Proof by contradiction – Year 13 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. "\(n\) is an odd integer" (or "\(n\) is not even").
   2. Assume \(n^2\) is even and \(n\) is odd, so \(n=2k+1\). Then \(n^2=4k^2+4k+1=2(2k^2+2k)+1\), which is odd. This contradicts \(n^2\) being even [2 marks]. Hence \(n\) is even [1 mark]. \(\blacksquare\)

   The proof must start by assuming the opposite of the conclusion, here that \(n\) is odd, while keeping the hypothesis that \(n^2\) is even. Writing the odd number as \(2k+1\) with \(k\) an integer is essential; a numerical example such as \(3^2=9\) proves nothing. The final mark needs a concluding statement.

2. Answer to Question 2 (4 marks)

   Assume \(\sqrt5=\dfrac ab\) where \(a,b\) are integers with no common factor [1 mark]. Then \(a^2=5b^2\), so \(5\mid a^2\) and therefore \(5\mid a\) [1 mark]. Write \(a=5k\): \(25k^2=5b^2\), so \(b^2=5k^2\), so \(5\mid b\) [1 mark]. Both are divisible by 5, which contradicts the assumption that the fraction was in lowest terms, so \(\sqrt5\) is irrational [1 mark]. \(\blacksquare\)

   The step \(5\mid a^2\Rightarrow5\mid a\) is valid because 5 is prime.

3. Answer to Question 3 (4 marks)

   Assume integers \(m,n\) exist with \(12m+18n=5\) [1 mark]. Then \(6(2m+3n)=5\) [1 mark], so \(2m+3n=\dfrac56\) [1 mark]. But \(2m+3n\) is an integer, and \(\dfrac56\) is not. This is a contradiction, so no such integers exist [1 mark]. \(\blacksquare\)

   Equivalently, the left-hand side is a multiple of 6 for all integers, while 5 is not a multiple of 6. The key idea is to factorise the highest common factor of 12 and 18; dividing by 2 only, to get \(6m+9n=\tfrac52\), works equally well since the left side is then an integer.

4. Answer to Question 4 (4 marks)

   1. \(N=p_1p_2\cdots p_k+1\).
   2. Each \(p_i\) divides \(p_1p_2\cdots p_k\), so dividing \(N\) by \(p_i\) leaves remainder 1. So no \(p_i\) divides \(N\) [2 marks].
   3. \(N\gt 1\) so \(N\) has a prime factor (it is prime or a product of primes). That prime factor is not on the list, but the list was supposed to contain all primes. Contradiction. \(\blacksquare\)

   Note that \(N\) itself need not be prime: \(2\times3\times5\times7\times11\times13+1=30031=59\times509\).

5. Answer to Question 5 (4 marks)

   1. Assume \(xy=\dfrac pq\) with integers \(p,q\), \(q\ne0\), and \(x=\dfrac rs\) with \(r\ne0\). Then \(y=\dfrac{ps}{qr}\), a ratio of integers, so \(y\) is rational. This contradicts \(y\) being irrational. Hence \(xy\) is irrational. \(\blacksquare\)
   2. \(\dfrac{\sqrt2}3=\dfrac13\times\sqrt2\) with \(\dfrac13\) rational and non-zero and \(\sqrt2\) irrational, so it is irrational, by part (a).

   The condition \(x\ne0\) is needed so that \(r\ne0\) and we may divide by \(x\); with \(x=0\), \(xy=0\) is rational.

Total : 20 marks