1.1 Proof
You should be able to use four different types of proof.
Proof by deduction
o A proof by deduction (or by reasoning) always has a few logical steps.
Proof by exhaustion
o In a proof by exhaustion you check all possible cases separately.
Disproof by counter-example
o When trying to find a counter-example, you may need to try several possibilities.
Proof by contradiction
o In a proof by contradiction you start with an assumption and then make some
deductions until you get to something that is wrong (a contradiction of the
assumption).
You can use the symbols ,  and  in proof questions.

means implies.
For example, x + 2 = 5  x = 3.

means is equivalent to.
2
For example, y  ( x  1)( x  2)  y  x  3x  2.
You can read this as
y  ( x  1)( x  2)  y  x 2  3x  2 and y  x 2  3x  2  y  ( x  1)( x  2).

means is implied by.
This is not often used, but an example is ( x  1)( x  4)  0  x  4 .
Examples
Example 1
Proof by deduction
Use completing the square to prove that n 2  6n  10 is positive for all values of n.
n 2  6n  10  n 2  6n  9  9  10
First complete the square.
= ( n  3) 2  1
(n  3)2  0 for all values of n.
Remember that squares cannot be
negative.
Therefore n 2  6n  10 is positive for all
values of n.
With a proof you need a conclusion.
Example 2
Proof by exhaustion
Prove that 2n < (n + 1)2 for all positive integers n < 6.
The only positive integers less than 6 are
1, 2, 3, 4 and 5.
Check that 2n < (n + 1)2 for each case.
n = 1:
2n = 21 = 2 and (n + 1)2 = (1 + 1)2 = 4;
2<4
It may be helpful to begin by explaining
what you will do to prove the statement.
In a proof by exhaustion, work through
all the possibilities in a systematic way
to make sure you don’t miss any.
n = 2:
2n = 22 = 4 and (n + 1)2 = (2 + 1)2 = 9;
4<9
n = 3:
2n = 23 = 8 and (n + 1)2 = (3 + 1)2 = 16;
8 < 16
n = 4:
2n = 24 = 16 and (n + 1)2 = (4 + 1)2 = 25;
16 < 25
n = 5:
2n = 25 = 32 and (n + 1)2 = (5 + 1)2 = 36;
32 < 36
2n < (n + 1)2 for all five possible cases
and this completes the proof.
Example 3
Always end your proof with a
conclusion.
Wemin states the following.
A number of the form n2 + 3n + 7 is a prime number for any
integer greater than 0.
Prove that Wemin is wrong by finding a counter-example.
Try n = 1
n 2  3n  7 = 12  3  1  7  11
This is prime.
You may have to try several values
before you find a counter-example.
Try n = 2
n 2  3n  7 = 22  3  2  7  17
This is prime.
Try n = 3
n 2  3n  7 = 32  3  3  7  25
This is not prime, as 25 can be divided by 5.
This proves that Wemin is wrong.
n = 3 is a counter-example.
Always end your proof with a
conclusion.
Note: You may have spotted that n = 7 is another counter-example.
When n = 7, each of the three terms of n2 + 3n + 7 is a multiple of 7 and so the
sum is also a multiple of 7. (72 + 3 × 7 + 7 = 11 × 7)
Example 4
Find a counter-example to disprove the statement
1 1
m  n   for integers m and n (m ≠ 0 and n ≠ 0).
m n
For m = 2 and n = –1,
1
1 1
and
 1.

n
m 2
So 2  1, but
1
 1.
2
You can pick any m > 0 and n < 0.
Always finish with a statement. You
could add ‘This proves that the
statement is not correct.’
You need to be familiar with two proofs by contradiction.
 Proof that there are infinitely many primes.

Proof of the irrationality of
Example 5
2 (an irrational number cannot be written as a fraction).
Proof by contradiction
Prove that there are an infinite amount of prime numbers.
Assume that there are a finite amount of
primes:
p1 , p2 , p3 , ... , pN
Start by stating your initial
assumption.
Multiply all these prime numbers and add 1.
P  p1  p2  p3  ...  pN  1
P is greater than any of the prime numbers
and cannot be prime.
This means one of the primes, pi , must
divide into P.
P cannot be prime because it would
mean you had found an extra
prime!
So pi must divide into
p1  p2  p3  ...  pN  1
pi divides into ( p1  p2  p3  ...  pN ) as
pi is in the product.
So pi must therefore also divide into 1,
which is a contradiction.
The assumption must be incorrect.
So there are infinitely many prime numbers.
Remember to explain how the
contradiction proves the statement.
Example 6
Proof by contradiction
Prove that
2 is irrational.
Assume that
fraction.
2 can be written as a
p
where the fraction is in its
q
simplest form, so p and q only have 1 as
a common factor.
So
2
2
p
p2
2 2
q
q
 2q 2  p 2
This is the same as assuming that
is a rational number.
If 2 is rational then it can always be
written as a fraction in its simplest
form.
Square both sides.
[1]
Multiply both sides by q2.
So p2 is even and therefore p is even.
Even × even = even
 p  2m  p  (2m)  4m
If p is even it can be written as 2m
where m is any integer.
2
2
2
2
Then, from [1]
2q 2  4m2  q 2  2m2
Divide both sides by 2.
So q2 is even and therefore q is even.
But this is a contradiction.
Both p and q are even, but the
assumption is that p and q only have 1
as a common factor. So the assumption
is incorrect and therefore 2 is
irrational.
You should explain what the
contradiction is and how it proves that
2 is irrational.
Note: The proof that 3 is irrational is similar. Replace the 2 in the proof by 3 and
replace ‘so p2 is even and therefore p is even’ by ‘so p2 can be divided by 3 and
therefore p can be divided by 3’.
Exercise
1.
Prove that x 2  8 x  21 is positive for all values of x.
2.
Prove that 3 x 2  6 x  5 is positive for all values of x.
3.
(a) Prove that n 2  6n has a factor 4 for all even numbers n.
(b) Prove by counter-example that n 2  6n does not have a factor 4 for all integers n.
4.
Find a counter-example to prove that n 2  n  11 is not prime for all positive integer values of n.
5.
3
3
2
2
Prove that x  y  ( x  y)( x  xy  y ) .
6.
Prove by exhaustion that there are exactly two natural numbers less than 101 that are both a
square and a cube integer. [Note: the natural numbers are the set {1, 2, 3, … }.]
7.
Find a counter-example to disprove the following statement.
y2  1  y  1
8.
Disprove the following statement.
m2  16  m  4  0
9.
Prove that n 2  3n  13 is not prime for all positive integer values of n.
10. Insert one of ,  or  into the following statement about integer n. Explain your answer.
n2 + 1 is even ……. n is odd
11. Insert one of ,  or  into the following statement.
m2 is an integer ……. m is an integer
12. Find a counter-example to disprove the following statement.
p and q are both irrational  pq is irrational
13. Find a counter-example to disprove the following statement.
p and q are both irrational  p  q is irrational
14. Prove that
3 is irrational.
15. Prove that
7 is irrational.
Answers
1.
x2  8x  21  x2  8x  16  16  21  ( x  4)2  5
(x + 4)2 ≥ 0 for all values of x, and therefore (x +4)2 + 5 > 0.
Hence x 2  8 x  21 is positive for all values of x.
2.
3x2  6 x  5  3( x  1)2  2 ;
( x  1)2  0, so 3( x  1)2  2  0
Therefore 3x 2  6 x  5  0 for all values of x.
3.
(a) n is even  n  2m where m is an integer.
n2  6n  (2m)2  6(2m)  4m2  12m  4(m2  3m) or 4m(m  3)
So n 2  6n is a multiple of 4.
(b) A counter-example is given by any odd value of n.
For example, for n = 1, n 2  6n  1  6  7 . This is not a multiple of 4.
4.
n = 11 is a counter-example.
n  n  11  11  11  11  143  11  13
2
2
You may have found a different
counter-example, such as n = 10.
Therefore n 2  n  11 is not always prime.
5.
( x  y)( x2  xy  y 2 )  x( x 2  xy  y 2 )  y( x 2  xy  y 2 )
 x3  x 2 y  xy 2  yx 2  xy 2  y 3
With a proof by deduction like this,
you start with the expression on one
side and keep working until you get
to the expression on the other side.
 x3  y 3
So ( x  y)( x 2  xy  y 2 )  x3  y3 , which is what you needed to prove.
6.
7.
The cubes less than 101 are 13 = 1, 23 = 8, 33 = 27 and 43 = 64. (53 = 125 > 101)
Of these four numbers, two are squares, 12 = 1 and 82 = 64.
So there are exactly two cubes between 0 and 101 that are also squares.
2
2
When y  2, y  4 . Therefore y  1  y  1 is not correct.
2
So y  1  y  1 is not correct.
8.
m  4 gives a counter-example. m2  (4)2  16, but m  4  4  4  8  0
This proves that the statement is false.
9.
n = 13 gives a counter-example.
n 2  3n  13  132  3  13  13  221  13  17.
Therefore n 2  3n  13 is not prime for all values of n.
10.  because
n2 + 1 is even  n2 is odd  n is odd
and n is odd  n2 is odd  n2 + 1 is even
11.  because when m is an integer m × m is an integer.
However m2 = 3 gives m =
12. p =
2 and q =
3 , so when m2 is an integer, m is not always an integer.
8 provide a counter-example. Both p and q are irrational.
But pq = 16 = 4, which is a rational number.
13. p =
2 and q = – 2 provide a counter example. Both p and q are irrational.
But p + q =
14. Assume that
So
3
3
2 –
2 = 0, which is a rational number.
3 can be written as a fraction.
p
where the fraction is in its simplest form, so p and q only have 1 as a common factor.
q
p2
 3q 2  p 2
2
q
So p2 can be divided by 3 and therefore p can be divided by 3.
 p  3m  p 2  (3m)2  9m2 , where m is an integer.
 3q 2  9m2  q 2  3m2
So q2 can be divided by 3 and therefore q can be divided by 3.
This is a contradiction. Both p and q can be divided by 3, but the assumption is that p and q only
have 1 as a common factor. So 3 is irrational.
15. Assume that
So
7
7
7 can be written as a fraction.
p
where the fraction is in its simplest form, so p and q only have 1 as a common factor.
q
p2
 7q 2  p 2
2
q
So p2 can be divided by 7 and therefore p can be divided by 7.
 p  7m  p 2  (7m)2  49m2 , where m is an integer.
 7q 2  49m2  q 2  7m2
So q2 can be divided by 7 and therefore q can be divided by 7.
This is a contradiction. Both p and q can be divided by 7, but the assumption is that p and q only
have 1 as a common factor. So 7 is irrational.