A-level Maths revision · Pure, year 1

Proof and counterexamples

The idea

Proof by deduction uses general algebra: an even number is 2n2n, an odd number 2n+12n + 1. Proof by exhaustion checks every case of a finite list. One counterexample is enough to disprove a statement.

  1. Write the numbers in general form, or list every case.
  2. Simplify into a form that shows the claim, such as 3(… )3(\dots) for a multiple of 33.
  3. Finish with a sentence that states what has been shown.

Watch out Checking a few examples does not prove a statement about all numbers.

See it worked

Worked example

Prove that the sum of any three consecutive integers is a multiple of 33.

  1. Let the integers be nn, n+1n + 1 and n+2n + 2.
  2. Their sum is 3n+3=3(n+1)3n + 3 = 3(n + 1).
  3. n+1n + 1 is an integer, so the sum is a multiple of 33.

Answer The sum is 3(n+1)3(n + 1), a multiple of 33

Your turn

Question 1Calculator allowed3 marks

Find the smallest positive integer nn that shows the claim 'n2+n+41n^2 + n + 41 is prime for every positive integer nn' is false.

Question 2Non-calculator4 marks

Prove by exhaustion that a square number leaves remainder 00 or 11 when divided by 44.

Write your answer, then show the working
  1. Every integer is 2k2k or 2k+12k + 1.
  2. (2k)2=4k2(2k)^2 = 4k^2: remainder 00.
  3. (2k+1)2=4k2+4k+1=4(k2+k)+1(2k + 1)^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1: remainder 11.
  4. Both cases are covered, so the claim holds.

Answer Every integer is even or odd: (2k)2=4k2(2k)^2 = 4k^2 and (2k+1)2=4(k2+k)+1(2k + 1)^2 = 4(k^2 + k) + 1

Question 3Non-calculator3 marks

Find the smallest positive integer nn for which 2n+12^n + 1 is not prime.

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