. pdf

Section 16.3 Residue Integration Method. pdf

Section 17.1 Conformal Mapping. pdf

Section 17.2 Linear Fractional Transformations. pdf

Section 18.1 Electrostatic Fields. pdf

Section 18.2 Using Conformal Mapping. Modeling. pdf

Section 18.3 Heat Problems. pdf

Section 18.5 Poisson's Integral Formula for Potentials. pdf

Section 18.6 General Properties of Harmonic Functions. pdf

Homework Assignments:
Homework #1 (due Thursday 09-09-10): Sections 9.1, 9.2, 9.3.
Homework #2 (due Thursday 09-16-10): Sections 9.4, 9.5, 9.7.
Homework #3 (due Thursday 09-23-10): Sections 9.8, 9.9, 10.1, 10.2.
Homework #4 (due Thursday 09-30-10): Sections 10.3, 10.4, 10.5, 10.6.
Homework #5 (due Thursday 10-07-10): Sections 10.7, 10.8, 10.9.

Homework #6 (due Thursday 10-21-10): Sections 13.1, 13.2.

Homework #7 (due Thursday 10-28-10): Sections 13.4, 13.5, 14.1, 14.2.

Homework #8 (due Thursday 11-04-10): Sections 14.3, 14.4, 15.1.

Homework #9 (due Thursday 11-11-10): Sections 15.2, 15.3, 15.4.

Homework #10 (due Tuesday 11-23-10): Sections 16.1, 16.2, 16.3.

Homework #11 (due Thursday 12-02-10): Sections 16.4, 17.1, 17.2

Homework #12 (Optional, due Tuesday 12-14-10): Sections 17.3, 18.1, 18.2, 18.3

Homework Solutions:
Homework #1 pdf
Homework #2 pdf
Homework #3 pdf
Homework #4 pdf
Homework #5 pdf

Exam Solutions:
Exam 1 (10-07-10): Solutions
Exam 2 (11-18-10): Solutions

## Complex conjugate

where a is the real component and b i is the imaginary component, the complex conjugate, z*, of z is:

The complex conjugate can also be denoted using z . Note that a + b i is also the complex conjugate of a - b i .

The complex conjugate is particularly useful for simplifying the division of complex numbers. This is because any complex number multiplied by its conjugate results in a real number:

Thus, a division problem involving complex numbers can be multiplied by the conjugate of the denominator to simplify the problem.

## Number Theory - Types of Math Numbers

Those ten simple symbols, digits, or numbers that we all learn early in life that influence our lives in far more ways than we could ever imagine. Have you ever wondered what our lives would be like without these 10 elegant digits and the infinite array of other numbers that they can create? Birthdays, ages, height, weight, dimensions, addresses, telephone numbers, license plate numbers, credit card numbers, PIN numbers, bank account numbers, radio/TV station numbers, time, dates, years, directions, wake up times, sports scores, prices, accounting, sequences/series of numbers, magic squares, polygonal numbers, factors, squares, cubes, Fibonacci numbers, perfect, deficient, and abundant numbers, and the list goes on ad infinitum. Engineers, accountants, store clerks, manufacturers, cashiers, bankers, stock brokers, carpenters, mathematicians, scientists, and so on, could not survive without them. In a sense, it could easily be concluded that we would not be able to live without them. Surprisingly, there exists an almost immeasurable variety of hidden wonders surrounding or emanating from these familiar symbols that we use every day, the natural numbers.

Over time, many of the infinite arrays, or patterns, of numbers derivable from the basic ten digits have been categorized or classified into a variety of number types according to some purpose that they serve, fundamental rule that they follow, or property that they possess. Many, if not all, are marvelously unique and serve to illustrate the extreme natural beauty and wonder of our numbers as used in both classical and recreational mathematics.

In the interest of stimulating a broader interest in number theory and recreational mathematics, this collection will endeavor to present basic definitions and brief descriptions for several of the number types so often encountered in the broad field of recreational mathematics. The number type descriptions that follow will not be exhaustive in detail as space is limited and some would take volumes to cover in detail. A list of excellent reading references is provided for those who wish to learn more about any specific number type or explore others not included. It is sincerely hoped that the material contained herein will stimulate you to read and explore further. I also hope that after reading, digesting, and understanding the material offered herein, that you will have enjoyed the experience and that you will never utter those terrible, unforgettable words, "I hate math."

Some basic definitions of terms normally encountered in the classroom are given first.

Integers - Any of the positive and negative whole numbers, . - 3, - 2, - 1, 0, + 1, + 2, + 3, . The positive integers, 1, 2, 3. are called the natural numbers or counting numbers. The set of all integers is usually denoted by Z or Z+

Digits - the 10 symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, used to create numbers in the base 10 decimal number system.

Numerals - the symbols used to denote the natural numbers. The Arabic numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are those used in the Hindu-Arabic number system to define numbers.

Natural Numbers - the set of numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17. that we see and use every day. The natural numbers are often referred to as the counting numbers and the positive integers.

Whole Numbers - the natural numbers plus the zero.

Rational Numbers - any number that is either an integer "a" or is expressible as the ratio of two integers, a/b. The numerator, "a", may be any whole number, and the denominator, "b", may be any positive whole number greater than zero. If the denominator happens to be unity, b = 1, the ratio is an integer. If "b" is other than 1, a/b is a fraction.

Fractional Numbers - any number expressible by the quotient of two numbers as in a/b, "b" greater than 1, where "a" is called the numerator and "b" is called the denominator. If "a" is smaller than "b" it is a proper fraction. If "a" is greater than "b" it is an improper fraction which can be broken up into an integer and a proper fraction.

Irrational Numbers - any number that cannot be expressed by an integer or the ratio of two integers. Irrational numbers are expressible only as decimal fractions where the digits continue forever with no repeating pattern. Some examples of irrational numbers are

Transcendental Numbers - any number that cannot be the root of a polynomial equation with rational coefficients. They are a subset of irrational numbers examples of which are Pi = 3.14159. and e = 2.7182818. the base of the natural logarithms.

Real Numbers - the set of real numbers including all the rational and irrational numbers.

Irrational numbers are numbers such as

Rational numbers include the whole numbers (0, 1, 2, 3, . ), the integers (. - 2, - 1, 0, 1, 2, . ), fractions, and repeating and terminating decimals.

Put on it all the whole numbers 1,2,3,4,5,6,7. etc

then put all the negatives of the whole numbers to the left of 0

Then put in all of the fractions .

Then put in all of the decimals [ some decimals aren't fractions ]

Now you have what is called the "real number line"

The way to get a number that is not "real" is to try to find the square root of - 1

can't be 1 because 1 squared is 1, not -1

can't be -1 because the square of -1 is 1, not -1

So there is no number on your number line to be

and new numbers would need to be put somewhere.

The number types currently entered, and/or planned to be entered, are listed below and will be updated as new entries are made in the future. When appropriate, and time permitting, some of the number definitions/descriptions will be expanded further to provide additional information.

Abundant, Algebraic, Amicable, Arrangement, Automorphic, Binary, Cardinal, Catalan, Complex, Composite, Congruent, Counting, Cubic, Decimal, Deficient, Even, Factor, Factorial, Fermat, Fibonacci, Figurate, Fractional, Friendly, Generating, Gnomon, Golden, Gyrating, Happy, Hardy-Ramanujan, Heronian, Imaginary, Infinite, Integers, Irrational, Mersenne, Monodigit, Narcissistic, Natural, Oblong, Octahedral, Odd, Ordinal, Parasite, Pell, Pentatope, Perfect , Persistent, Polygonal, Pronic, Pyramidal , Pythagorean, Quasiperfect, Random, Rational, Real, Rectangular, Relatively Prime, Semi-perfect, Sequence, Sociable, Square, Superabundant, Tag, Tetrahedral, Transcendental, Triangular, Unit Fraction, Whole.

Several of the numbers form unique patterns that are often used in the solution of mathematical problems. When distinct patterns are applicable, the first ten numbers of the patterns will be given along with specific relationships, or equations, that will enable you to find any number in the pattern.

A number n for which the sum of divisors σ(n)>2n, or, equivalently, the sum of proper divisors (or aliquot sum) s(n)>n.

An abundant number is a number n for which the sum of divisors σ(n)>2n, or, equivalently, the sum of proper divisors (or aliquot sum) s(n)>n.

Abundant numbers are part of the family of numbers that are either deficient, perfect, or abundant.

Abundant numbers are numbers where the sum, Sa(N), of its aliquot parts/divisors is more than the number itself Sa(N) > N or S(N) > 2N. (In the language of the Greek mathematicians, the divisors of a number N were defined as any whole number smaller than N that, when divided into N, produced whole numbers. The factors/divisors of a number N, less the number itself, are referred to as the aliquot parts, aliquot divisors, or proper divisors, of the number.) Equivalently, N is also abundant if the sum, S(N), of "all" its divisors is greater than 2N.

Sa(N)-->1..1..1..3..1. 6. 1. 7. 4. 8. 1. 16. 1. 10. 9. 15. 1. 21. 1. 22..11..12. 1. 36

. 12,18,20, and 24 are abundant.

It can be readily seen that using the aliquot parts summation, sa(24) = 1+2+3+4+6+8+12 = 36 > N = 24 while s(24) = 1+2+3+4+6+8+12+24 = 60 > 2N = 48, making 24 abundant using either definition.

Only 21 of the numbers from 1 to 100 are abundant. The number 945 is the first odd abundant number.

Every even integer greater than 46 is expressible by the sum of two abundant numbers.

Every integer greater that 83,159 is expressible by the sum of two abundant numbers.

Any multiple of an abundant number is abundant.

A prime number or any power of a prime number is deficient. The divisors of a perfect or deficient number is deficient.

The abundant numbers below 100 are 20, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90 and 96.

(See perfect, semi-perfect, multiply-perfect, quasi-perfect, deficient, least deficient, super abundant)

Algebraic numbers are the real or complex number solutions to polynomial equations of the form:

The coefficients a, b, c, d, . p, q, are integers or fractions. All rational numbers are algebraic while some irrational numbers are algebraic.

An aliquot part is any divisor of a number, not equal to the number itself. The divisors are often referred to as proper divisors. The aliquot parts of the number 24 are 1, 2, 3,4, 6, 8 and 12..

An almost perfect number is typically applied to the powers of 2 since the sum of the aliquot parts is

or just 1 short of being a perfect number. It follows that any power of 2 is a deficient number

Alphametic numbers form cryptarithms where a set of numbers are assigned to letters that usually spell out some meaningful thought. The numbers can form an addition, subtraction, multiplication or division problem. One of the first cryptarithms came into being in 1924 in the form of an addition problem the words being intended to represent a student's letter from college to the parents. The puzzle read SEND + MORE = MONEY. The answer was 9567 += 1085 = 10,652. Of course, you have to use logic to derive the numbers represented by each letter..

Amicable numbers are pairs of numbers, each of which is the sum of the others aliquot divisors. For example, 220 and 284 are amicable numbers whereas all the aliquot divisors of 220, i.e., 110, 55, 44, 22, 10, 5, 4, 2, 1 add up to 284 and all the aliquot divisors of 284, i.e., 142, 71, 4, 2, 1 add up to 220. Also true for any two amicable numbers, N1 and N2, is the fact that the sum of all the factors/divisors of both, Sf(N1 + N2) = N1 + N2. Stated another way, Sf(220 + 284) = 220 + 284 = 504. Other amicable numbers are:

There are more than 1000 known amicable pairs. Amicable numbers are sometimes referred to as friendly numbers.

Several methods exist for deriving amicable numbers, though not all, unfortunately. An Arabian mathematician devised one method. For values of n greater than 1, amicable numbers take the form:

given that x, y, and z are prime numbers. n = 2 produces x = 11, y = 5 and z = 71 which are all prime and therefore result in the amicable number pair of 220/284. n = 3 produces the x = 23, y = 11 and z = 287 but 287 is composite, being 7x41. n = 4 produces x = 47, y = 23 and z = 1151, all prime, and thereby resulting in the amicable pair of 17,296 and 18,416. In reviewing the few amicable pairs shown earlier, it is obvious that this method does not produce all amicable pairs.

If the number 1 is not used in the addition of the aliquot divisors of two numbers, and the remaining aliquot divisors of each number still add up to the other number, the numbers are called semi-amicable. For example, the sum of the aliquot divisors of 48, excluding the 1, is 75 while the sum of the aliquot divisors of 75, excluding the 1, is 48.

As a matter of general information:

The sum of all the factors/divisors of a number N is given by

which when applied to an example of 60 = 2^2x3x5 results in

. Sf(60) = (2^3 - 1) x (3^2 - 1) x (5^2 - 1) = 7 x 4 x 6 = 168.

Stated another way, the sum of the factors of a number N is given by

which when applied to an example of

As noted earlier, the sum of the aliquot factors/divisors is the sum of all the factors/divisors minus the number itself.

The apocalypse number, 666, often referred to as the beast number, is referred to in the bible, Revelations 13:18.

While the actual meaning or relevance of the number remain unclear, the number itself has some surprisingly interesting characteristics.

The sum of the first 36 positive numbers is 666 which makes it the 36th triangular number.

The sum of the squares of the first seven prime numbers is 666.

Multiplying the sides of the primitive right triangle 12-35-37 by 18 yields non-primitive sides of 216-630-666.

Even more surprising is the fact that these sides can be written in the Pythagorean Theorem form:

Arrangement numbers, more commonly called permutation numbers, or simply permutations, are the number of ways that a number of things can be ordered or arranged. They typically evolve from the question how many arrangements of "n" objects are possible using all "n" objects or "r" objects at a time. We designate the permutations of "n" things taken "n" at a time as nPn and the permutations of "n" things taken "r" at a time as nPr where P stands for permutations, "n" stands for the number of things involved, and "r" is less than "n". To find the number of permutations of "n" dissimilar things taken "n" at a time, the formula is nPn = n! which is "n" factorial which means:

Example: How many ways can you arrange the letters A & B. Clearly 2 which is 2 x 1 = 2, namely AB and BA.

How many ways can you arrange the letters A, B & C in sets of three? Clearly 3P3 = 3 x 2 x 1 = 6, namely ABC, CBA, BAC, CAB, ACB, and BCA.

How many ways can you arrange A, B, C & D in sets of four? Clearly 4P4 = 4 x 3 x 2 x 1 = 24.

To find the number of permutations of "n" dissimilar things taken "r" at a time, the formula is:

Example: How many ways can you arrange the letters A, B, C, and D using 2 at a time? We have 4P4 = 4 x (4-2+1) = 4 x 3 = 12 namely AB, BA, AC, CA, AD, DA, BC, CB, BD, DB, CD, and DC.

How many 3-place numbers can be formed from the digits 1, 2, 3, 4, 5, and 6, with no repeating digit? Then we have 6P3 = 6 x 5 x (6-3+1) = 6 x 5 x 4 = 120.

How many 3-letter arrangements can be made from the entire 26 letter alphabet with no repeating letters? We now have 26P3 = 26 x 25 x (26-3+1) = 26 x 25 x 24 = 15,600.

Lastly, four persons enter a car in which there are six seats. In how many ways can they seat themselves? 6P4 = 6 x 5 x 4 x (6-4+1) = 6 x 5 x 4 x 3 = 360.

Another permutation scenario is one where you wish to find the permutations of "n" things, taken all at a time, when "p" things are of one kind, "q" things of another kind, "r' things of a third kind, and the rest are all different. Without getting into the derivation,

Example, how many different permutations are possible from the letters of the word committee taken all together? There are 9 letters of which 2 are m, 2 are t, 2 are e, and 1 c, 1 o, and 1 i. Therefore, the number of possible permutations of these 9 letters is:

Automorphic numbers are numbers of "n" digits whose squares end in the number itself. Such numbers must end in 1, 5, or 6 as these are the only numbers whose products produce 1, 5, or 6 in the units place. For instance, the square of 1 is 1 the square of 5 is 25 the square of 6 is 36.

What about 2 digit numbers ending in 1, 5, or 6? It is well known that all 2 digit numbers ending in 5 result in a number ending in 25 making 25 a 2 digit automorphic number with a square of 625. No other 2 digit numbers ending in 5 will produce an automorphic number.

Is there a 2 digit automorphic number ending in 1? We know that the product of 10A + 1 and 10A + 1 is 100A 2 + 20A + 1. "A" must be a number such that 20A produces a number whose tens digit is equal to "A". For "A" = 2, 2 x 20 = 40 and 4 is not 2. For "A" = 3, 3 x 20 = 60 and 6 is not 3. Continuing in this fashion, we find no 2 digit automorphic number ending in 1.

Is there a 2 digit automorphic number ending in 6? Again, we know that the product of 10A + 6 and 10A + 6 is 100A 2 + 120A + 36. "A" must be a number such that 120A produces a number whose tens digit added to 3 equals "A". For "A" = 2, 2 x 120 = 240 and 4 + 3 = 7 which is not 2. For "A" = 3, 3 x 120 = 360 and 6 + 3 = 9 which is not 3. Continuing in this manner through A = 9, for "A" = 7, we obtain 7 x 120 = 840 and 4 + 3 = 7 = "A" making 76 the only other 2 digit automorphic number whose square is 5776.

By the same process, it can be shown that the squares of every number ending in 625 or 376 will end in 625 or 376.

The sequence of squares ending in 25 are 25, 225, 625, 1225, 2025, 3025, etc. The nth square number ending in 25 can be derived directly from N(n) 2 = 100n(n - 1) + 25. (This expression derives from the Finite Difference Series of the squares.)

Binary numbers are the natural numbers written in base 2 rather than base 10. While the base 10 system uses 10 digits, the binary system uses only 2 digits, namely 0 and 1, to express the natural numbers in binary notation. The binary digits 0 and 1 are the only numbers used in computers and calculators to represent any base 10 number. This derives from the fact that the numbers of the familiar binary sequence, 1, 2, 4, 8, 16, 32, 64, 128, etc., can be combined to represent every number. To illustrate, 1 = 1, 2 = 2, 3 = 1 + 2, 4 = 4, 5 = 1 + 4, 6 = 2 + 4, 7 = 1 + 2 + 4, 8 = 8, 9 = 1 + 8, 10 = 2 + 8, 11 = 1 + 2 + 8, 12 = 4 + 8, and so on. In this manner, the counting numbers can be represented in a computer using only the binary digits of 0 and 1 as follows.

 Number B I N A R Y S E Q U E N C E 128 64 32 16 8 4 2 1 1 1 B 2 1 0 I 3 1 1 N 4 1 0 0 A 5 1 0 1 R 6 1 1 0 Y 7 1 1 1 8 1 0 0 0 N 9 1 0 0 1 O 10 1 0 1 0 T 11 1 0 1 1 A 12 1 1 0 0 T 13 1 1 0 1 I 14 1 1 1 0 O 15 1 1 1 1 N 76 1 0 0 1 1 0 0 157 1 0 0 1 1 0 1 1

As you can see, the location of the ones digit in the binary representation indicates the numbers of the binary sequence that are to be added together to yield the base 10 number of interest.

A cardinal number is a number that defines how many items there are in a group or collection of items. Typically, an entire group of items is referred to as the "set" of items and the items within the set are referred to as the "elements" of the set. For example, the number, group, or set, of players on a baseball team is defined by the cardinal number 9. The set of 200 high school students in the graduating class is defined by the cardinal number 200. (See ordinal numbers and tag numbers.)

Catalan numbers are one of many special sequences of numbers that derive from combinatorics problems in recreational mathematics. Combinatorics deals with the selection of elements from a set of elements typically encountered under the topics of probability, combinations, permutations, and sampling. The specific Catalan numbers are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16,796 and so on deriving from

This particular set of numbers derive from several combinatoric problems, one of which is the following.

Given "2n" people gathered at a round table. How many person to person, non-crossing, handshakes can be made, i.e., no pairs of arms crossing one another across the table? A few quick sketches of circles with even sets of dots and lines will lead you to the first three answers easily. Two people, one handshake. Four people, two handshakes. Six people, 5 handshakes. With a little patience and perseverance, eight people will lead you to 14 handshakes. Beyond that, it is probably best to rely on the given expression.

Choice numbers, more commonly called combination numbers, or simply combinations, are the number of ways that a number of things can be selected, chosen, or grouped. Combinations concern only the grouping of items and not the arrangement of those items. They typically evolve from the question how many combinations of "n" objects are possible using all "n" objects or "r" objects at a time? To find the number of combinations of "n" dissimilar things taken "r" at a time, the formula is:

which can be stated as "n" factorial divided by the product of "r" factorial times (n - r) factorial.

Example: How many different ways can you combine the letters A, B, C, and D in sets of three? Clearly,

namely ABC, ABD, ACD, and BCD. (Note that ACB, BAC, BCA, CAB and CBA are all the same combination just arranged differently. In how many ways can a committee of three people be selected from a group of 12 people? We have:

How many handshakes will take place between six people in a room when they each shakes hands with all the other people in the room one time? Here,

Notice that no consideration is given to the order or arrangement of the items but simply the combinations.

Another way of viewing combinations is as follows. Consider the number of combinations of 5 letters taken 3 at a time. This produces:

Now assume you permute (arrange) the r = 3 letters in each of the 10 combinations in all possible ways. Each group would produce r! permutations. Letting x = 5C3 for the moment, we would therefore have a total of x(r!) different permutations. This total, however, represents all the possible permutations (arrangements) of n things taken r at a time, which is shown under arrangement numbers and defined as nPr. Therefore,

Using the committee of 3 out of 12 people example from above,

Consider the following: How many different ways can you enter a 4 door car? It is clear that there are 4 different ways of entering the car. Another way of expressing this is:

If we ignore the presence of the front seats for the purpose of this example, how many different ways can you exit the car assuming that you do not exit through the door you entered? Clearly you have 3 choices. This too can be expressed as:

Carrying this one step further, how many different ways can you enter the car by one door and exit through another? Entering through door #1 leaves you with 3 other doors to exit through. The same result exists if you enter through either of the other 3 doors. Therefore, the total number of ways of entering and exiting under the specified conditions is:

Another example of this type of situation is how many ways can a committee of 4 girls and 3 boys be selected from a class of 10 girls and 8 boys? This results in:

See ARRANGEMENT NUMBERS for determining the number of possible arrangements between items.

A circular prime number is one that remains a prime number after repeatedly relocating the first digit of the number to the end of the number. For example, 197, 971 and 719 are all prime numbers. Similarly, 1193, 1931, 9311 and 3119 are all prime numbers. Other numbers that satisfy the definition are 11, 13, 37, 79, 113, 199 and 337.

Primes of two or more digits can only contain the digits 1, 3, 7 because, If 0, 2, 4, 5,6, or 8 were part of the number, in the units place, the number would be divisible by 2 or 5.

It is thought that there are an infinite number of circular primes but has not yet been proven.

Complex numbers are formed by the addition of a real number and an imaginary number, the general form of which is a + bi where i =

= the imaginary number and a and b are real numbers. The "a" is said to be the real part of the complex number and b the imaginary part.

Probably the easiest number to define after prime numbers.

The Fundamental Theorem of Arithmetic states that every positive integer greater than 1 is either a prime number or a composite number. As we know, a prime number "p" is any positive number the only divisors of which are 1 and p (or -1 and -p). Thus, by definition, any number that is not a prime number must be a composite number.

A composite number is any number having 3 or more factors/divisors and is the result of multiplying prime numbers together. Most of the positive integers are the product of smaller prime numbers.

Examples: 4, 6, 8, 10, 12, 14, 15, 16, 18, 20, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, etc., are all composite numbers, each being divisible by lower prime numbers. Every number divisible by 2, the only even prime, is composite.

Every composite number can be broken down to a single unique set of prime factors and their exponents.

Examples: 210 = 2 x 3 x 5 x 7 495 = 3 2 x 5 1 x 11 1 or 4500 = 2 x 2 x 3 x 3 x 5 x 5 x 5 = 2 2 x 3 2 x 5 3 . This is the one and only possible factorization of the number 210.

If a positive number N is evenly divisible by any prime number less than

, the number N is composite.

Wilson's Theorem states that for every prime number "p", [(p + 1)! + 1] is evenly divisible by "p". The converse was shown to also be true in that every integer "N" that evenly divides [(N + 1)! + 1] is prime. Combining these leads to the famous general theorem that a necessary and sufficient condition that an integer "N" be prime is that "N" evenly divide [(n + 1)! + 1]. Conversely, if "N" does not divide [(N + 1)! + 1], "N" is composite.

Unfortunately, the practical use of this method is minimal due to the large numbers encountered with high N's.

A number N is said to be congruent if there are two integers, x and y, that result in the expressions x 2 + Ny 2 and x 2 - Ny 2 being perfect squares. The smallest known congruent number is 5 which satisfies 41 2 + 5(12 2 ) = 49 2 and 41 2 - 5(12 2 ) = 31 2 . The use of the square of a negative number results in another solution of 2 2 + 5(1 2 ) = 3 2 and 2 2 - 5(1 2 ) = (-1) 2 . While there are many congruent numbers, finding them is an arduous task. The expressions x 2 + Ny 2 and x 2 - Ny 2 are often useful in solving many problems in recreational mathematics.

The counting numbers are the familiar set of whole numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. that we see and use every day. (The 0 is sometimes included.) The set of counting numbers is often referred to as the natural numbers.

A cubic number is the third power of a number as in a x a x a = a 3 . Those familiar with the evolution of the squares from adding successive odd numbers might not be too surprised to discover how the cubes evolve from summing odd numbers also. Clearly, the nth cube is simply n 3 . Cubes can be derived in other ways also:

The cube of any integer, "n", is the sum of the series of odd numbers beginning with (n 2 - n + 1) and ending with (n 2 + n - 1). Example: For n = 6, (n 2 - n + 1) = 31 and 31 + 33 + 35 + 37 + 39 + 41 = 216 = 6 3 .

 n 3 (n) 3 n 3 (n 2 - n + 1) (n 2 + n - 1) 1 1 3 = 1 1 = 1(1) 2 2 3 = 8 3 + 5 = 2(1 + 2 + 1) 3 3 3 = 27 7 + 9 + 11 = 3(1 + 2 + 3 + 2 + 1) 4 4 3 = 64 13 + 15 + 17 + 19 = 4(1 + 2 + 3 + 4 + 3 + 2 + 1) 5 5 3 = 125 21 + 23 + 25 + 27 + 29 = 5(1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1)

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21

Cross out every third number giving us

1. 2. 4. 5. 7. 8. 10. 11. 13. 14. 16. 17. 19. 20

1. 3. 7. 12. 19. 27. 37. 48. 61. 75. 91. 108. 127..147

Cross out every second number giving us

The last line of numbers are the perfect cubes.

The first ten cubes are 1, 8, 27, 64, 125, 216, 343, 512, 729 and 1000.

The sum of the first n cubes starting with 1, is 1 3 + 2 3 + 3 3 +. + n 3 .

which, quite surprisingly, is the square of the nth triangular number, defined by Tn = n(n+1)/2.

The sum of the cubes of the first n odd numbers is 2n 4 - n 2 = n 2 (2n 2 - 1).

The sum of the cubes of the first n even numbers is 2n 4 + 4n 3 + 2n 2 = 2n 2 (n + 1) 2 .

The sum of the first n cubes, 1 3 + 2 3 + 3 3 + 4 3 +. + n 3 is equal to the square of the sum of the first n integers. Thus, 1 3 + 2 3 + 3 3 + 4 3 +. + n 3 = (1 + 2 + 3 + 4 +. + n) 2 .

The cube of any integer is the difference of the squares of two other integers.

Every cube is either a multiple of 9 or right next to one.

A perfect cube can end in any of the digits 0 through 9.

Three digit numbers that are the sum of the cubes of their digits: 153, 370, 371, 407.

The smallest number that is the sum of 2 cubes in two different ways. 1729 = 1 3 + 12 3 = 10 3 + 9 3 .

What are the dimensions of two cubes with integral sides that have their combined volume equal to the combined length of their edges. What are the dimensions of the cubes? x = 2 and y = 4.

The sum of any two cubes can never be a cube.

The sum of a series of three or more cubes can equal a cube.

11 3 + 12 3 + 13 3 + 14 3 = 20 3

1134 3 + 1135 3 + . 2133 3 = 16,830 3

The cube root of a number N is the number "a" which, when multiplied by itself twice, results in the number N or N = axaxa. There is no formula for extracting the cube root of a number. It can be obtained by means of a long division method or a simple estimation method.

The sum of "n" terms of an arithmetic progression with the first term equal to the sum of the first "n" natural numbers and a common difference of "n" is n 3 .

First, a method for approximating the cube root of a number to several decimal places which is usually sufficient for everyday use.

1--Make an estimate of the cube root of N = n lying between successive integers a and b.

2--Compute A = N - a 3 and B = b 3 - N

Example: Find the cube root of 146

1--With 146 lying between 125 and 216, let a = 5 and b = 6.

3--n = 5 + (6 x 21)/[6 x 21 + 5 x 70] = 5 + 126/476 = 5 + .264 = 5.2647

4--The cube root of 146 is 5.2656

A more exact method for determining integer cube roots.

Note that each cube of the numbers 1 through 10 ends in a different digit:

n 3 . 1. 2. 3. 4. 5. 6. 7. 8. 9. 10

n 3 . 1. 8. 27. 64. 125. 216. 343. 512. 729. 1000

Also note that the last digit is the cube root for all cases except 2, 3, 7 and 8. A quick review of these exceptions leads to the fact that these four digits are the difference between 10 and the cube root, i.e., 8 = 10 - 2, 7 = 10 - 3, 3 = 10 - 7 and 2 = 10 - 8. How can this information be used to determine the cube root of a number?

Given the cube of a number between 1 and 100, say 300,763.

The last digit tells us that the last digit of the cube root is 10 - 3 = 7.

Eliminating the last 3 digits of the cube leaves the number 300.

The number 300 lies between the cubes of 6 and 7 in our listing above.

The first digit of our cube root will be the lowest of these two numbers, in this case 6.

Therefore, the cube root of 30,763 becomes 67.

The last digit of 4 is the last digit of the cube root.

The number 592 lies between 512 and 729, the cubes of 8 and 9.

Therefore, the cube root of 592,704 becomes 84.

Of course, this is only useful if you know ahead of time that the cube is a perfect cube, i.e., having an integral cube root.
CYCLIC NUMBERS

A cyclic number is a number of "n" digits that when multiplied by 1, 2, 3. n, results in the same digits but in a different order. For example, the number 142,857 is a cyclic number since 142,857 x 2 = 285,714, 142,857 x 3 = 428,571, 142,857 x 4 = 571,428, and so on. It is not known just how many cyclic numbers exist.

Decimal numbers are numbers expressed through the decimal, or base 10, number system where each digit represents a multiple of some power of 10. The term applies primarily to numbers that have fractional parts so indicated by a decimal point. A number less than 1 is called a decimal fraction, e.g., .673. A mixed decimal is one consisting of an integer and a decimal fraction, e.g., 37.937.

Rational numbers can be expressed in the form of a fraction, 1/2, or as a decimal, .50, 1/8, or as a decimal, .125. From experience, we know that a fraction expressed in decimal form will either terminate without a remainder such as 3/8 = 0.375 or 7/8 = 0.875, repeat the same digit endlessly such as 1/3 = .3333333. or 2/3 = .6666666. repeat a series of different digits repeatedly such as 1/27 = .037037037. or 1/7 = .142857142857. or repeat a series of digits after some non repeating digits such as 1/12 = .0833333.

All prime denominators produce repeating decimals. Fractions with the same denominator often produce decimals with the same period and period length but with the digits starting with a different number in the period. For instance, 1/7 = .142857142857142857. 2/7 = .285714285714285714. 3/7 = .428571428571428571. 4/7 = .571428571428571428. 5/7 = .714285714285714285. and 6/7 = .857142857142857142. Other denominators produce two or more repeating periods in different orders. Investigate those with prime denominators of 11, 13, 17, 19, etc. and then 9, 12, 14, 16, 18, etc.and see what results. What do you notice?

Another interesting property of repeating decimals of even period length is illustrated by the following. Take the decimal equivalent of 2/7 = .285714285714. The repeating period is 285714. Split the period into two groups of three digits and add them together. The result is 999. Do the same with any other repeating decimal period and the result will always be a series of nines. Take 1/17 = .0588235294117647. Adding 05882352 and 94117647 gives you 99999999.

All repeating decimals, regardless of the period and length, are rational numbers. This simply means that it can be expressed as the quotient of two integers. A question that frequently arises is how to convert a repeating decimal, which we know to be rational, back to a fraction.

Rational decimal fractions may be converted to fractions as follows:

Given the decimal number N = 0.078078078.

Multiply N by 1000 or 1000N = 78.078078078.

. 999N = 78 making N = 78/999 = 26/333.

Given the decimal number N = .076923076923.

Muliply N by 1,000,000 or 1,000,000N = 76,923.0769230769230.

Subtracting N = .0769231769230.

. 999,999N = 76,923 making N = 76,923/999,999 = 1/13

An easier way to derive the fraction is to simply place the repeating digits over the same number of 9's. For example, the repeating decimal of .729729729729 converts to the fraction of 729/999 = 27/37.

Another trick where zeros are involved is to place the repeating digits over the same number of 9's with as many zeros following the 9's as there are zeros in the repeating decimal. For example, .00757575 leads to 075/9900 = 1/132.

Deficient numbers are part of the family of numbers that are either deficient, perfect, or abundant.

Deficient numbers, dN, are numbers where the sum of its aliquot parts (proper divisors), sa(N), is less than the number itself sa(N) < N. (In the language of the Greek mathematicians, the divisors of a number N were defined as any whole numbers smaller than N that, when divided into N, produced whole numbers. The factors/divisors of a number N, less the number itself, are referred to as the aliquot parts, or aliquot divisors, of the number.) Equivalently, N is also deficient if the sum, s(N), of all its divisors is less than 2N. It can be readily seen that using the aliquot parts summation, sa(16) = 1+2+4+8 = 14 < N = 16 while using all of the divisors, s(16) = 1+2+4+8+16 = 31 < 2N = 32, making the number 16 deficient under either definition.

Sa(N)-->1..1..1..3..1. 6. 1. 7. 4. 8. 1. 16. 1. 10. 9. 15. 1. 21. 1. 22..11..12. 1. 36

1,2,3,4,5,7,8,9,10,11,13,14,15,16,17,19,21,22,23 are all deficient.

A prime number or any power of a prime number is deficient. The divisors of a pefect or deficient number is deficient.

The digital root of a number is the single digit that results from the continuous summation of the digits of the number and the numbers resulting from each summation. For example, consider the number 7935. The summation of its digits is 24. The summation of 2 and 4 is 6, the digital root of 7935. Digital roots are used to check addition and multiplication by means of a method called casting out nines. For example, check the summation of 378 and 942. The DR of 378 is 9, 3+7+8=18, 1+8=9.. The DR of 942 is 6, 9+4+2=15, 1+5=6. Adding 9 and 6 produces 15, The DR of 15 is 6, 1+5=6. The summation of 378 and 942 is 1320. The DR of 1320 is 6. With the two final DR's are equal, the addition is correct.

Egyptian fractions are the reciprocals of the positive integers where the numerator is always one. They are often referred to as unit fractions. They were used exclusively by the Egyptians to represent all forms of fractions. The two fractions that they used that did not have a unit fraction was 2/3 and 3/4. The only other fractions that they seemed to have a strong interest in were those of the form 2/n where n was any positive odd number. The Rhind papyrus contains a list of unit fractions representing a series of 2/n for odd n's from 5 to 501. It is unclear as to why they found these 2/n fractions so important. These and other 2/n fractions may be derived from 2/n = 1/[n(n+1)/2] + 1/[n+1)/2].

In the year 1202, Leonardo Fibonacci proved that any ordinary fraction could be expressed as the sum of a series of unit fractions in an infinite number of ways. He used the then named greedy method for deriving basic unit fraction expansions. He described the greedy method in his Liber Abaci as simply subtracting the largest unit fraction less than the given non unit fraction and repeating the process until only unit fractions remained. It was later shown that the greedy method, when applied to any fraction m/n, results in a series of no more than "m" unit fractions. An example will best illustrate the process.

Reduce the fraction 13/17 to a sum of unit fractions. Dividing the fraction yields .7647. Of the unit fractions 1/2, 1/3, 1/4, 1/5, etc., 1/2 is the largest that is smaller than 13/17 so we compute 13/17 - 1/2 = 9/34 making 13/17 = 1/2 + 9/34. Repeating with 9/34, we have 9/34 - 1/4 = 2/136 = 1/68 making 13/17 = 1/2 + 1/4 + 1/68. Alternatively, divide the numerator into the denominator and use the next highest integer as the new denominator. 17/13 = 1.307 making 2 the denominator of the fraction to be subtracted from 13/17.

Reduce the fraction 23/37 to a sum of unit fractions. 23/37 - 1/2 = 9/74 - 1/16 = 70/1184 - 1/17 = 6/20,128 - 1/3355 = 1/33,764,720 making 23/37 = 1/2 + 1/16 + 1/17 + 1/3355 + 1/33,764,720.

There are many methods or algorithms that derive the unit fractions for any fraction m/n. Much more information regarding unit fractions can be found at

Algorithms for Egyptian Fractions at http://www1.ics.uci.edu/

The unit fractions derived by means of the method shown, or any other method, can be further broken down into other unit fractions by means of the identity 1/a = 1/(a+1) + 1/a(a+1), also known to Fibonacci. For example, 1/2 = 1/(2+1) + 1/2(2+1) = 1/3 + 1/6. Further still, 1/3 = 1/(3+1) + 1/3(3+1) = 1/4 + 1/12 and 1/6 = 1/(6+1) + 1/6(6+1) = 1/7 + 1/42 yielding 1/2 = 1/4 + 1/7 + 1/12 + 1/42. While the number of unit fractions derivable for any given fraction is therefore infinite, there is apparently no known procedure for deriving a series with the least number of unit fractions or the smallest largest denominator. .

A shape is called equable if its area equals its perimeter. There are exactly five equable Heronian triangles: the ones with side lengths (5,12,13), (6,8,10), (6,25,29), (7,15,20), and (9,10,17).

Equivalent numbers are numbers where the aliquot parts (proper divisors other than the number itself) are identical. For instance, 159, 559 and 703 are equivalent numbers since their aliquot parts sum to 57.

f(159) = 1, 3, 53 and 159 where 1 + 3 + 53 = 57.

f(559) = 1, 13, 43 and 559 where 1 + 13 + 43 = 57.

f(703) = 1, 19, 37 and 703 where 1 + 19 + 37 = 57.

Probably the easiest number to define, an even number is any number that is evenly divisible by 2.

The nth even number is given by Ne = 2n.

The sum of the set of "n" consecutive even numbers beginning with 2 is given by Se = n(n + 1).

The sum of the set of m consecutive even numbers starting with n1 and ending with n2 is given by Se(n1-n2) = n2^2 - n1^2 + (n1 + n2) or (n1 + n2)(1 + n1 - n2).

The sum of the squares of the even numbers starting with 2^2 is given by Se^2 = (4n^3 + 6n^2 + 2n)/3.

An even number multiplied by any number, or raised to any power, results in another even number.

## 18.1: Complex numbers - Mathematics

The last topic in this section is not really related to most of what we’ve done in this chapter, although it is somewhat related to the radicals section as we will see. We also won’t need the material here all that often in the remainder of this course, but there are a couple of sections in which we will need this and so it’s best to get it out of the way at this point.

In the radicals section we noted that we won’t get a real number out of a square root of a negative number. For instance, (sqrt < - 9>) isn’t a real number since there is no real number that we can square and get a NEGATIVE 9.

Now we also saw that if (a) and (b) were both positive then (sqrt = sqrt a ,sqrt b ). For a second let’s forget that restriction and do the following.

Now, (sqrt < - 1>) is not a real number, but if you think about it we can do this for any square root of a negative number. For instance,

So, even if the number isn’t a perfect square we can still always reduce the square root of a negative number down to the square root of a positive number (which we or a calculator can deal with) times (sqrt < - 1>).

So, if we just had a way to deal with (sqrt < - 1>) we could actually deal with square roots of negative numbers. Well the reality is that, at this level, there just isn’t any way to deal with (sqrt < - 1>) so instead of dealing with it we will “make it go away” so to speak by using the following definition.

[ equire box[2pt,border:1px solid black] <>>]

Note that if we square both sides of this we get,

[ equire box[2pt,border:1px solid black] <<= - 1>>]

It will be important to remember this later on. This shows that, in some way, (i) is the only “number” that we can square and get a negative value.

Using this definition all the square roots above become,

These are all examples of complex numbers.

The natural question at this point is probably just why do we care about this? The answer is that, as we will see in the next chapter, sometimes we will run across the square roots of negative numbers and we’re going to need a way to deal with them. So, to deal with them we will need to discuss complex numbers.

So, let’s start out with some of the basic definitions and terminology for complex numbers. The standard form of a complex number is

where (a) and (b) are real numbers and they can be anything, positive, negative, zero, integers, fractions, decimals, it doesn’t matter. When in the standard form (a) is called the real part of the complex number and (b) is called the imaginary part of the complex number.

Here are some examples of complex numbers.

The last two probably need a little more explanation. It is completely possible that (a) or (b) could be zero and so in 16(i) the real part is zero. When the real part is zero we often will call the complex number a purely imaginary number. In the last example (113) the imaginary part is zero and we actually have a real number. So, thinking of numbers in this light we can see that the real numbers are simply a subset of the complex numbers.

The conjugate of the complex number (a + bi) is the complex number (a - bi). In other words, it is the original complex number with the sign on the imaginary part changed. Here are some examples of complex numbers and their conjugates.

Notice that the conjugate of a real number is just itself with no changes.

Now we need to discuss the basic operations for complex numbers. We’ll start with addition and subtraction. The easiest way to think of adding and/or subtracting complex numbers is to think of each complex number as a polynomial and do the addition and subtraction in the same way that we add or subtract polynomials.

There really isn’t much to do here other than add or subtract. Note that the parentheses on the first terms are only there to indicate that we’re thinking of that term as a complex number and in general aren’t used.

a (left( < - 4 + 7i> ight) + left( <5 - 10i> ight) = 1 - 3i)

b (left( <4 + 12i> ight) - left( <3 - 15i> ight) = 4 + 12i - 3 + 15i = 1 + 27i)

c (5i - left( < - 9 + i> ight) = 5i + 9 - i = 9 + 4i)

Next let’s take a look at multiplication. Again, with one small difference, it’s probably easiest to just think of the complex numbers as polynomials so multiply them out as you would polynomials. The one difference will come in the final step as we’ll see.

1. (7ileft( < - 5 + 2i> ight))
2. (left( <1 - 5i> ight)left( < - 9 + 2i> ight))
3. (left( <4 + i> ight)left( <2 + 3i> ight))
4. (left( <1 - 8i> ight)left( <1 + 8i> ight))

So all that we need to do is distribute the 7(i) through the parenthesis.

Now, this is where the small difference mentioned earlier comes into play. This number is NOT in standard form. The standard form for complex numbers does not have an () in it. This however is not a problem provided we recall that

[7ileft( < - 5 + 2i> ight) = - 35i + 14left( < - 1> ight) = - 14 - 35i]

We also rearranged the order so that the real part is listed first.

In this case we will FOIL the two numbers and we’ll need to also remember to get rid of the ().

[left( <1 - 5i> ight)left( < - 9 + 2i> ight) = - 9 + 2i + 45i - 10 = - 9 + 47i - 10left( < - 1> ight) = 1 + 47i]

[left( <4 + i> ight)left( <2 + 3i> ight) = 8 + 12i + 2i + 3 = 8 + 14i + 3left( < - 1> ight) = 5 + 14i]

Here’s one final multiplication that will lead us into the next topic.

[left( <1 - 8i> ight)left( <1 + 8i> ight) = 1 + 8i - 8i - 64 = 1 + 64 = 65]

Don’t get excited about it when the product of two complex numbers is a real number. That can and will happen on occasion.

In the final part of the previous example we multiplied a number by its conjugate. There is a nice general formula for this that will be convenient when it comes to discussing division of complex numbers.

So, when we multiply a complex number by its conjugate we get a real number given by,

Now, we gave this formula with the comment that it will be convenient when it came to dividing complex numbers so let’s look at a couple of examples.

1. (displaystyle frac<<3 - i>><<2 + 7i>>)
2. (displaystyle frac<3><<9 - i>>)
3. (displaystyle frac<<8i>><<1 + 2i>>)
4. (displaystyle frac<<6 - 9i>><<2i>>)

So, in each case we are really looking at the division of two complex numbers. The main idea here however is that we want to write them in standard form. Standard form does not allow for any (i)'s to be in the denominator. So, we need to get the (i)'s out of the denominator.

This is actually fairly simple if we recall that a complex number times its conjugate is a real number. So, if we multiply the numerator and denominator by the conjugate of the denominator we will be able to eliminate the (i) from the denominator.

Now that we’ve figured out how to do these let’s go ahead and work the problems.

Notice that to officially put the answer in standard form we broke up the fraction into the real and imaginary parts.

This one is a little different from the previous ones since the denominator is a pure imaginary number. It can be done in the same manner as the previous ones, but there is a slightly easier way to do the problem.

First, break up the fraction as follows.

Now, we want the (i) out of the denominator and since there is only an (i) in the denominator of the first term we will simply multiply the numerator and denominator of the first term by an (i).

The next topic that we want to discuss here is powers of (i). Let’s just take a look at what happens when we start looking at various powers of (i).

Can you see the pattern? All powers if (i) can be reduced down to one of four possible answers and they repeat every four powers. This can be a convenient fact to remember.

We next need to address an issue on dealing with square roots of negative numbers. From the section on radicals we know that we can do the following.

[6 = sqrt <36>= sqrt = sqrt 4 ,sqrt 9 = left( 2 ight)left( 3 ight) = 6]

In other words, we can break up products under a square root into a product of square roots provided both numbers are positive.

It turns out that we can actually do the same thing if one of the numbers is negative. For instance,

[6i = sqrt < - 36>= sqrt ight)left( 9 ight)> = sqrt < - 4>,sqrt 9 = left( <2i> ight)left( 3 ight) = 6i]

However, if BOTH numbers are negative this won’t work anymore as the following shows.

[6 = sqrt <36>= sqrt ight)left( < - 9> ight)> e sqrt < - 4>,sqrt < - 9>= left( <2i> ight)left( <3i> ight) = 6 = - 6]

We can summarize this up as a set of rules. If (a) and (b) are both positive numbers then,

Why is this important enough to worry about? Consider the following example.

If we were to multiply this out in its present form we would get,

Now, if we were not being careful we would probably combine the two roots in the final term into one which can’t be done!

So, there is a general rule of thumb in dealing with square roots of negative numbers. When faced with them the first thing that you should always do is convert them to complex number. If we follow this rule we will always get the correct answer.

So, let’s work this problem the way it should be worked.

[left( <2 - sqrt < - 100>> ight)left( <1 + sqrt < - 36>> ight) = left( <2 - 10i> ight)left( <1 + 6i> ight) = 2 + 2i - 60 = 62 + 2i]

The rule of thumb given in the previous example is important enough to make again. When faced with square roots of negative numbers the first thing that you should do is convert them to complex numbers.

There is one final topic that we need to touch on before leaving this section. As we noted back in the section on radicals even though (sqrt 9 = 3) there are in fact two numbers that we can square to get 9. We can square both 3 and -3.

The same will hold for square roots of negative numbers. As we saw earlier (sqrt < - 9>= 3,i). As with square roots of positive numbers in this case we are really asking what did we square to get -9? Well it’s easy enough to check that 3(i) is correct.

However, that is not the only possibility. Consider the following,

and so if we square -3(i) we will also get -9. So, when taking the square root of a negative number there are really two numbers that we can square to get the number under the radical. However, we will ALWAYS take the positive number for the value of the square root just as we do with the square root of positive numbers.

## Solutions

Solution: 1 Factoring a polynomial

We are trying to find solutions to the equation $x^3 = 1$ which are the roots of the equation $x^3-1=0$. One solution is $x = 1$. This means that $x-1$ divides $x^3 -1$ with no remainder. Performing long division, we find $x^3-1 = (x-1)(x^2+x+1).$ To find the roots of the polynomial $x^2 +x+ 1$ we can use the quadratic formula which says that the roots are $frac<-1 pm sqrt<1-4>> <2>= frac<-1> <2>pm frac><2>.$ We know that $i^2 = -1$ so $(sqrt<3>i)^2 = -3$ and this means that $sqrt <-3>= sqrt<3>i$. So our two roots of $x^2 - x + 1$ are $-frac<1> <2>pm frac><2>i$

We have found 3 solutions to $x^3-1 = 0$, namely $x = 1, x = -frac<1> <2>+ fraci><2>,$ and $x = -frac<1> <2>- fraci><2>$.

We can apply the same technique to solve the equation $x^4 = 1$ or $x^4 - 1 = 0$. We can first write $x^4 -1$ as a difference of squares: $x^4-1 = (x^2 + 1)(x^2 - 1).$ We have $x^2 -1 = (x+1)(x-1)$. Similarly we have $x^2 + 1 = (x+i)(x-i)$. So this gives $x^4 + 1 = (x+i)(x-i)(x+1)(x-1)$ and the solutions to $x^4-1 = 0$ are $x = -i,i,-1,1$.

Solution: 2 Using Geometry

Suppose $z$ is a complex number which can be written in polar form as follows: $z = re^.$ Here $r$ represents the distance from $z$ to 0 and $heta$ is the measure of the directed angle made by the non-negative real numbers and the ray from 0 through $z$. One important property of the polar form is that it facilitates calculating powers:

Using the polar form for a complex number $z$ and its properties indicated above, $z = re^$ is a cube root of 1 when $z^3 = 1$ or $r^3e^ = 1.$ This means that $r^3 = 1,$ so $r = 1$ since $r$ is a real number and $rgeq0$. Moreover $3 heta$ is an integer multiple of $2pi$. This means we can have $heta = 0$, $heta = frac<2pi><3>$, or $heta = frac<4pi><3>$. These are the solutions for $heta$ that we find by choosing our integer multiple of $pi$ to be 0, 1, and 2. If we choose different integer multiples we will find angles that differ from the three listed above by an integer multiple of 2$pi$. The three cube roots of 1 are pictured below: Â

Notice that the three numbers whose cube is 1 evenly divide the unit circle into 3 equal pieces.

Writing $z = re^$, the solutions of $z^4 = 1$ are the same as the solutions to $r^4e^ = 1.$ This means that $r^4 = 1,$ so $r=1$ since $r$ is a real number and $r geq 0$. In addition $4 heta$ must be an integer multiple of $2pi$. Choosing multiples of 0, 1, 2, and 3 gives $heta= 0, frac<2>, pi, frac<3pi><2>$. These correspond to the four numbers 1, $i$, -1, and $-i$, as pictured below:

The fourth roots of 1 divide the unit circle evenly into four sections.

Solution: 3 Solving equations

1. Writing $(a+bi)^3=(a^3-3ab^2)+(3a^2b-b^3)i=1+0i$ and equating real and imaginary parts of $(a+bi)^3=1$ gives $a^3 - 3ab^2 = 1$ and $3a^2b - b^3 = 0$ Factoring the second equation as $b(3a^2-b^2)=0$, we see that either $b=0$ or $3a^2=b^2$. If $b=0$, then $a^3=1$, giving the obvious cube root of 1. If $b eq 0$, then $3a^2=b^2$, and substituting this into $a^3-3ab^2=1$ gives $a^3-9a^3=1$, so $a=-frac<1><2>$, and then $b=pmfrac><2>$.
2. Similarly, if we write $(a+bi)^4=((a^2-b^2)+2abi)^2= (a^4-6a^2b^2+b^4)+(4a^3b-4ab^3)i$ then equating imaginary parts in $(a+bi)^4=1$, gives $(4a^3b-4ab^3)=0.$ Factoring the left-hand side as $4ab(a^2-b^2)=0$, we see that any 4th root of 1 must have $a=0$, $b=0$, or $a^2=b^2$. If $a=0$, then the equation $(bi)^4=1$ quickly gives $b=pm 1$. Similarly, if $b=0$, then the equation $a^4=1$ again gives $a=pm 1$ (remember that $a$ and $b$ are real). Finally, if $a=pm b$, then equating the real parts of $(apm ai)^4=1$ gives $(a^4-6a^2a^2+a^4)=1$ or $-4a^4=1$. Since there are no real values of $a$ for which this is true, the only solutions to these equations are $(a,b)=(pm 1,0)$ and $(a,b)=(0,pm 1)$, corresponding to the four roots $pm 1$ and $pm i$.

## Thinking about Science with David Hukins

It may seem that this post is about very abstract mathematics. But the ideas explained here are very useful for explaining oscillations and waves (post 18.11), especially when combined with the concept of the number e (post 18.15).

Counting is like moving along the line in the picture above – from left to right. When we haven’t moved, we are at position 0. We can move to position 1. If we move the same distance again, we get to 2, then to 3, 4, 5 and so on. These are the positive whole numbers or positive integers. There are also numbers like 0.3, 1.27, 3.85. Together with the integers, these numbers may up the set of positive rational numbers. There are other numbers, like π (post 17.11) and e (post 18.15), whose positions we can’t fix exactly – but we can calculate them with the precision that we need – these are the irrational numbers. If we move to 4 twice, we arrive at 4 x 2 = 8 (4 multiplied by 2). Notice that 2 x 2 = 4, 3 x 3 = 9, 4 x 4 =16 we say that 4 = 16 ½ (post 18.2) is the square root of 16, and that 3 = 9 ½ is the square root of 9. Numbers like 4, 9, 16, 25, 36, 49…. have square roots that are integers. Others, like 2, 3, 5,6, 8… have squares roots that are not integers – they can be rational or irrational numbers.

If we move from 0 towards the left, we meet the negative numbers, like -1, -2, -3, -0.25, -3.5, -1/3… that can be rational or irrational. Moving -4 twice we get to 2 x (-4) = -8 we have multiplied -4 by 2. Multiplying -4 by -2 moves us in the opposite (positive) direction, so that (-2) x (-4) = 8. Multiplying a positive number by a negative number gives us a negative answer: multiplying a negative number by another negative number gives us a positive number.

The ideas explained in the previous paragraph mean that the number 1 has two square roots, 1 and -1. If you think about it, you will see that all positive numbers have two square roots – one positive: the other negative. But we can’t think of any number that, when multiplied by itself gives -1, or any other negative number, because -1 x -1 = 1 to get -1 we need to multiply -1 by +1 and the two numbers are not the same. So, we might conclude that negative numbers don’t have square roots. This is what I was taught before I went to university.

But this is a very unsatisfactory conclusion it means that numbers behave completely differently depending on whether we move to the right or to the left from 0, in the picture above. As a result, the equation x 2 = 1 has two solutions (1 and -1) and x 2 = -1 has no solutions. To overcome this problem, we invent the number i that is defined by i 2 = -1. Now -1 has two square roots i and –i (because –i x –i =1). And all the other negative numbers have two square roots – the square roots of -16 are 4i and -4i. We call the numbers that have i as a factor (3i, -27i, 2.236i), imaginary numbers numbers that don’t have i as a factor (5, 1.32, π) are called real numbers. Now all numbers have two square roots. Some people (mostly engineers) represent the square root of -1 by j instead of i it doesn’t make any difference to the ideas expressed here.

Real numbers belong on the line at the beginning of this post: imaginary numbers don’t.

There are also numbers, like 2 + 3i that have a real part (2) and an imaginary part (3i) they are called complex numbers. Adding to the real part (by adding real numbers) doesn’t affect the value of the imaginary part and adding to the imaginary part (by adding further imaginary numbers) doesn’t affect the value of the real part. The real and imaginary parts are independent of each other they are orthogonal (appendix 2 of post 16.50). We can’t represent complex numbers along a single line but we can represent them on two lines that are perpendicular to each other – like the axes of an orthogonal Cartesian coordinate system (appendix 2 of post 16.50), as shown in the picture below.

So, the values of real numbers lie along a line the values of complex numbers lie in a plane – the Argand plane.

The picture above shows the position of the complex number z = a + ib on the Argand plane. We define │z│ (the modulus of z) as the length of z in the Argand plane. According to Pythagoras’ theorem (appendix 1 of post 16.50) z 2 = a 2 + b 2 . We define z* (the complex conjugate of z) by z* = aib. Now zz* = (a + ib) x (aib) = a 2 +a(-ib) + (ib)(a) + (ib)(-ib) = a 2 –abi +abii 2 b 2 = a 2 + b 2 = │z│.

Notice (also in the picture) above, that z is associated with an angle, θ, called its argument. You can see that a = zcosθ and b = zsinθ (appendix 3 of post 16.50), so that a/b = tanθ (appendix 5 of post 16.50).

If you read post 18.11, you will see that │z│ behaves just like the amplitude of a wave or oscillator and θ behaves just like ωt, where ω is an angular frequency and t represents time. So, complex numbers are often used to represent the behaviour of oscillators and waves. To pursue this further, we need to think about some more mathematical ideas. You may also have noticed that complex numbers could be considered as an alternative way of representing two-dimensional vectors (post 17.2). But be careful because i or j here have completely different meanings to the way they are used in vector algebra (see post 17.14).