# Proof A level

```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.
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.
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 &lt; (n + 1)2 for all positive integers n &lt; 6.
The only positive integers less than 6 are
1, 2, 3, 4 and 5.
Check that 2n &lt; (n + 1)2 for each case.
n = 1:
2n = 21 = 2 and (n + 1)2 = (1 + 1)2 = 4;
2&lt;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&lt;9
n = 3:
2n = 23 = 8 and (n + 1)2 = (3 + 1)2 = 16;
8 &lt; 16
n = 4:
2n = 24 = 16 and (n + 1)2 = (4 + 1)2 = 25;
16 &lt; 25
n = 5:
2n = 25 = 32 and (n + 1)2 = (5 + 1)2 = 36;
32 &lt; 36
2n &lt; (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 &times; 7 + 7 = 11 &times; 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 &gt; 0 and n &lt; 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).
Prove that there are an infinite amount of prime numbers.
Assume that there are a finite amount of
primes:
p1 , p2 , p3 , ... , pN
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,
The assumption must be incorrect.
So there are infinitely many prime numbers.
Remember to explain how the
Example 6
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 &times; 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.
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.
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 &gt; 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,
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 &gt; 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 &times; 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.
```