WJEC Maths for A2 – Pure

1 Proof 12 Answer 1 ∴ ȍ3 k Ȏ 2 = 3 b 2 ∴ 9 k 2 = 3 b 2 ∴ 3 k 2 = b 2 ∴ b 2 has a factor 3 ∴ b has a factor 3 a and b have a common factor (i.e. 3) This is a contradiction and means that the initial assumption is wrong. Hence √ 3 is irrational. Note that in Section 1.2 Example 1 we used the fact that if a 2 has a factor 3, then a has a factor 3. This result was established in Example 1. In future work you may use (without proof) that if a 2 has a factor p , p being a prime , then a has a factor p . For example, if p 2 is divisible by 7 then p is divisible by 7. 2 Complete the following proof by contradiction to show that √ 7 is irrational. Assuming that √ 7 is rational. Then √ 7 may be written in the form a b where a , b are integers having no common factors. ∴ a b = √ 7 ∴ a = √ 7 b ∴ a 2 = 7 b 2 , has a factor 7. ∴ a , has a factor 7 so that a = 7 k , where k is an integer. Answer 2 Now a 2 = 7 b 2 ∴ ȍ7 k Ȏ 2 = 7 b 2 ∴ 49 k 2 = 7 b 2 ∴ 7 k 2 = b 2 , so b 2 has a factor 7. ∴ b has a factor 7 a and b have 7 as a common factor This is a contradiction and means that the initial assumption is wrong. Hence √ 7 is irrational. Here we substitute a = 3 k , into a 2 = 3 b 2 . a and b should have no common factors. This is because a and b were assumed to have no common factors and we have proved the existence of a common factor. Make sure that you mention the word contradiction and say exactly what the contradiction is. BOOST G ra d e ⇪⇪⇪⇪