WJEC Maths for A2 – Pure

1.3 Proof of the irrationality of √2 11 Answer 2 4 x + 9 x < 12 Multiplying both sides by x gives 4 x 2 + 9 < 12 x Subtracting 12 x from both sides gives 4 x 2 − 12 x + 9 < 0 Hence ȍ2 x − 3Ȏȍ2 x − 3Ȏ < 0 So ȍ2 x − 3Ȏ 2 < 0 For ȍ2 x − 3Ȏ 2 to be negative (i.e. < 0), 2 x − 3 would have to be an imaginary number. This contradicts that 2 x − 3 and x are real. Hence 4 x + 9 x ≥ 12 1.3 Proof of the irrationality of √2 The Greeks discovered that the diagonal of a square of side 1 unit has a diagonal whose length is not rational ( i.e. it cannot be expressed as the fraction a b ) . They used Pythagoras’ theorem to ϐind that the length of the diagonal was √ 2 so this meant √ 2 was irrational. Using proof by contradiction to prove the irrationality of √2 We start off by supposing that √ 2 is rational and can therefore be expressed as a fraction in the form a b . So, we have √ 2 = a b where a and b do not have any common factors. Squaring both sides, we obtain 2 = a 2 b 2 Rearranging this equation gives 2 b 2 = a 2 and this means a 2 is even as it has a factor of 2. For a 2 to be even, a has to be even because if you square an even number you always obtain an even number and if you square an odd number you always get an odd number. If a is even then a 2 must be divisible by 4. This means that b 2 and hence b must be even. Now if a and b are both even, this is a contradiction to the assumption that a and b do not have any common factors. Hence √ 2 cannot be rational so it is therefore irrational. Examples 1 Complete the following proof by contradiction to show that √ 3 is irrational. Assume that √ 3 is rational. Then √ 3 may be written in the form a b where a and b are integers having no common factors. ∴ a 2 = 3 b 2 ∴ a 2 has a factor 3. ∴ a 2 has a factor 3 so that a = 3 k where k is an integer. Note we can multiply both sides by x without worrying about reversal of the inequality as x is a positive number. The left-hand side of the inequality is a quadratic function and can therefore be factorised. For any real and positive value of x , (2 x − 3) 2 will always be greater than or equal to zero. It can never be negative if 2 x − 3 is real. It is not easy to think up the proof for the irrationality of surds such as √ 3 or the inϐinity of primes. It is therefore worth spending time memorising them. BOOST G ra d e ⇪⇪⇪⇪ Note that the symbol ∴ means therefore.