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BRIDGE OF DON ACADEMY – DEPARTMENT OF MATHEMATICS
ADVANCED HIGHER: UNIT 2 – NUMBER THEORY AND METHODS OF PROOF
Number Theory and Methods of Proof
We are going to look at some of the methods used to determine the absolute truths of mathematics.
We will start with a proposition or an implication and determine whether it is true or false.
A statement, P, can be made and we can claim that it implies that another statement, Q, is true.
We then either have to prove whether this proposition is true or false.
Throughout this topic we will have to very careful about defining what we are talking about. There
are different types of numbers. Some of the more notable groups are:







Natural numbers, denoted by N  {1, 2, 3, 4, 5, 6 .....}
Whole numbers, denoted by W  {0, 1, 2, 3, 4, 5 .....}
Integers, denoted by Z  {.....  3,  2,  1, 0, 1, 2, 3 .....}
Rational numbers, denoted by Q  { a : a  Z , b  N}
b
Irrational numbers, which will denoted as ~Q  {a : a  Q} [~Q is pronounced “not Q”]
Real numbers, denoted by R  {x :    x  }
Prime numbers, which we will denote as P  { p : p has exactly 2 factors}
The Fundamental Theorem of Arithmetic (first proven by Carl Friedrich Gauss) states that any
integer greater than 1 can be expressed as the product of prime numbers in only one way.
For example,
5 = 5 as it is a prime
35  5  7 , a product of two primes
1785  3 5  7 17 , a product of four primes.
This gives us the following rules:
1. p divides a means that p divides a without a remainder (definition).
2. a prime number p is an integer greater then 1, which has only 1 and p as its factors
(definition).
3. if a prime number p divides ab, where a and b are integers, then either p divides a or p
divides b. Also, if a prime number p divides ab, where a and b are primes, then either p = a
or p = b.
4. Every integer greater than 1 can be expressed as the product of prime numbers.
Axioms are statements that are obvious or accepted as being true by definition.
For example 1 + 1 = 2 as this is the definition of what we mean by the number two.
Now do this:
Read pages 15, 16 and 17 of the textbook.
Page 1 of 11
BRIDGE OF DON ACADEMY – DEPARTMENT OF MATHEMATICS
ADVANCED HIGHER: UNIT 2 – NUMBER THEORY AND METHODS OF PROOF
Perhaps the first theorem that we came across was that of Pythagoras.
He proposed that
statement A: “If a triangle is right-angled”
implied that
statement B: “the square on the hypotenuse is equal to the sum of the squares on the other two
sides”.
A B
He proved this was true (or at least somebody did and he’s been given the credit ever since).
However, we also know that
B A
i.e. the converse of the Theorem of Pythagoras is true: that “if the square on the hypotenuse is equal
to the sum of the squares on the other two sides then the triangle is right-angled”.
This means we have an equivalence
A B
i.e. a two-way implication.
This is because statement B is true “if and only if” statement B is true.
Is this always the case?
In one of my classes, all the boys have blue eyes.
Statement C: If I pick a “boy from my class”
implies
statement D: “the pupil has blue eyes”.
CD
TRUE
Is the converse true? Does D  C ?
If “the pupil has blue eyes” does this mean that they are a “boy from my class”?
D C
For x  R , decide whether the following implications are true or false:
1.
If x  5 then x 2  25 .
2.
If x > 0 then x 2  0 .
3.
If u . v  0 and u  0 and v  0 , then u is perpendicular to v .
4.
If g ( x)  x 2 and f ( x)  x  1 then f  g( x)   x2 1 .
For the above implications where A  B is true, identify which of the converse implications,
B  A are true and, therefore A  B .
Now try:
Page 5, Exercise 1B – questions 1a, 1b, 2, 4
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BRIDGE OF DON ACADEMY – DEPARTMENT OF MATHEMATICS
ADVANCED HIGHER: UNIT 2 – NUMBER THEORY AND METHODS OF PROOF
Direct Proof
This is usually the most obvious method when you see it and the easiest to follow the logic of.
However, it often requires a moment of inspiration, a “Eureka!” moment, to think of the necessary
steps.
All the proofs that you have seen so far, throughout S1-S5 and so far in S6), have been direct
proofs: Theorem of Pythagoras, Sum of n terms of arithmetic and geometric sequences,
sin 2   cos 2   1 etc.
Examples:
1.
The sum to n terms of a geometric series where a  3 and r  3 :
Sn  3  9  27  81  ............  3n
3Sn 
9  27  81  ............  3n  3n1
2S n  3S n  S n
 3n 1  3


 Sn  1 3n 1  3
2
 3 3n  1
2


a(1  r n )
, but we only have this because
1 r
somebody did the above work for a general case and gave us this formula.
We would use our ‘magical’ formula, Sn 
2.
c
a

Pythagoras gives us
c2  a 2  b2 .
We know that
sin    a and cos    b
c
c
tan    a .
b
and that
b
Therefore,
   
sin 2   cos 2   a
c
2
 b
c
2
2
 a2  b 2
c
c
2
2
 a 2 b
c
2
 c2
c
1
2
and
sin    a  b
cos   c c
c
 a 
c b
 a
b
 tan  
Page 3 of 11
BRIDGE OF DON ACADEMY – DEPARTMENT OF MATHEMATICS
ADVANCED HIGHER: UNIT 2 – NUMBER THEORY AND METHODS OF PROOF
3.
Prove that a  b  ab .
2
4.
(a)
By considering the the Fundamental Theorem of Arithmetic, find the prime number,
p, such that 19p + 81 is a perfect square.
(b)
Find two non-prime whole numbers, x, which make 19x + 81 a perfect square.
Now try:
Page 10, Exercise 2A – questions 3, 5
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BRIDGE OF DON ACADEMY – DEPARTMENT OF MATHEMATICS
ADVANCED HIGHER: UNIT 2 – NUMBER THEORY AND METHODS OF PROOF
It can be easy to disprove a conjecture: all we need is a “counter example”. In mathematics there
are only absolute truths so if we can give an example for which a conjecture does not hold true then
the conjecture is false.
Examples:
5.
6.
If a  b then a 2  b 2 , a, b  R .
Let a  2 and b  5
 a b.
However, a 2  4 and b2  25
 a 2  b2 .
For all natural numbers m, if m 2 is divisible by 4 then m is divisible by 4.
Page 5 of 11
BRIDGE OF DON ACADEMY – DEPARTMENT OF MATHEMATICS
ADVANCED HIGHER: UNIT 2 – NUMBER THEORY AND METHODS OF PROOF
Proof by Contradiction
Often, proving a proposition, P  Q , is pretty difficult, if not impossible, to prove using the
method of direct proof.
We can often prove that the contrapositive of P  Q i.e. ~ P  ~ Q (pronounced “not P implies
not Q). This is called proof by contradiction.
We have to be careful to correctly identify the contrapositive of a statement. For example, if we
state that “all boys in a class have blue eyes” then the contrapositive is “some boys in the class do
not have blue eyes”. Clearly only one of these statements can be true and if one is false then the
other must be true. Either P  Q or ~ P  Q .
Examples:
7.
Prove that 3 is irrational.
Assume that 3 is rational
 3  a where there exists a, b  N such that a and
b
b have no common factors (i.e. a cannot be
b
simplified).
3a
b
 b 3 a

 3b 2  a 2
 a 2 is a multiple of 3
(this could only be, if and only if, a 2  (3  a1  a2  .....  an )2
where 3, a1, a2..... an are the prime factors of a (see the
Fundamental Theorem of Arithmetic).
 a 2 is a multiple of 3  a is a multiple of 3
Let a  3k where k  N
 3b 2  (3k ) 2
 3b 2  9k 2
 b 2  3k 2
 b 2 is a multiple of 3  b is a multiple of 3
Therefore a and b have a common factor of 3 which is a contradiction.
Hence 3 is not rational 
3 is irrational .
Page 6 of 11
BRIDGE OF DON ACADEMY – DEPARTMENT OF MATHEMATICS
ADVANCED HIGHER: UNIT 2 – NUMBER THEORY AND METHODS OF PROOF
8.
Prove by contradiction that if n is odd then n 2  1 is even, n  N .
9.
Prove by contradiction that if n 3 is odd then n is odd.
Page 7 of 11
BRIDGE OF DON ACADEMY – DEPARTMENT OF MATHEMATICS
10.
ADVANCED HIGHER: UNIT 2 – NUMBER THEORY AND METHODS OF PROOF
Prove by contradiction that if ( x  y ) is irrational where x, y  R then either “x or y” or “x
and y” are irrational.
Now try:
Page 14, Exercise 3A – questions 1, 2, 13
Page 17, Exercise 3B – question 2
Page 8 of 11
BRIDGE OF DON ACADEMY – DEPARTMENT OF MATHEMATICS
ADVANCED HIGHER: UNIT 2 – NUMBER THEORY AND METHODS OF PROOF
Proof by Induction
This is a very powerful and logical method of proof. It is done in four stages:
Step 1 – Show that it is true for a particular value (usually the initial value)
Step 2 – Assume that is true for a “general value”
Step 3 – Show that if it is true for a “general value” then it is also true for a “(general value) + 1”
Step 4 – Draw your conclusion (this is the clever bit – see below).
This method is best illustrated by example:
11.
Prove by induction that 23n  1 is divisible by 7, n  N .
Step 1:
For n  1 ,
231  1  8  1  7 , which is divisible by 7
 true for n  1
Step 2:
Assume true for n  k
 23k  1  7q, where q  N
 23k  7q  1
Step 3:
For n  k 1 ,

23(k 1)  1  23k  3  1
 23k  23  1
 (7q  1)  23  1
 8(7q  1)  1
 56q  8  1
 56q  7
 7(8q  1)
Step 4:
Hence, if true for n  k then also true for n  k 1 and since
true for n  1 , it must be true for all values of n  N .
Page 9 of 11
BRIDGE OF DON ACADEMY – DEPARTMENT OF MATHEMATICS
ADVANCED HIGHER: UNIT 2 – NUMBER THEORY AND METHODS OF PROOF
Examples:
12.
Prove by induction that 32n  1 is divisible by 8, n  N .
Page 10 of 11
BRIDGE OF DON ACADEMY – DEPARTMENT OF MATHEMATICS
ADVANCED HIGHER: UNIT 2 – NUMBER THEORY AND METHODS OF PROOF
n
13.
Prove by induction that
 2r  n2 (2n  2), n  N .
r 1
Now try:
Page 20, Exercise 4 – questions 7a, 9(b, c), 11, 12
Page 24, Exercise 5 – questions 1, 3
Page 25, Review – questions 1 to 6
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