mast 10008 week 4

Lecture 13Proofs

In mathematics there are rigorous rules of deduction needed toprove a fact.

We must state definitions and assumptions, which often requirestranslating sentences to mathematical language.

Example 13.1 Prove that the square of every non-zero numberis positive.

Proof

I.13.1

Example 13.2 Prove Pythagoras theorem.

Proof

I.13.2

Example 13.3 Prove thatp2 is not rational.

Fact 1: Any fraction can be expressed in the form pq where atleast one of p and q is odd.

Fact 2: If the square of an integer is even, then the integer itselfis even.

Fact 3: The square of an even integer is a multiple of 4.

To prove the theorem by contradiction, we will assume that thereis a rational number for which the square is 2 and use this as-sumption to derive a contradiction.

I.13.3

Example 13.4 Prove that the number of primes numbers is in-finite.

Fact: For n > 0, the only common factor of n and n + 1 is 1.The following proof is famous, given by Euclid in 300 BC.

To prove the theorem by contradiction, we will assume that thereare only finitely many primes.

I.13.4

Example 13.5 The set Q of rational numbers is countable whilethe set R of real numbers is not countable.

I.13.5

Example 13.6 Let A be a symmetric matrix. Prove that thereexists a direction preserved by A, i.e. there exists a vector vsuch that Av = v for some scalar .

You can attempt questions 9 to 23 in problem sheet 3 ofyour workbook.

I.13.6

Lecture 14Mathematical induction

Suppose we have a proposition/assertion about any natural num-ber n. For example, the sum of the first n positive integers isn(n+1)/2.

This proposition can be made for any n = 1,2, ... and we notateit P (n). For example, P (1) is the sum of the first 1 positiveintegers is 1.2/2.

It is important to note that P (n) may be true or false.

Example 14.1 P (n) is the proposition n is even.

Thus P (n) is true half the time and false half the time.

I.14.1

The principle of mathematical induction

Let P (n) be a proposition about natural numbers.

1. If P (1) is true, and

2. P (k)) P (k +1),

then by the principle of mathematical induction P (n) is true forevery natural number n.

Example 14.2

1+ 2+ ...+ n =1

2

n(n+1)

P (1) :

I.14.2

P (k)) P (k +1)

I.14.3

The principle of mathematical induction relies on the least ele-ment property of the natural numbers: Any non-empty subset ofthe natural numbers has a least element.

Given a proposition P (n), let

S = {n 2 N | P (n) is false}.

If we can prove S is empty, then we have proven P (n) is truefor all n 2 N.

If P (1) is true then S is not all of N.

If S is non-empty then it possesses a least element m, say.Thus m 1 /2 S so P (m 1) is true.

If we know P (k) ) P (k + 1) then P (m 1) is true impliesP (m) is true, but this contradicts m 2 S!

We conclude that S has no least element, hence it is empty. I.14.4

Example 14.3 The sum of the interior angles of an n-sidedpolygon is (n 2).

Example 14.4

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

8

I.14.5

Notice we can adjust the principle of mathematical induction:


1. If P (n0

) is true, and

2. P (k)) P (k +1),

then by the principle of mathematical induction P (n) is true forevery natural number n n

0

.

Example 14.5 When is n3 < 2n?

I.14.6

Another version of mathematical induction.



2. P (1), P (2), ..., P (k)) P (k +1),


Example 14.6 Every positive integer factors into a product ofprimes.

You can attempt questions 24 and 25 in problem sheet3 of your workbook.

I.14.7

Lecture 15Mathematical induction continued: inequalities



2. P (k)) P (k +1),

(or 20. P (1), P (2), ..., P (k)) P (k +1))


I.15.1

Example 15.1 If n 2 N is odd then n(n2 1) has a factor of24.

I.15.2

Example 15.2 The number of ways to cut a stick of length ninto pieces, each of integer length, is 2n1.

1 = 1, 2 = 21+ 1

, 3 = 32+ 1

1+ 2

1+ 1+ 1+ 1

, 4 = 43+ 1

1+ 3

2+ 2

2+ 1+ 1

1+ 2+ 1

1+ 1+ 2

1+ 1+ 1+ 1+ 1

I.15.3

Inequalities

Example 15.3 Find n0

such that for n n0

, n3 > (n+1)2.

Add 3k2 + 3k +1 to both sides of k3 > (k +1)2.

I.15.4

In the last example we compared a polynomial with a polyno-mial. Now compare a factorial to a polynomial.


such that for n n0

, n! > n3.

Reduce this to the polynomial case:

multiply both sides of k! > k3 by (k +1).

I.15.5

Compare a power to a polynomial.


such that for n n0

, 2n > n3.

Reduce this to the polynomial case:

multiply both sides of 2k > k3 by 2.

You can attempt questions 26 and 27 in problem sheet3 of your workbook.

I.15.6

Lecture 16Complex Numbers

Some simple equations do not admit any solutions in R. Forexample, x2 + 1 = 0. To solve such equations we define theimaginary number i =

p1.

Definition 16.1 (Complex number) A complex number is ofthe form z = x+ iy, x, y 2 R .

The set of all complex numbers is C = {x+ iy : x, y 2 R} .

The real and imaginary parts of z are:

x = Re(z), y = Im(z).

Example 16.1 Let z = 4 7i.

Re(z) = Im(z) = I.16.1

Two complex numbers are equal if and only if they have thesame real and imaginary parts:

a+ ib = c+ id, a = c and b = d.

Addition

(a+ ib) + (c+ id) = (a+ c) + i(b+ d)

Multiplication THIS IS IMPORTANT

(a+ ib)(c+ id) = (ac bd) + i(ad+ bc)

The modulus of z = a+ bi is |z| =q

a2 + b2.

The complex conjugate of z = a+ bi is z = a bi.

To divide complex numbers we multiply by the complex conju-gate of the denominator.

Example 16.2 Simplify4+ i

3 2i I.16.2

The Argand plane/complex plane

A complex number, z 2 C, may be represented by a point orvector in a coordinate plane, the Argand or Complex plane,

R2 = {(x, y) : x, y 2 R},since each complex number, z = x + iy, determines a uniqueordered pair (x, y) and vice-versa.

Many properties of complex numbers have geometric interpre-tations.

I.16.3

Real numbers are complex numbers with a zero imaginarypart. i.e. z = x = x+0y. R C.

Purely imaginary numbers are complex numbers with a zeroreal part. i.e. z = iy = 0+ iy.

Unlike real numbers, complex numbers are not ordered!

Example 16.3 Represent the following in the complex plane.

1. z1

2. |z1

|

3. |z1

z2

|

4. z1

I.16.4

Regions of the complex plane

Example 16.4 Sketch the region given by

{z : |z| 4} \ {z :

6

< arg z

3

} .

Example 16.5 Sketch the curve given by z z = 9.

I.16.5

Example 16.6 Sketch the curve |z i| = |z 2|.

Example 16.7 Sketch the region |z (2 i)| 3.

I.16.6

Properties of complex arithmetic

z1 + z2 2 Cz1

z2

2 C (closure)

0 = 0+ 0i1 = 1+ 0i

) z +0 = zz 1 = z (zero and unit)

z1 + z2 = z2 + z1z1

z2

= z2

z1

(commutative)

(z1 + z2) + z3 = z1 + (z2 + z3)(z

1

z2

)z3

= z1

(z2

z3

)

(associative)

z1

(z2

+ z3

) = z1

z2

+ z1

z3

(distributive)

z1

z2

= 0) z1

= 0 or z2

= 0

I.16.7

Properties of Complex Numbers

Let w and z be complex numbers.

|z| = |z|

|zw| = |z| |w|

z = z

z + w = z + w

zw = z w

zz = |z|2

I.16.8

Polar form

A complex number z = x + iy may be represented by polarcoordinates, (r, ) where x = r cos and y = r sin .

The polar form of a complex number z is

z = r(cos + i sin )

where r = |z| =q

x2 + y2 and tan =y

x.

The angle is the argument of z, or arg z. It is not unique! Wesometimes choose such that < .

The (real) exponential function ex has the properties

e0 = 1, exey = ex+y,d

dxekx = kekx

I.16.9

Properties of cos + i sin :

If = 0, cos 0 + i sin 0 =

(cos + i sin )(cos+ i sin)

dd(cos + i sin )

The complex exponential is

ei := cos + i sin

Thus we write polar form as

z = r(cos + i sin ) = rei

Example 16.8 Write z =p3 i and w = 1 +

p3i in polar

form.

Products and division in polar form

Let z = r1

ei and w = r2

ei. Then

zw = r1

eir2

ei = r1

r2

ei+i = r1

r2

ei(+)

z

w=

r1

ei

r2

ei=

r1

r2

eii =r1

r2

ei()

I.16.10

Find

(p3 i)(1 +

p3i)

p3 i

1+

p3i

(1 + i)2(1+ ip3)

6

You can attempt questions 1 to 12 in problem sheet 4 ofyour workbook.

I.16.11

mast 10008 week 4

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