Algebra 1X Notes v3aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

7/28/2019 Algebra 1X Notes v3aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

http:///reader/full/algebra-1x-notes-v3aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa 1/79

1X

Algebra

2011-2012



Contents

1 Numbers 11.1 Number Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.1 Geometric Interpretation of . . . . . . . . . . . . . . . . . . . . 101.2.2 Regions in the Complex Plane . . . . . . . . . . . . . . . . . . . . 13

1.2.3 Multiplication of Complex Numbers . . . . . . . . . . . . . . . . . 14

2 Methods of Proof 172.1 Direct Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Examples and Counterexamples . . . . . . . . . . . . . . . . . . . . . . . 182.3 Proof By Contradiction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.4 Converse and Contrapositive . . . . . . . . . . . . . . . . . . . . . . . . . 202.5 Proof By Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.6 Variations of the Principle of Induction . . . . . . . . . . . . . . . . . . . 25

3 Sequences and Series 283.1 Standard Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.2 Arithmetic Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.3 Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.4 Two Special Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4 The Binomial Theorem and Applications 374.1 Permutations and Combinations . . . . . . . . . . . . . . . . . . . . . . . 374.2 The Binomial Theorem and Applications . . . . . . . . . . . . . . . . . . 404.3 Application to Trigonometric Calculations . . . . . . . . . . . . . . . . . 46

5 Polynomials and roots in the polar form 485.1 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485.2 Roots in the Polar Form . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

6 Matrices 556.1 Scalars, Matrices and Arithmetic . . . . . . . . . . . . . . . . . . . . . . 556.2 Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . 616.3 Inverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 686.4 Transpose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 726.5 Transformations of the Plane . . . . . . . . . . . . . . . . . . . . . . . . 746.6 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

i



Chapter 1

Numbers

1.1 Number Systems

Number systems are sets of numbers closed under addition and multiplication.For example,

= set of natural numbers (positive integers) = {1, 2, 3, 4, . . .}.This set has the following properties:

1. is closed under addition, i.e. if , then + .2. is closed under multiplication, i.e. if , then .

Notice that

1. is not closed under substraction, e.g. 1 2 = 1 .2. is not closed under division, e.g. 1

2 .

To get more closure properties working we need to enlarge the number system. So, welook at

= set of integers = {. . . , 2, 1, 0, 1, 2, 3, 4, . . .}. is closed under addition, subtraction and multiplication but not division since, for ex-

ample,1

2 .

Let , be integers. We write if divides , i.e. if = for someinteger . We write if does not divide . For example,

4 20 (because 20 = 5 4),(4) 20,

3 3,3 8.

Example 1.1. Show that if 3

and 3

then 3

( + ).

1



Solution. Suppose 3 and 3 . Then = 3 and = 3 for some integers , . Itfollows that

+ = 3 + 3 = 3( + ),

where + is an integer. Therefore 3 ( + ) as required.

Note. It is crucial that in our arguments , and + are integers. To illustrate byexample: 3 8 because there is no integer such that 8 = 3. There is however a number

such that 8 = 3, namely =8

3, but is not an integer.

Special Types of Integers

A number is called even if 2 , i.e. if = 2 for some .

A number is called odd if 2 . This is the same as having the presentation = 2 + 1 for some . Equivalently, every odd number is 1 more than an evennumber.

A number is called prime if is greater than 1 and has no positive integerdivisors other than 1 and . The first few primes are 2, 3, 5, 7, 11, 13, 17, 19, . . .

Question. Are there infinitely many primes? (Euclid c300B.C.)

Greatest Common Divisor

We say is a common divisor of , if and . The greatest commondivisor (g.c.d.) of and is denoted gcd(, ), e.g. gcd(8, 12) = 4. We say two numbers,, , are coprime if gcd(, ) = 1.

Division Algorithm

For given , there exists unique , with 0 and > 0 such that

= +

where is called the remainder and is called the quotient.

Indeed, this can be seen as follows. Let be such that

< + 1 ( is

uniquely determined by these inequalities). Then < + . Define = .Then we have 0

< as required.

2



Euclidean Algorithm (by example)

This algorithm allows us to find the g.c.d. of two integers and also to express the g.c.d.as a combination of these two integers. Thus it consists of two steps.

1. Find the g.c.d. of 96 and 36.We have

96 = 36 2 + 24 division of 96 by 36 with the remainder (1.1)36 = 24 1 + 12 division of 36 by the remainder from (1.1) (1.2)

24 = 12 2 + 0. continue and stop when the remainder is 0 (1.3)The g.c.d. is the number we divide by in the last equation, i.e. 12.

2. Represent the greatest common divisor of 96 and 36 as gcd(96 , 36) = 96 + 36.

We start with the second to the last line (equation (1.2)) and express the remainderin the circle as

12 = 36 24 1. (1.4)

Then we use the previous equation (1.1) to express the remainder in the box as

24 = 96 36 2,and we plug these into equation (1.4) to get

12 = 36 (96 36 2) 1.Upon collecting the terms with 36 and 96 we get

12 = 36 3 96which is the required form with = 3 and = 1.

The Euclidean algorithm allows us to establish an important property of prime num-bers (Proposition 1.1 below). Let us first state the following Lemma.

Lemma 1.1. Let ,, be such that and . Then ( + ) for any, .Proof. We have = , = for some , . Then

+ = + = ( + ).

Since + we get ( + ) as required.

Proposition 1.1. Let , , and let be prime. Suppose that . Then atleast one of and is divisible by .

3



Proof. Suppose that . Then gcd(, ) = 1. By the Euclidean algorithm there exists, such that 1 = + . Hence = + . We have and , henceby Lemma 1.1 .

A corollary of this Proposition is the Fundamental Theorem of Arithmetic.

Theorem 1.2 (Fundamental Theorem of Arithmetic). Let > 1 be an integer. Then

= 11 22 . . .

(1.5)

where , , are prime such that 1 < 2 < .. . < , and the presentation of (1.5)is unique.

Exercise 1. Prove the Fundamental Theorem of Arithmetic.

Rational Numbers

In order to have the division we enlarge our number system from to

=set of rational numbers=

{

, and = 0}.( stands for quotients).Recall that for , = 0 we have

=

if and only if = .

e.g.

15

12=

5

4because 15 4 = 12 5

.

Also,

1= for any . Therefore and clearly .

The set of rational number has the following properties:

1. is closed under addition.Indeed,

+

= +

,

and + is an integer if ,,, are integers, also is a nonzero integer if , are.For example,

3

5+

1

2=

6 + 5

10=

11

10,

also3

5+

2

4=

12 + 10

20=

22

20=

11

10(as expected!).

4



2. is closed under subtraction.Indeed,

=

and

is an integer if ,,, are, also is a nonzero integer if , are.

3. is closed under multiplication.Indeed,

=

and is an integer if , are integers; is a nonzero integer if , are.

4. is closed under division by non-zero numbers.Indeed,

=

=

and is an integer, is a non-zero integer where ,,, are integers and , and are non-zero.

But does not have all the solution of equations that we need. For example, althoughthe equation 2 = 4 has the solutions = 2 in , the equation 2 = 2 has no solutionsin . This is because

2 is irrational (i.e. it cannot be expressed as

with , .

We will prove this later). So we add irrational numbers and enlarge our number systemfrom to

= set of real numbers.

stands for reals; it is usually pictured as a line.

Figure 1.1Illustration of as a line. Here we marked

2, , as some examples of well known irrational numbers.

Calculations With Real Numbers

We would normally like to present numbers in the simplest way. For the fractionsinvolving roots this normally means rationalizing the denominators, i.e. getting rid ofirrational numbers in the denominators.

Example 1.2. Simplify2 +

3

4 3.

5



Solution. First we get rid of the

3 in the denominator. We have

2 +

3

4

3

=2 +

3

4

3

4 +

3

4 +

3

(Multiplying by

= 1 leaves the value unchanged but allows us to use the difference of

two squares formula, ( )( + ) = 2 2, to simplify the denominator). So wecontinue,

(2 +

3)(4 +

3)

(4 3)(4 + 3) =8 + 3 + 6

3

16 3 =11 + 6

3

13.

Example 1.3. Simplify

7232

.

Solution.

7232

=98

4

8=

94

=32

.

Example 1.4. Show that

5

9 45 = 2.

Solution. Recall that = if , are positive.Consider

5 2. We have

(

5 2)2 = 5 + 4 4

5 = 9 4

5. (1.6)

Notice that

5 2 > 0 (as 5 > 4) and 9 45 > 0 (as 9 45 = (5 2)2). Therefore5 2 =

9 45, on taking the positive square root of both sides of (1.6). Hence

result.

6



Recall the formula and derivation for the solution of quadratic equation. Let ,, , = 0. Then

2 + + = 0 2 +

+

= 0

+ 22 2

42+

= 0 (completing the square)

+

2

2=

2

42

+

2

2=

2 442

+ 2

=

2 42

=

2 42

Example 1.5. Solve the equation 22 + 7 4 = 0.

Solution. We have =7 81

4=

7 94

, so the solutions are = 4 and = 12

.

It is easy to construct a quadratic equation with the required roots.

Example 1.6. Find a quadratic equation whose roots are = 3 and 7.

Solution. An equation is ( 3)( 7) = 0, i.e. 2

10 + 21 = 0.

Example 1.7. Find a quadratic equation whose roots are = 3 7.Solution. An equation is

( 3

7)( 3 +

7) = 0,

i.e. ( 3)2 7 = 0

(here we used the difference of two squares formula ( )( + ) = 2 2). Equiv-alently, the equation is 2 6 + 2 = 0.

Notice that not all quadratic equations have real solutions. Indeed if 2 4 < 0then there are no real solutions. What do we do then?

Trick: Introduce =1. This helps solving quadratic equations. For example,

2 + 3 = 0 can be solved by

=

3 =

(1)3 =

1

3 =

3.

7


http:///reader/full/algebra-1x-notes-v3aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa 10/79

More generally the solution to 2 + + = 0 where 2 4 < 0 is given by

= 2 4

2=

(1)(4 2)2

= 4 2

2

and now 4 2 > 0 so the square root 4 2 is defined.

Example 1.8. Solve the equation 2 10 + 40 = 0.

Solution.

=10 100 160

2=

10 602

=10 4 (1) 15

2=

10 41152

=

=10

2

1

15

2 = 5 115 = 5 15 where = 1.

Is this trick justifiable? Yes: this is in our next section.

1.2 Complex Numbers

The set ofcomplex numbers, is a number system that contains and a special number

such that 2

= 1. consists of all numbers of the form + where , .The number system has the following properties.

1. is closed under addition.Indeed,

( + ) + ( + ) = ( + ) + ( + )

and + and + are real if,,, are real. (For example, (3 + 4) + ( 7 2) =10 + 2). Note that is closed under subtraction by similar reasons.

2. is closed under multiplication. Indeed,

( + )( + ) = + ( + ).

Here we used 2 = 1, and , + are both real if ,,, are real.For example,

(2 + )(5 + 3) = 10 3 + (6 + 5) = 7 + 11,(2 + 3)(5 4) = 10 + 12 + (8 + 15) = 22 + 7,

(7

) = 7 + 1 = 1 + 7.

8



3. is closed under division by non-zero numbers.Indeed,

+

+ =

( + )( )( + )(

)

= + + ( )

2 + 2=

+

2 + 2+

2 + 2

and 2+2

, 2+2

, provided ,,, and 2 + 2 = 0, i.e. provided , areboth nonzero, i.e. + = 0, as required.

By simplifying a complex number we normally mean rearranging it to the form + where , and , are presented in the simplified form.

Example 1.9. Simplify2 +

3 4 .

Solution.

2 + 3 4 =

2 + 3 4

3 + 43 + 4

= 6 4 + 119 + 16

= 2 + 1125

= 225

+ 1125

.

If = + where , , then:

is called the real part of , denoted Re(),

is called the imaginary part of , denoted Im(),

is called the complex conjugate of , denoted by .

Example 1.10. Find the real and imaginary parts of3 +

(7 3)(1 + 2).

Solution. First we simplify the denominator, then we arrive to the number in the form + . We have,

3 +

(7 3)(1 + 2) =3 +

7 + 6 + 11=

3 +

13 + 11=

3 +

13 + 11 13 11

13 11 =39 + 11 20

169 + 121=

=50

20

290 =5

2

29 =5

29 + 229 .

Therefore the real part is5

29and the imaginary part is 2

29.

Now we consider how a simple equation for a complex number can be solved.

Example 1.11. Solve for the equation

3 = 1 + .

Solution. We consider two different methods of the solution.

9



Method 1. The equation is

= (1 + )(3 ) = 3 + 3 2+ = 3 + 3

(2 + ) = 3 + 3

=

3 + 3

3 + ,

so that,

=3(1 + )

2 + 2

2 =3(2 + + 1)

4 + 1=

3(3 + )

5=

9

5+

3

5.

Method 2. The equation is = (1 + )(3 ). Let = + , where , . We have

+ = (1 + )(3 ) + = 3 + 3 + + = (3 + ) + (3 ).

Now, the LHS equates the RHS which means that the real part of LHS equals thereal part of the RHS and the imaginary part of the LHS equals the imaginary partof the RHS. Thus, we obtain

= 3 + and = 3 ,

i.e.

2 = 3, + 2 = 3.

By multiplying the first equation by 2 and adding the second equation we get

5 = 9, i.e. =9

5. Substituting this in to the first equation we get =

3

5.

Therefore =9

5+

3

5.

1.2.1 Geometric Interpretation of

Complex numbers can be represented as the points in a plane by identifying the number + with the point with coordinates (, ).

10



Figure 1.2

An example of various complex numbers represented as points on the plane.

The horizontal axis, OX, is called the real axis; it contains the real numbers. Thevertical axis, OY, is called the imaginary axis; it contains the numbers of the form ,where ; these numbers are called the imaginary numbers. The non-zero imaginarynumbers are called purely imaginary. When a plane is used this way it is called thecomplex plane or Argand diagram.

Note. A formal proof of the existence of might start by taking a plane and then

defining addition and multiplication of points in the plane in such a way that the planebecomes .

Let = + where , . Then

2 + 2 is called the modulus of , denoted. It is the distance of the point from the origin in the complex plane.

For example,

3 + 2 =

32 + 22 =

13,

= 1,

1 + 3 = 10 (see Figure 1.3).

Note also that the conjugate, = , is the mirror image of in the real axis (seeFigure 1.4). In particular = if and only if is on the real axis, i.e. = if and onlyif is real. We also have = 2. Indeed,

= ( + )( ) = 2 + 2 = 2.

Any point in the plane has unique cartesian coordinates (, ) (named after Descartes1596

1650). It can also be represented by its polar coordinates, (, ) (see Figure 1.5)

11



that are defined by:

=

2 + 2, (1.7)

tan =

, (1.8)

= cos , = sin . (1.9)

So the complex number = + can also be written as = (cos + sin ) where = is the modulus of , and is usually called the argument of and is denotedarg(). This is called the polar form of the complex number . Note that 0 and isdefined up to 2, , and it can be chosen such that 0 < 2.

Figure 1.3Complex numbers and their moduli.

Figure 1.4Complex conjugation.

Figure 1.5Polar coordinates.

Example 1.12. Find the modulus and argument of the complex numbers 1 + , 1 ,, 1, 3 .Solution.

(a) = 1 + has the modulus

2 and argu-

ment

4.

12



(b) = 1 has the modulus 2 and argu-ment

4

(or7

4

).

(c) = has the modulus 1 and argument

2.

(d) = 1 has the modulus 1 and argument.

(e)

= 3 has modulus = 3 + 1 =2 and tan = 13 = 13 . Note also that is in the third quadrant as = cos < 0

and = sin < 0. Hence = +

6=

7

6.

Thus = 3 has modulus 2 andargument

7

6.

1.2.2 Regions in the Complex Plane

Sometimes it is convenient to define various regions in the plane using complex numbers.Recall that =

2 + 2 is distance of from the origin. More generally, if1 = 1 +1

and 2 = 2 + 2 (1, 1, 2, 2 ) then

2 1 =

(2 1)2 + (2 1)2 = distance from 1 to 2.

Example 1.13. Sketch the regions in the complex plane given by

13



(a) { : 3 < 1} , (b) { : 3 > } ,(c)

: 0 < arg <

4

.

Solution.

(a) (b)

(c)

1.2.3 Multiplication of Complex Numbers

Polar coordinates are particularly useful when multiplying complex numbers. Let =(cos + sin ) and = (cos + sin ) be complex numbers in the polar form. Then

= (cos cos sin sin + (sin cos + cos sin )) = (cos( + ) + sin( + )).So, when complex numbers are multiplied,

moduli are multiplied, arguments are added.

By repeated application of this principle we also get De Moivres Theorem.

Theorem 1.3 (De Moivres Theorem). Let = ( + sin ) be a complex number inthe polar form. Then

= (cos + sin )

for every positive integer .

14



Now we consider a few examples on multiplication and taking powers of complexnumbers.

Example 1.14. By calculating (1 + )(1 +

3) in two different ways show that

cos7

12 =

2

6

4 and sin7

12 =

2 +

6

4 .

Solution. We have

(1 + )(1 +

3) = 1

3 + (1 +

3). (1.10)

But also,

1 + =

2(cos

4+ sin

4) (from Example 1.12)

and

1 +

3 = 2(cos

3+ sin

3) (similar to Example 1.12).

So

(1 + )(1 +

3) = 2

2

cos

4+

3

+ sin

4

+

3

= 2

2

cos

7

12+ sin

7

12

. (1.11)

From (1.10) and (1.11) we obtain

(1

3) + (1 +

3) = 2

2

cos

7

12+ sin

7

12

.

Equating real and imaginary parts we obtain

1

3 = 2

2cos7

12and 1 +

3 = 2

2sin

7

12.

Thus

cos7

12 =1

3

22 =

2

6

4 , and sin7

12 =1 +

3

22 =

2 +

6

4

as required.

Example 1.15. Find the real and imaginary parts of (3 )8.

Solution. From Example 1.12

3

= 2cos

7

6

+ sin7

6 .15



Therefore, by De Moivres theorem

(

3 )8 = 28

cos8 7

6+ sin

8 76

= 28

cos

4

3+ sin

4

3

as8

7

6 =28

3 = 8 +4

3 . Hence

(

3 )8 = 28

12

3

2

= 27 27

3 = 128 128

3,

so that the real and imaginary parts are 126 and 1263, respectively.

Example 1.16. For which positive integers is (1 + 3) real?

Solution. We have 1 + 3 = 2 (cos 23 + sin 23 ) (Check this). Therefore by deMoivres theorem

(1 +

3) = 2

cos2

3+ sin

2

3

.

This is real if and only if sin2

3= 0, i.e. if and only if

2

3is an integer, i.e. if and only

if is divisible by 3.

Now we mention a few more properties in connection with polar coordinates and de

Moivres theorem. Let . Note firstly that (cos + sin )(cos sin ) = 1, hence,1

cos + sin = cos sin .

Equivalently1

= when = 1.

Then we note the following relation

(cos sin ) = cos sin (1.12)for any

. Indeed

(cos sin ) = (cos() + sin())

and by de Moivres theorem

(cos() + sin()) = cos () + sin() = cos sin as required. Finally, note that

(cos + sin ) =

1

cos + sin

= (cos sin ) = cos sin

for any .

16



Chapter 2

Methods of Proof

2.1 Direct Proof

A direct proof is a style of proof laid out as in the following:

hypothesis logical argument conclusion.

Example 2.1. Prove that if , are even integers then so is 3 + 5.

Solution. We have that , are even integers. Then = 2 and = 2 for someintegers , . It follows that 3 + 5 = 3 2 + 5 2 = 2(3 + 5) and 3 + 5 is an integer.Therefore 3 + 5 is even.

The the following propositions are proven by direct proofs.

Proposition 2.1. Let . Prove that(a) if is even then 2 is even,

(b) if is odd then 2 is odd.

Proof. (a) Let be even. Then = 2 for some . It follows that 2 = 42 =2(22) and 22 . Therefore 2 is even.

(b) Let be odd. Then = 2 + 1 for some . It follows that2 = (2 + 1)2 = 42 + 4 + 1 = 2(22 + 2) + 1,

and 22 + 2 . Therefore 2 is odd.

Proposition 2.2. Let ,, . Prove that if and then .

Proof. We have and hence = 1 for some 1 . We also have so = 2for some

2 . Therefore =

2=

(

1

2) with

1

2 . Thus

as required.

17



Proposition 2.3. Show that if , then = .

Solution. Let , . Then = + and = + for some ,,, . Itfollows that

= ( + )( + ) = ( ) + ( + ),and = ( ) ( + ).

On the other hand

= ( )( ) = ( ) ( + )

so = as required.

2.2 Examples and Counterexamples

We often use examples to illustrate results. For instance, 3 is odd and 32 = 9 is odd isan example that illustrates the general result that if is odd then 2 is odd but doesnot prove that result. A proof must deal with all , not just = 3 (see the proof ofProposition 2.1(b) above).

However, if we want to prove that a general statement is false, one counterexample isenough.

Example 2.2. Show that the statement if is odd then 2 is even is false.

Solution. = 3 is a counterexample. Indeed 3 is odd and 3 2 = 9 is not even.

Example 2.3. Show that the following statement is false: the sum of two primes isnever prime.

Solution. 2 + 3 = 5 is a counterexample as 2, 3, 5 are all primes.

2.3 Proof By Contradiction

Proof by contradiction is a style of proof showing that a proposition is true. The formatis as follows:

Assume that a statement is false logical argument a contradiction.Therefore we can conclude is true.

18



We will demonstrate proof by contradiction in a number of proofs below. Note firstthat any rational number can always be expressed in its lowest terms, i.e. in the form

where , are integers having no common factor grater than 1. This can always

be achieved by canceling all common factors of the numerator and denominator, e.g.15

20 =3

4 , in lowest terms.

Proposition 2.4. Let be such that 2 is odd. Then is odd.

Proof. Suppose the statement is false, that is we have an integer such that 2 is oddbut is even. Then = 2 for some . Then 2 = (2)2 = 2 (22), and we have22 so 2 is even. This is a contradiction hence the statement is true.

Proposition 2.5. Let be such that 2 is even. Then is even.

Proof. We have , 2 is even. Suppose that is odd. Then = 2 + 1 for some . Hence

2

= (2 + 1)2

= 42

+ 4 + 1 = 2(22

+ 2) + 1. We have 22

+ 2 thus 2 is odd. This is a contradiction, hence is even.Recall that a number is called irrational if , i.e. cannot be represented

as =

where , .

Example 2.4. Prove that

2 is irrational.

Solution. Suppose

2 is rational. Then we can express it in its lowest terms, i.e.2 =

for some integers , having no common factors greater than 1. Then =

2

so that 2 = 22. So 2 is even. Therefore is even (see Proposition 2.5). So = 2

for some . Then (2)2

= 22

, i.e. 42

= 22

, i.e. 2

= 22

. So 2

is even.Therefore is even. But this means that and have 2 as a common factor, which isa contradiction. Therefore

2 is irrational.

Example 2.5. Prove that if is irrational then + 2 is irrational.

Solution. Let be irrational. Let us assume that + 2 is rational. Then the equality = ( + 2) 2 expresses as the difference of two rational numbers which is rationalsince is closed under subtraction. Hence is rational, which is a contradiction. Thus

+ 2 must be irrational.

Now we consider two statements on the prime numbers.

Proposition 2.6. Every positive integer greater than 1 has a prime factor.

Proof. Let be a positive integer greater than 1. Let be the smallest integer greaterthan 1 that divides (there will always be such an integer: if all else fails, divides ).Suppose is not prime. Then = , where > , 2, , . We have and hence (see Proposition 2.2), but this is not possible by the choice of , so wehave a contradiction. Hence must be prime.

19



Theorem 2.1 (Euclid). There are infinitely many prime numbers.

Proof. Suppose there is only a finite number of primes, say 1, 2, . . . , . Let =12 . . . + 1. must have a prime factor (from Proposition 2.6), i.e. for some. But 12 . . . , i.e. ( 1). By Lemma 1.1 ( ( 1)) i.e. 1, which isimpossible. Therefore there must be an infinite number of primes.

2.4 Converse and Contrapositive

Consider the statement

If then (also written , spoken: implies )where and are some statements.

Its converse is the statement

If then (also written ).Examples below show that whether the converse is true or not is completely independentof whether the original statement is true or not. We assume that , in the examplesbelow.

Example A

[Statement] If 4 then 42. (True: prove it)Notice that in this case is the statement 4 and is the statement 42.

[Converse] If 4

2 then 4

. (False: = 2 is a counterexample)

Example B

[Statement] If is even then 2 is even. (True: see Proposition 2.1)

[Converse] If 2 is even then is even. (Also true: see Proposition 2.5)

Example C

[Statement] If 2 > 9 then > 3. (False: = 4 is a counterexample)[Converse] If > 3 then 2 > 9. (True)

Thecontrapositive

to the statement if then is the statementIf (not ) then (not ). (also written (not )(not ))

Contrapositives of the statements given in examples A, B and C are

Example A

[Contrapositive] If 4 2 then 4 (True)

Example B

[Contrapositive] If 2 is not even then is not even.Equivalent formulation is If 2 is odd then is odd. (True)

20



Example C

[Contrapositive] If > 3 then 2 > 9.Equivalent formulation is If 3 then 2 9 (False: = 4 is a counterexam-ple)

Exercise 2. Explain why the contrapositive is true if and only if the original statementis true.

2.5 Proof By Induction

Firstly we introduce some notations on sequences and series.

Definition 2.1. A sequence is a list of numbers 1, 2, 3, . . . (sometimes denoted as{} or just {}). is called the -th term of the sequence. A series is a sum ofterms of a sequence.

notation is used for summing up successive terms of a sequence, e.g.

25=4

= 4 + 5 + . . . + 25,

where {} is some sequence. Another example is6

=3 = 3 + 4 + 5 + 6 = 18.Now proof by induction is a proof using the following principle (or its variations).

Principle of Mathematical Induction

Let () denote some statement. That is the statement involves a parameter .Suppose (1) is true. Suppose also that () is true for = + 1 whenever it is truefor = (here ). Then () is true for any .

Why does this principle work? We have that (1) is true. Then it follows from ourassumptions that (2) is true. Hence also (3) is true. By repeating the process we get() is true for any .

Thus in order to establish that some statement () is true for all it is sufficientto check that (1) is true and to establish ( +1) assuming () is true. The statement(1) is called the base of induction. When proving (+1), assuming () is true, we aredoing the step of induction. The statement () is then called the inductive hypothesis.

Proposition 2.7. For all positive integers ,

=1

2 =1

6( + 1)(2 + 1).

Proof. Let () denote the statement that

=12 =

1

6( + 1)(2 + 1).

21



When = 1,

LHS =1

=1

2 = 12 = 1,

RHS =1

6 1 2 3 = 1 = LHS.So (1) is true.

Suppose now that () is true when = , i.e. suppose

=1

2 =1

6( + 1)(2 + 1).

Then

+1

=1 2 =

=1 2 + ( + 1)2 =1

6

( + 1)(2 + 1) + ( + 1)2 by the inductive hypothesis,

=1

6( + 1) [(2 + 1) + 6( + 1)] =

1

6( + 1)(22 + 7 + 6) =

=1

6( + 1)( + 2)(2 + 3) =

1

6(( + 1)( + 1) + 1) (2( + 1) + 1) ,

so that () is true when = + 1.Thus (1) is true and ( + 1) is true whenever () is true. Therefore by the

principle of mathematical induction () is true for all positive integers .

Example 2.6. A sequence of numbers 1, 2, 3, . . . is defined by

1 = 2 and +1 = 2 for all 1.Find 2, 3, 4. Guess the general formula for and prove it by induction.

Solution. We have

1 = 2,

2 = 21 1 = 3,3 = 22 2 = 4,4 = 23 3 = 5.

This suggests that = +1 for all . Let () denote the statement = +1. When = 1 LHS= 2 (as we are given 1 = 2) and RHS= 2 =LHS. So () is true when = 1.

Suppose now that () is true when = , i.e. suppose = + 1.

Then

+1 = 2 by the definition of the sequence,hence +1 = 2( + 1) by the inductive hypothesis.Thus +1 = + 2 = ( + 1) + 1.

22



We get () is true when = + 1.So (1) is true and ( + 1) is true whenever () is true. Therefore () is true

for all positive integers by the principle of mathematical induction.


1 = 1 and =1

21 + 1 for all 2.

Find 2 and 3. Prove that, for all positive integers n,

= 2 121

. (2.1)

Solution. We have 2 =1

21 + 1 =

3

2, 3 =

1

22 + 1 =

3

4+ 1 =

7

4(which is in agreement

with (2.1)).

Let () denote the statement (2.1). When = 1, LHS= 1 = 2 and RHS= 2 120

=

2 1 1 =LHS. So () is true when = 1.Suppose now that () is true when = , i.e. suppose

= 2 121

.

Then

+1 =1

2 + 1 by the definition of the sequence,

hence +1 =1

2

2 1

21

+ 1 by the inductive hypothesis.

Thus +1 = 1 12

+ 1 = 2 12

= 2 12(+1)1

so that () is true when = + 1.So (1) is true and ( + 1) is true whenever () is true. Therefore () is true

for all positive integers .

Example 2.8. Recall that ! = ( 1) ( 2) . . . 3 2 1 so that 2! = 2 1,3! = 3 2 1, 4! = 4 3 2 1, etc. Calculate

=1

(!)

when = 1, 2, 3, guess the general formula and prove it by induction.

23



Solution.

1=1

(!) = 1 (1!) = 1 1 = 1,2

=1

(!) = 1 (1!) + 2 (2!) = 1 1 + 2 2 = 5,3

=1

(!) = 1 (1!) + 2 (2!) + 3 (3!) = 1 1 + 2 2 + 3 6 = 23.

This suggests that

=1

(!) = ( + 1)! 1. (2.2)

Let () denote the statement (2.2). When = 1,

LHS =1

=1

(!) = 1 (1!) = 1 1 = 1 and

RHS = 2! 1 = 2 1 = 1 = LHS.So () is true when = 1.


=1 (!) = ( + 1)! 1.

Then

+1=1

(!) =

=1

(!) + ( + 1) ( + 1)! by the definition of the sum,

hence+1=1

(!) = ( + 1)! 1 + ( + 1) ( + 1)! by the inductive hypothesis.

Thus

+1=1

(!) = ( + 2) ( + 1)! (+2)(+1)(1)...321

1 = ( + 2)! 1 = (( + 1) + 1)! 1

so that () is true in the case = + 1.So, (1) is true and ( + 1) is true whenever () is true. Therefore () is true


Example 2.9. Prove that, for all positive integers ,

72 + 3 is divisible by 4.

24



Solution. Let () denote the statement that 72 + 3 is divisible by 4. When = 1,72 + 3 = 72 + 3 = 52, which is divisible by 4. So () is true when = 1.


72 + 3 is divisible by 4,

so that 72 + 3 = 4 for some . Then

72(+1) + 3 = 72+2 + 3 = 72 72 + 3 = 49(4 3) + 3= 4 49 147 + 3 = 4 49 144.

Thus 72(+1) + 3 = 4(49 36),

which is divisible by 4 since 49 36 , so that () is true when = + 1.So (1) is true and ( + 1) is true whenever () is true. Therefore () is true


2.6 Variations of the Principle of Induction

We consider two other versions of the principle of mathematical induction.

A. Different Starting Point

Here we suppose that some statement (), , is known to be true for = 0 where0 . We also suppose that ( + 1) is true whenever () is true where , 0. Then the modified principle of mathematical induction states that () is truefor all such that 0.

This is a generalization of the principle of mathematical induction from Section 2.5where we had the same principle with 0 = 1.

Example 2.10. Prove that 2 > 2 for all integers 5.

Note. The result is true when = 1 but not true when = 2, 3, 4.

Solution. Let () denote the statement 2 > 2. When = 5, 2 = 25 = 32,2 = 52 = 25 and 32 > 25 so that () is true when = 5.

Suppose now that () is true (and 5), i.e. suppose

2 > 2.

Then

2+1 = 2

2 > 2

2.

25



So to prove that 2+1 > ( + 1)2 it will be sufficient to prove that 22 ( + 1)2.22 ( + 1)2 22 ( + 1)2 0 2 2 1 0 ( 1)2 1 1 0 ( 1)2 2 1

2

1 + 2.Since 5 this is true and hence ( + 1) is true.

So (5) is true and ( + 1) is true whenever () is true for 5. Therefore ()is true for all integers 5.

B. Inductive Step Involving More Than One Smaller Value

This version of the principle can be formulated as follows. Suppose that a statement

() is true for = 1. Suppose also that ( + 1) is true whenever () is true for all such that 1 . Then the modified principle of mathematical induction states that() is true for any .

As you see the difference with the principle of mathematical induction from Section2.5 is that we allow ourselves to assume at the inductive step that () is true not justfor = but for all (or some) smaller values of as well.


1 = 1, 2 = 1 and = 1 + 62 for all 3.Prove that, for all positive integers ,

=1

5(3 (2)). (2.3)

Solution. Let () denote the statement (2.3).

When = 1,

LHS = 1 = 1 and RHS =1

5

(3

(

2)) =

1

5 5 = 1 = LHS.

When = 2,

LHS = 2 = 1 and RHS =1

5(32 (2)2) = 1

5(9 4) = 1 = LHS.

So () is true when = 1 and when = 2. Consider now (), where 3 andsuppose that () is true for all values of less than . We have for the LHS of ()by definition

= 1 + 62 =1

5 (31

(

2)1) + 6

1

5 (32

(

2)2)

26



by the inductive hypothesis. Thus

=1

5

[(31 + 6 32) ((2)1 + 6 (2)2)] =

=

1

5 [(31 + 2 31) ((2)1 3 (2)1)] ==

1

5

[3 31 ((2) (2)1)] = 1

5(3 (2)) = RHS of ()

so that () is true.So (1) and (2) are true and, for 3, () is true whenever () is true for all

positive integer values of < . Therefore () is true for all positive integers .

27



Chapter 3

Sequences and Series

3.1 Standard Series

Let us repeat some definitions from the beginning of Section 2.5. A sequence is a list ofnumber 1, 2, 3, . . . (sometimes denoted {}). The n-th term of the sequence is .A series is a sum of terms from a sequence, e.g.

25=2 = 2 + 3 + 4 + . . . + 24 + 25.

The series,

=1

= 1 + 2 + 3 + . . . + ,

=1

2 = 12 + 22 + 32 + . . . + 2,

=1

3 = 13 + 23 + 33 + . . . + 3,

are sometimes called the standard series. The following formulae for them can be provenby induction

=1

=1

2( + 1),

=1 2 =

1

6( + 1)(2 + 1),

=1

3 =1

42( + 1)2

(we did

=1 2 in Proposition 2.7).

28



Remark 3.1. Notice that the formula for

3 is the square of the formula for

.Does anything similar happen for

=1

5?

Using the sums of the standard series more series can be summed up.

Example 3.1. Find the sum of the first terms of the sequence 2 3, 3 4, 4 5, . . ..Solution. The -th term is ( + 1)( + 2). So the required sum is

=1

( + 1)( + 2) =

=1

(2 + 3 + 2) =

=

=1

2 +

=1

3 +

=1

2 =

=1

2 + 3

=1

+

=1

2 =

=1

6( + 1)(2 + 1) +

3

2( + 1) + 2 =

1

6 [( + 1)(2 + 1) + 9( + 1) + 12] =

=16

(22 + 12 + 22) =13

(2 + 6 + 11)

Example 3.2. The triangular numbers are the numbers of dots in the followingdiagrams:

etc.1 = 1 2 = 3 3 = 6 4 = 10

Find a formula for the -th triangular number and also for the sum of the first triangular numbers.

Solution.

= 1 + 2 + 3 + . . . + =

=1

=1

2( + 1).

Sum of the first triangular numbers is

1 + 2 + . . . =

=1

=

=1

1

2( + 1) =

1

2

=1

(2 + ) =

=1

2

1

6( + 1)(2 + 1) +

1

2( + 1)

=

1

12( + 1) [(2 + 1) + 3] =

=1

12( + 1)(2 + 4) =

1

6( + 1)( + 2).

29



3.2 Arithmetic Series

An arithmetic progression is a sequence of the form

, + , + 2, + 3 , . . . (3.1)

where , in general. The first element is called the initial term.The differencebetween any two successive elements is and it is called the common difference. The-th term of an arithmetic progression (3.1) is then given by = + ( 1).

An arithmetic series is a sum of terms from an arithmetic progression. Let be thesum of the first terms of the arithmetic series (3.1). Then

= + ( + ) + ( + 2) + . . . + ( + ( 1)) =1=0

( + ).

Proposition 3.1. The sum of the first terms of the arithmetic progression (3.1) is

=

2(2 + ( 1)).

Equivalently, equals times the average of the first and the -th terms of the pro-gression.

Proof. Let us consider

2 =1=0

( + ) +1=0

( + ).

The second sum runs over = 0, . . . , 1. Let us replace index with = 1 ,then the range for is = 0, 1, . . . , 1. Thus

2 =1=0

( + ) +1=0

( + ( 1 )).

Let us now rename to this is just a summation index so it does not matter how we

call it. We get,

2 =1=0

( + ) +1=0

( + ( 1 )) =1=0

( + + + ( 1 )) =

=1=0

(2 + ( 1)) = (2 + ( 1))

since the terms in the last sum do not depend on and there are terms in the sum.

Hence =1

2(2 + (

1)) as required.

30



Notice that what happened in the proof was that we just represented as

= +( + ) + . . . + + ( 1), (3.2)and = + ( 1) + + ( 2) + . . . +. (3.3)

Then the pairwise summation (that is the first element in (3.2) with the first elementin (3.3), then the second element in (3.2) with the second element in (3.3), etc.) gives

2 = (2 + ( 1)) + (2 + ( 1)) + . . . + (2 + ( 1)) = (2 + ( 1)),as required. It is however important to be able to use freely summation notations as wedo in the proof above.

Exercise 3. Prove Proposition 3.1 by proving firstly that

=1 =12

( + 1).

Example 3.3. Find the sum 2 + 5 + 8 + 11 + . . . + 59.

Solution. This is an arithmetic series with (in the usual notation) = 2 and = 3.To find the number of terms we note that 59 = + ( 1) = 2 + 3( 1) so that 1 = 57

3= 19 and hence = 20. Thus the required sum is

= 20 =

2[2 + ( 1)] = 10[4 + 19 3] = 610.

Example 3.4. The sum of the first terms of an arithmetic progression with initial

term 0 and common difference 32 is 1760. Find .

Solution. In the usual notation

1760 =

2[2 + ( 1)] =

2[32( 1)] = 16( 1)

hence ( 1) = 110. Therefore = 11.

Example 3.5. (Higher Maths 1916) Find the sum of the first 6 terms of an arithmetic

progression of which is the 3rd term and is the 5th term.Solution. In the usual notation for an arithmetic progression the -th term is = + ( 1), so the 3rd term equals + 2 and the 5th term equals + 4. Thus

+ 2 = (3.4)

+ 4 = . (3.5)

(3.5)(3.4) gives

2 =

, i.e. =

2

31



and 2(3.4)(3.5) gives = 2 .

(3.6)

Hence the sum of the first 6 terms equals

6 = 3(2 + 5) = 3

2(2 ) + 5

2

= 3

8 4 + 5 5

2

=

3

2(3 + ).

3.3 Geometric Series

A geometric progression is a sequence of the form,,2, 3, . . . , (3.7)

where , in general. Then is called the initial term and is called the commonratio it is the ratio of any two successive terms. The -th term is then given by =

1.We assume = 1 as the case = 1 is trivial. A geometric series is a sum of terms

from a geometric progression. Let be the sum of the first terms of the geometricprogression (3.7). Then

= + + 2 + 3 + . . . + 1.

Proposition 3.2. The sum of the first terms of the geometric progression (3.7) equals

=(1 )

1 .

Proof. We have

=1=0

.

Also consider =

1=0

=1=0

+1 =

=1

as we introduced the summation index = + 1; changes in the range 1, . . . , . Hence

=1=0

=1

=

0 +

1=1

1=1

+

=

= 0 = (1 ).

Thus =(1 )

1 as required.

32



Example 3.6. Find the sum of the first terms of the series 1 +1

2+

1

22+

1

23+ . . ..

Solution. This is a geometric series. In the usual notation = 1, =1

2and the

required sum is

=(1 )

1 =1 (1 12 )

1 12

= 2

1 12

.

Idea of Convergence

Notice that as increases to infinity the sum approaches 2 since the term1

2ap-

proaches 0. We write 2 as . We say that the series has sum to infinity 2and we also write = 2.

In general, let 1 + 2 + 3 + . . . be any series and let be the sum of the first terms, i.e. =

=1 . If tends to a fixed number as tends to ( as

) we say that a series converges and has sum to infinity , and we write = and

=1 = . On the other hand, if does not tend to any fixed number as we say that the series diverges, does not have a sum to infinity and that

=1 doesnot exist.

Proposition 3.3. For < 1, the geometric series

+ + 2

+ 3

+ . . .

converges and has sum to infinity 1

.

Sketch of Proof. If < 1 it can be shown that 0 as . So

=(1 )

1

1 as .

A representation of rational numbersRational numbers are often given in the form of periodic decimal expressions. Forinstance, the number 3. . . ., where ,,,, are some integers such that0 ,,,, 9, can also be denoted in the form 3. where the dots above and mean that the part between and , that is the part , is a periodic (infinitelyrepeating) part.

Sums of geometric series can be used to transfer the rational numbers from periodicdecimal forms into the form of a ratio of two integers.

Example 3.7. Express the repeating decimal 0.123(= 0.12323232323 . . .) as a rationalnumber in its lowest terms.

33



Solution. We have

0.123 = 0.123232323 . . . =1

10+

23

103+

23

105+

23

107+ . . . =

1

10+

23

103

1 +

1

102+

1

104+ . . .

=

=

1

10 +

23

1000

=0

(102

)

=

1

10 +

23

1000 11 1100 =1

10 +

23

990 =

122

990 =

61

495 .

A Special Case of the geometric series

Consider the series

1 + + 2

+ 3

+ . . . (3.8)

i.e. a geometric series with = 1 and = . Then

=1=0

= 1 + + . . . + 1 = 1 1

by above. This can be rewritten as:

1 = ( 1)(1 + . . . + + 1).

So, for example,

3 1 = ( 1)(2 + + 1),4 1 = ( 1)(3 + 2 + + 1),5 1 = ( 1)(4 + 3 + 2 + + 1).

Also, provided < 1, the geometric series (3.8) has sum to infinity 11 , i.e.

1

1

= 1 + + 2 + 3 + . . . .

Replacing by we obtain1

1 + = 1 + 2 3 + . . . (provided < 1).

If we were allowed to integrate infinite series term by term we would have got a series forlogarithm:

ln(1 + ) = 12

2 +1

33 1

44 + . . . (provided < 1).

This can actually be justified see later courses.

34



3.4 Two Special Series

1. Consider the series

=11

2=

1

12+

1

22+

1

32+

1

42+ . . .

Let be the sum of the first terms. Then

=1

12+

1

22+

1

32+

1

42+ . . . +

1

2

and

=

=1

1

2< 1 +

=2

1

( 1) as1

21

=02+1

=2+11

>

1

=02+1

=2+11

2+1=

1

=0(2+1 2) 1

2+1=

1

=01

2=

2

as the sum from 2 + 1 to 2+1 contains 2+1 2 = 2 terms. Thus 2 > 2

. So as

gets arbitrarily large so too does the sum, i.e. as . Thus, theseries diverges, i.e.

=11

does not exist.

36



Chapter 4

The Binomial Theorem andApplications

4.1 Permutations and Combinations

Suppose we have different objects (here ). If we choose of them (0 ),the objects are called a combination of objects from . If instead we choose of the objects and place them in a definite order (so that we have a first object, a second object,etc.) the objects are called a permutation (or an ordered combination) of objects from.

The number of permuations of objects from is denoted . The number of

combinations of objects from is denoted or more usually

(spoken choose

).

Proposition 4.1. The number of permutations is = ( 1)( 2) . . . ( +1).

Proof. To form a permutation of objects from , we need to choose a first object. Thiscan be any of the objects. So we have choices. For each of these choices we can thenchoose our second objects from the 1 remaining objects, i.e. we have 1 choices.At the third step we have 2 choices, etc. Thus the total number of choices for stepsis

= ( 1)( 2) . . . ( + 1).

For example, 73 = 7 6 5 = 210.Recall now that for one defines ! = 1 2 . . . and by definition 0! = 1. The

number ! is pronounced as factorial.

Corollary 4.1. The number of permutations of objects from is = !.

Corollary 4.2. One can rearrange the number of permutations as =!

( )!.

37



Proposition 4.2. The number of combinations of objects from ( , ,0 ) is given by

=

!

!(

)!

=( 1)( 2) ( + 1)

(

1)(

2)

1

.

(That is we take the product of successive numbers down from and divide it by !).

Proof of Proposition 4.2. To form a permutation of objects from we have to choosethe objects and put them into order. We separate these two stages. First choose the objects, i.e. form a combination of objects from , then put them into order, i.e. forma permutation of objects from . This tells us that

=

.

Thus

=

= ( 1) . . . ( + 1)( 1) . . . 1

by Proposition 4.1 and Corollary 4.2, which gives us one of the required forms. This canalso be rearranged as

=

( 1) . . . ( + 1) ( )!( 1) . . . 1 ( )! =

!

!( )!as needed.

For example,

The number of combinations of 3 out of 7 is7

3

=

7 6 53 2 1 = 35.

The number of combinations of 6 balls from 49 is49

6

=

49 48 47 46 45 446 5 4 3 2 1 = 13, 983, 816.

So, by the way, the chance of winning the lottery is about 1 in 14 million.

Now we discuss some properties of the number of combinations

.

Proposition 4.3. For any , , 0 ,

=

.

Proof.

=

!

!( )! ,

=!

( )!( ( ))! =!

( )!! .

Hence result.

38



An alternative proof which does not require any calculations is as follows. When weremove objects from a collection of we leave behind (and viceversa). So thenumber of ways of choosing is the same as the number of ways of choosing . ThusProposition 4.3 holds.

Proposition 4.3 can simplify calculations. For example,21

19

=

21 20 19 . . . 319 18 17 . . .

can also be represented as

21

19

=

21

2

=

21 202 1 = 210

which is easier to compute. Note also that

0

=

= 1 (4.1)

for any . That is there is only one way of choosing objects from (or, equivalently,of choosing none objects).

Proposition 4.4. Let . Then

+ 1

=

+

1

for any such

that 1 .

Proof.

RHS =!

!( )! +!

( 1)!( + 1)! =( + 1) !

! ( + 1) ( )! + !

( 1)!( + 1)!=

( + 1) !!( + 1)! +

!!( + 1)! =

( + 1 + ) !!( + 1)! =

( + 1) !!( + 1)! =

( + 1)!

!( + 1 )!= LHS.

Exercise 4. Give a combinatorial argument that proves Proposition 4.4 withoutcalculations.

Let us also formally define

= 0 if < 0 or +1. Then it is easy to check that

Proposition (4.4) is true for any , . Propositions 4.4 and formula (4.1) implythat the number of combinations can be read off the Pascal triangle. In the followingdiagramme we present Pascals triangle up to and including row 4.

39



1 Row 0

1

!!CC

CCCC

CCC add 1

}}{{{{{{{{{ Row 1

1

@@

@@@@

@@add 2

}}{{{{{{{{{

!!CC

CCCC

CCC add 1

~~~~~~~~~~

Row 2

1 3 3 1 Row 3

1 4 6 4 1 Row 4

In general, the -th row consists of the numbers 0,1,2, . . . ,.4.2 The Binomial Theorem and Applications

Theorem 4.1 (The Binomial Theorem). For any , , one has the expansion

( + ) =

=0

. (4.2)

Equivalently

( + ) =

0

+

1

1 +

2

22 + . . . +

1

1 +

. (4.3)

Combinatorial Proof. We need to expand the brackets in (+). The result is the sum ofmonomials

=0

for some coefficients 0, . . . , . Indeed the monomial

appears in the expansion when we choose in brackets and in the remaining( ) brackets when expanding the power; and no other monomials can arise. Thismonomial appears as many times as we can choose objects (in this case theexpressions + ) out of . Thus = (

) as required.Algebraic Proof. We prove the theorem by induction. Let () be the statement givenby Equation (4.2). For = 1 we have

+ =1

=0

1

1 = +

so (1) is true. Suppose now that () is true. That is

( + ) =

0 +

11 + . . . +

, for some

.

40



Then

( + )+1 = ( + )( + ) = ( + )

0

+

1

1 + . . . +

=

= 0+1 + 1 + 0 + 2 + 1 12 + . . .. . . +

+

1

+

+1 =

=

+ 1

0

+1 +

+ 1

1

+

+ 1

2

12 + . . . +

+ 1

+

+ 1

+ 1

+1,

where we used Proposition 4.4 and the formulas

0= 1 =

+ 1

0 ,

= 1 =

+ 1

+ 1.

Thus the statement ( + 1) is true. Since (1) is true and ( + 1) is true whenever() is true we conclude that () is true for any by the principle of mathematicalinduction.

Remark 4.1. The equivalent formulation is ( + ) =

=0

. Indeed, let

us replace the summation index with in equation (4.2). Then varies from 0to , so varies from 0 to and we get

( + ) =

=0

=

=0

by Proposition 4.3.

Remark 4.2. Due to the Binomial theorem the numbers of combinations

are also

called the binomial coefficients.

Note that the coefficients in the RHS of equation (4.2) are the numbers in row ofPascals Triangle.

Example 4.1. Expand ( + )4.

Solution. Row 4 of Pascals Triangle is 1 4 6 4 1. So

( + )4 = 4 + 43 + 622 + 43 + 4.

Example 4.2. Expand

24

.

41



Solution. By modifying the expression from Example 4.1 we get

2

2= 4 + 43

2

+ 62

2

2+ 4

2

3+

2

4=

= 4

23 +

3

222

1

23 +

1

164.

Example 4.3. Find the term independent of in the expansion of

3 +

2

2

33.

Solution. By the binomial theorem

3 + 2

233

=33=0

33

(3)33

22

=33=0

33

3332333.

The term independent of is the term with 33 3 = 0, i.e. = 11, and therefore thisterm equals

33

11

322211.

Binomial theorem can also be used to simplify various expressions. We start with thefollowing proposition.

Proposition 4.5. For any , one has the expansion

(1 + ) =

=0

. (4.4)

Proof. This follows immediately from the Binomial Theorem at = 1, = .

Example 4.4. Find the sum

1 +1

3

7

1

+

1

9

7

2

+ . . . +

1

37

7

7

.

Solution. Notice that the required sum equals

1 +

1

3

7(see Proposition 4.5 with

= 1/3, = 7). Thus, the sum equals

4

3

7=

16384

2187.

42



Let us put = 1 and = 1 in Proposition 4.5, formula (4.4). We obtain thefollowing relations

0+

1+ . . . +

= 2,

0

1

+

2

3

+ . . . + (1)

= 0.

Thus the sum of the entries in the -th row of Pascals triangle is 2 and the alternatingsum of these entries is zero. For example, for = 4 we get

1 + 4 + 6 + 4 + 1 = 16 = 24,

and 1 4 + 6 4 + 1 = 0.

Let us further explore Proposition 4.5, to derive more identities with binomial coeffi-

cients. By differentiating (4.4) we get

(1 + )1 =

=0

1. (4.5)

Let us put = 1 in (4.5). Note that the LHS of (4.5) equals 0 at = 1 under theextra assumption 2. We get the following result.Proposition 4.6. Let . Then

=1

= 21,and

=1

(1)

= 0 if 2.

Remark 4.3. Proposition 4.5 is also useful for some approximations. Indeed, in the

expansion(1 + ) =

0

+

1

+

2

2 + . . . +

the terms on the RHS get smaller and smaller if 1. So the first few terms may givea reasonable approximation to (1 + ) in this case.

Example 4.5. Write down the binomial theorem for (1+ ), where , . Dif-ferentiate the result to show that

=1

2

= (+1)22. Show that

=1

(1)2

=

0 if

3.

43



Solution. We have

(1 + ) =

=1

(4.6)

by the Binomial Theorem. By differentiating (4.6) we get

(1 + )1 =

=1

1, (4.7)

and by differentiating (4.7) we get

(

1)(1 + )2 =

=1 ( 1)

2. (4.8)

Now we put = 1 and sum (4.8) and (4.7):

21 + ( 1)22 =

=1

2

,

which is the required relation as

21

+ ( 1)22

= 22

(2 + 1) = ( + 1)22

.

Now let us put = 1 and consider the difference of (4.7) and (4.8). Assuming 3we get

0 =

=1

(1)1

=1

( 1)

(1)2 =

=

=1

(1)1

(1 ( 1)(1)) =

=1

(1)12

. (4.9)

By multiplying (4.9) by (1) we get the required relation.

Example 4.6. Simplify

=1

3

where .

Solution. We have (1 + ) =

=0

by the binomial theorem. Differentiation

44



gives

(1 + )1 =

=0

1, (4.10)

( 1)(1 + )2 =

=0

( 1)

2, (4.11)

( 1)( 2)(1 + )3 =

=0

( 1)( 2)

3 =

=

=0

(3 32 + 2)

3. (4.12)

Let us put = 1 and let us add the equations (4.11) multiplied by 3 to the equation(4.12) (so that 2 terms cancel). We get

( 1)( 2)23 + 3( 1)22 =

=0

(3 )

. (4.13)

Let us now add (4.10) and (4.13) (so that the terms cancel). We have

( 1)23( 2 + 3 2) + 21 =

=0

3

.

Thus

=0

3

= ( 1)23( + 4) + 21 = 23(2 + 3 4 + 4) = 2( + 3)23.

Example 4.7. Prove that

=1

(

1)() = 0 if() is a polynomial of degree atmost 1. (For the definition of the degree of a polynomial see Section 5.1 below).

Sketch of Solution. Write down the Binomial Theorem for (1+ ) and its first deg derivatives. Substitute = 1 to get 0 in the lefthand sides of the identities. Showthat there is a combination of the obtained identities which gives

=0

(1)()

in

the righthand side for any given polynomial ().

45



4.3 Application to Trigonometric Calculations

We are going to obtain some trigonometric identities with the help of complex numbers.Recall from Section 1.2.3 that

cos + sin = (cos + sin )

by De Moivres Theorem. This theorem can be used to express cosine and sine of amultiple angle through cosine and sine of the angle.

Example 4.8. Express cos 4 as a polynomial in cos .

Solution. By De Moivres Theorem

cos4 + sin4 = ( + sin )4.

By the binomial theorem

(cos + sin )4 = cos4 + 4 cos3 ( sin ) + 6 cos2 ( sin )2 + 4 cos ( sin )3 + ( sin )4 =

= cos4 + 4 cos3 sin 6cos2 sin2 4 cos sin3 + sin4 == (cos4 6cos2 sin2 + sin4 ) + (4cos3 sin 4cos sin3 ).

By equating the real parts we obtain

cos4 = cos4 6cos2 sin2 + sin4 = cos4 6cos2 (1 cos2 ) + (1 cos2 )2 == cos4

6cos2 + 6 cos4 + 1

2cos2 + cos4 = 8 cos4

8cos2 + 1

Note that by taking imaginary parts we can also express sin4 as a polynomial of cos and sin . Further we can reverse the process in order to express powers of cosine andsine of an angle as a combination of the trigonometric functions of the multiple angles.

Recall from Section 1.2.3 that if = cos + sin , then

= cos + sin ,

1

= cos sin

(see de Moivres theorem and relation (1.12)) for any . This implies that

+1

= 2 cos , (4.14)

1

= 2 sin . (4.15)

Example 4.9. Express cos4 in the form cos4 + cos2 + .

46



Solution. Let = cos + sin . Then using (4.14) and the binomial theorem we get

(2cos )4 =

+

1

4= 4 + 43

1

+ 62

1

2+ 4

1

3+

1

4=

= 4 + 42 + 6 +4

2 +1

4 =4 + 14 + 42 + 12 + 6.

Hence

16 cos4 = 2cos 4 + 8 cos 2 + 6,

and

cos4 =1

8cos4 +

1

2cos2 +

3

8.

Example 4.10. Express sin5 in the form sin5 + sin3 + sin .

Solution. Let = cos + sin . Then using (4.15) and the binomial theorem we get

(2 sin )5 =

1

5= 5 + 54

1

+ 103

1

2+ 102

1

3+ 5

1

4+

1

5=

= 5 53 + 10 10 + 53 15 = 5 15 53 13 + 10 1 .Hence

32 sin5 = 2 sin5 10 sin3 + 20 sin ,

and

sin5 =1

16sin5 5

16sin3 +

5

8sin .

Exercise 5. Simplify the sum = 1 + cos 2 + cos 4 + cos 6 + cos 8 + cos 10.Hint. Find a combination of and where

= 1 + sin 2 + sin 4 + sin 6 + sin 8 + sin 10.

47



Chapter 5

Polynomials and roots in the polarform

5.1 Polynomials

Definition 5.1. A polynomial () is a function of the form

() = + 1

1 + . . . + 1 + 0

where and for = 0, 1, . . . , . We say that a polynomial () is real if for all = 0, 1, . . . , . The polynomial () has degree if = 0, this is denotedas deg() = . In this case is called the leading term of ().

Remark 5.1. In the case () 0, i.e. all of the coefficients = 0, we say () is azero polynomial. We do not define a degree of in this case.

Remark 5.2. Polynomials of degree 1 are also called linear polynomials.

We considered division of integers with remainder in Section 1.1. There is a version

of a division with remainder for polynomials.

Euclidean Algorithm for Polynomials

Let () and () be polynomials where () is not a zero polynomial. Then there existpolynomials () and () such that () = () () + (), with deg () < deg (),or () is a zero polynomial.

To obtain these we use long division, e.g. let us take () = 24 + 3 + 2 3 2,and () = 2 + 2 + 3. We have the following process:

48



We stop as deg(4 5) = 1 < 2 = deg (). So () = 22 3 + 1 and () = 4 5(and indeed () = () () + ()).Definition 5.2. A number is called a root of a polynomial () if () = 0.

We have the following Theorem as a corollary from the Euclidean Algorithm.

Theorem 5.1. A number is a root of the nonzero polynomial () if and only if() is divisible by , i.e. () = ( )() for some polynomial ().Proof. By the Euclidean algorithm () = ( )() + () where deg () < 1 or() is a zero polynomial, i.e. in both cases () = constant = . Then () = , so() = 0 = 0.

Recall now that if = + , where , then = is the complex conjugateof . Note also that

+ = 2 = 2Re, = 2 + 2 = 2.Also for any , we have

+ = + , = . (5.1)

By iterating the last equality we also have

= (5.2)

for any

. Furthermore, the following lemma takes place.

Lemma 5.2. Let () be a real polynomial. Then () = () for any .

Proof. Let () have the form () =

=0

where . By applying the properties

(5.1) and (5.2) we get

() = + 11 + . . . + 1 + 0 = + 1 1 + . . . + 1 + 0 ==

+ 11 + . . . + 1 + 0 = ()

as required.

49



As a corollary from this lemma we get the following.

Proposition 5.1. Let () be a real polynomial. If is a root of () then isalso a root of ().

Proof. Let be a root of (), so () = 0. By Lemma 5.2 we have () = 0 a srequired.

This leads us to the following proposition.

Proposition 5.2. Let be a nonreal root of a real polynomial (). Then() = (2 2Re() + 2)() (5.3)

where () is some real polynomial.

Proof. By Theorem 5.1 we can represent () = ( )() where () is a polynomial.We have () = ( )() = 0 by Proposition 5.1. As we have = 0, so() = 0. By Theorem 5.1 () = ( )() for some polynomial (). Thus

() = ( )( )() = (2 ( + ) + ))()as required.

Exercise 6. Show that the polynomial () from (5.3) is a real polynomial.

Example 5.1. Given that 2 + 3 is a root of the equation4 + 72

12 + 130 = 0,

find all the roots of the equation.

Solution. Since 2 + 3 is a root, so is 2 3 and2 2Re(2 + 3) + 2 + 32 = 2 + 4 + 13

is a factor of the polynomial. Hence

4 + 72 12 + 130 = (2 + 4 + 13)(2 4 + 10).

So the remaining roots are given by 2

4 + 10 = 0, i.e. =

4 16 402

=4 24

2=

4 4612

=4 26

2= 2

6.

Therefore the required roots are 2 3, 2 6.

We conclude by stating the following important result.

Theorem 5.3 (Fundamental Theorem of Algebra). Any polynomial () has a root in.

50



We do not prove Theorem 5.3 in this course. The theorem has the following immediateconsequences.

1. The number system is large enough to deal with the solutions of polynomialequations.

2. Every complex polynomial of degree can be factorised into linear factors. (Ex-ercise).

3. Every real polynomial can be factorised into a product of real linear factors (cor-responding to real roots) and real quadratic factors (corresponding to pairs , ofcomplex roots as in Proposition 5.2).

5.2 Roots in the Polar Form

Let . One can define exponents of imaginary numbers and it appears that the

following formula takes place:

= cos + sin . (5.4)

While we do note prove the relation (5.4) we show that it is consistent with a number offamiliar properties of exponential and trigonometric functions.

1. Let = 0. Then = 0 = 1 and cos + sin = 1, so (5.4) indeed holds in thiscase.

2. The known relation cos sin = 1cos + sin

can be rearranged using (5.4)

into = 1

which should hold for the exponential function.

3. De Moivres theorem in the form (cos + sin ) = cos + sin is rearrangedusing (5.4) into () = which is one of the index laws.

4. De Moivres theorem in the form

(cos + sin )(cos + sin ) = cos( + ) + sin( + )

takes the form = (+) which is a familiar property of exponential functions.

One way to establish relation (5.4) would be to use the power series for the functions, cos and sin . It will be discussed in the later courses that the following expansiontake place:

==0

!,

so for = ,

=

=0()

!= 1 +

2

2!

3

3!+

4

4!+

5

5!+ . . . (5.5)

51



And also

cos = 1 2

2!+

4

4!

6

6!+ . . . (5.6)

and

sin = 3

3!+

5

5!

7

7!+ . . . (5.7)

Then formula (5.4) can be seen by comparing (5.5) with the sum of the series (5.6) andthe series (5.7) multiplied by .

Example 5.2. 2 = cos 2 + sin2 = 1. Similarly, 2 = 1 for any . Alsoclearly the opposite holds: if = 1 for some then = 2 for some .

Formula (5.4) is useful for finding the roots of complex numbers. Let = (cos +

sin ) =

.Proposition 5.3. For all positive integers , the -th roots of = are =

1 (

+2 ), where = 0, 1, 2, . . . , 1.

Remark 5.3.

1. In Proposition 5.3 1 means the unique real nonnegative -th root of the real

nonnegative number .

2. If = 0, that is = 0, the only -th root is 0.

3. Proposition 5.3 states in particular that any complex number has distinct -throots in unless = 0.

Proof of Proposition 5.3. Let be the -th root of, that is = . Let have modulus and argument , so = . By de Moivres theorem has modulus and argument. So:

= = and = + 2 for some

= 1

and =

+ 2

= 1

( +2

)

.

Also we note that for = 0 we have

= (+2

) = (+2 ) 2 () = 1

.

Hence all the roots 0, . . . , 1 are pairwise different, and any -th root, , satisfies = for some , 0 1. (We just divide by , that is = + where, , 0 1.)

Let us now sketch the -th roots on a diagram. Let

= 1 (

+2 ),

= 0.

52



Figure 5.1A sketch of general complex roots on the complex plane; =

1

( +2

). The figure shows a circle ofradius

1 with its centre located at the origin.

Figure 5.1 shows that 0, 1, . . . , 1 are distinct roots and that for other values of gives repetitions of these roots.

Example 5.3. Solve the equation 2 = 4.

Solution. Note that 4 = 4

2 in the polar form. By Proposition 5.3 the square rootsof 4 are

= 1

2 (

2 +2

2 ), = 0, 1.

That is

0 = 24 and 1 = 2

54 .

Thus

0 = 2cos 4 + sin

4 = 2

2

2

+

2

2 =

2(1 + ).

Similarly, 1 = 2

cos

5

4+ sin

5

4

= 2(1+ ). Thus the roots are 2(1+ ).

Example 5.4. Find the cube roots of unity, i.e. solve the equation 3 = 1.

Solution. We have 1 = 10 in the polar form. So the cube roots of unity are

= 113 (

0+23 ), = 0, 1, 2.

53



So we have

0 = 1, 1 = 23 = cos

2

3+ sin

2

3= 1

2+

3

2,

and

2 = 43 = cos

4

3+ sin

4

3= 1

2

3

2.

Note. The two nonreal cubic roots of 1 are conjugates of each other. This can also beseen algebraically because 3 = 1, i.e. 3

1 = 0, is a real polynomial equation so that

Theorem 5.1 applies. Geometrically the cubic roots can be seen sketched on Figure 5.2.

Figure 5.2A diagram of the cube roots of unity on the complex plane.

Proposition 5.4. The sum of all -th roots of unity is zero.

Proof. The -th roots of unity are 1, , 2, . . . , 1 where = 2 . Notice that

1 + + 2 + . . . + 1 = 1 1 (5.8)

because the LHS is the sum of the first terms of a geometric series. Since = 1, theRHS of (5.8) equals 0 as required.

54



Chapter 6

Matrices

6.1 Scalars, Matrices and Arithmetic

Scalar is an alternative name for a number. It is used when we want to restrict all thenumbers in our considerations to be of a specific type. For example our scalars will bereal numbers means that all the numbers in the calculations are assumed to be real.A matrix is a rectangular array of scalars. An matrix (spoken by ) is amatrix with rows and columns. Such a matrix contains scalars. For example,

3 0 12 5 4

is a 2 3 matrix; it has 6 scalars.

For an matrix , is called the type (or the size) of the matrix, the rowsare numbered from the top and the columns are numbered from the left. The scalar inthe -th row and the -th column is called the (, )-th entry, it is denoted or ().

For example, if = 3 0 12 5 4 then 12 = 0, 22 = 5, 21 = 2, etc. Two matrices and are said to be equal ( = ), if they have same type and all the corre-sponding entries are equal: = for all = 1, . . . , and = 1, . . . , .

There are some important operations which we can perform with matrices.

Addition

Matrices of the same type can be added entry by entry.

Definition 6.1. Let and be matrices. Then + is defined to be the matrix with the entries( + ) = + for all and , 1 , 1 .

For example,

3 0 12 5 4

+

1 2 44 1 0

=

4 2 56 6 4

.

Remark 6.1. If and are not of the same type then + is not defined.

55



Scalar Multiplication

To multiply a matrix by a scalar we multiply all the entries by that scalar.

Definition 6.2. If is an matrix and is a scalar then is the matrixwith the entries

() = for all and , 1 , 1 .

For example,

3

3 0 12 5 4

=

9 0 36 15 12

.

Subtraction

Having matrix addition and scalar multiplication we also get matrix subtraction. Indeedwe define = + (1), provided and have the same type . The matrixelements can then be found as

( ) = for all , , 1 , 1 .

For example, 3 0 12 5 4

1 2 44 1 0

=

2 2 3

2 4 4

.

Rules of Arithmetic

The operations of addition and multiplication by scalars satisfy the following rules:

(a) Commutativity of addition: + = + ,

(b) Associativity of addition: ( + ) + = + ( + ),

(c) ( + ) = + ,

(d) () = (),

where ,, are matrices of the same type and , are scalars.

We prove property (a), proofs of the other properties are similar.

Proof of (a). Note that + and + are both defined. We have

( + ) = + = + = ( + ),

so all the corresponding matrix entries in + and + are equal, hence + = + .

56



By using the rules (a) (d) further properties can be obtained. For example, we have

( + ) = ( ) (6.1)if , and are matrices of the same type. It is easy to verify (6.1) directly, howeverlet us derive it from the previous properties. By the definition of subtraction, we have

( + ) = + (1)( + ) = + ((1) + (1))by (c). Using associativity (b), we get

+(

(1) + (1)) = ( + (1)) + (1) = ( ) by the definition of subtraction, thus (6.1) is established.

Remark 6.2. Due to associativity (b), we can omit brackets when considering expres-sions with a few matrix additions.

Example 6.1. Simplify 2(3 ) + 3 ( + 2), where the matrices and have thesame type.

Solution. We have

2(3 ) + 3( + 2) = 2(3) + 2((1)) + 3 + 3(2) == 6 + (2) + 3 + 6 = 9 + 4.

Zero Matrix

The zero matrix (denoted , or just ) is the matrix with all entrieszero. For example, 2,3 =

0 0 00 0 0

. The zero matrix has the following important

properties:

+ , = = , + ,

0 = , (here in the LHS 0 is a zero scalar),

, = ,,

for all matrices and all scalars .

Just as a zero matrix has all entries 0, a zero row of a matrix means a row all of whoseentries are zero, and a nonzero row means a row not all of whose entries are zero (like-

wise for columns). So, for example, in the matrix

1 2 1 30 0 0 01 2 0 00 0 0 1

, row 2 is a zero row

Algebra 1X Notes v3aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Documents

Transcript of Algebra 1X Notes v3aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa