WJEC Mathematics for AS Level - Pure

1.5 Proof by deduction 13 Answer 4 If a = 1, b = −2 and c = 1, then the quadratic equation becomes x 2 − 2 x + 1 = 0 Factorising this equation, we obtain ( x − 1)( x − 1) = 0 Solving, gives x = 1, which is a positive root. Hence the root is positive, and the counter-example proves that the statement is incorrect. 1.5 Proof by deduction Proof by exhaustion cannot be used if there is a large number or even inϐinite number of values to test, so another type of proof is needed. Proof by deduction draws a conclusion from something known or assumed. When we solve a simple equation such as 10 x = 20 we ϐirstly assume that 10 x = 20 and that it is possible to divide both sides of an equation by a non-zero number (i.e. 10 in this case) to arrive at the deduction that x = 2. Proof by deduction uses algebra to decide whether a particular statement is true or false. Examples 1 Prove that for any integer value n , if n is odd then the value of n 2 will also be odd. Answer 1 If p is an integer then 2 p will always be even and 2 p + 1 will always be odd. Now we can set n = 2 p + 1 so n 2 = (2 p + 1) 2 = 4 p 2 + 4 p + 1 As 4 p 2 + 4 p will always be even (as it can be divided exactly by 2) adding one to it will make it odd. Hence 4 p 2 + 4 p + 1 will be odd. So if n is odd then the value of n 2 will also be odd. Hence the statement has been proved. 2 Use proof by deduction to prove the statement ‘if a and b are even integers, the sum and difference of a and b are divisible by 2’. Answer 2 For integers less than 6, this was proved by exhaustion earlier, but the general case can be proved by deduction in the following way: As a and b are even integers they both have 2 as factors and can be written in the following ways: a = 2 p and b = 2 q where p and q are integers. Hence a + b = 2 p + 2 q = 2( p + q ) which has 2 as a factor and so must be even. And a − b = 2 p − 2 q = 2( p − q ) which has 2 as a factor and so must be even. So the statement has been proved. There are lots of numbers which could be used and this counter-example is only one of many possible answers. Note that the method of exhaustion cannot be used here as there are inϐinitely many different odd integers for n .