Mathematics · Pearson Edexcel 9MA0 · Topic 1

Proof

Proving statements are always true (by deduction, exhaustion or contradiction) and disproving them with a counter-example.

  • 1 sub-topic
  • Taught in Year 13
  • 24 practice questions
  • 1 animated lesson
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What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 1.1Proof by deduction, exhaustion, counter-example and contradictionY13

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Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 24 questions in this topic, free, no account needed.

Proof

Easy

Proof by deduction · Year 12 · 3 marks

  1. Prove that n2−nn^2 - n is even for every integer nn.
    (3)
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  1. Factorise to see the structure of the expression.n2−n=n(n−1)n^2 - n = n(n - 1)
  2. Of any two numbers next to each other, one is a multiple of 2.nn and n−1n - 1 are consecutive integers, so one of them is even.
  3. Finish with a clear statement of what has been proved.An even number times any integer is even, so n2−nn^2 - n is even.

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Proof

Medium

Proof by exhaustion · Year 12 · 4 marks

  1. Prove that the square of any integer is either a multiple of 3 or one more than a multiple of 3.
    (4)
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  1. Split into every possible case: the remainder on dividing by 3 is 0, 1 or 2.Every integer is 3k3k, 3k+13k + 1 or 3k+23k + 2
  2. A multiple of 3.(3k)2=3(3k2)(3k)^2 = 3(3k^2)
  3. One more than a multiple of 3.(3k+1)2=3(3k2+2k)+1(3k + 1)^2 = 3(3k^2 + 2k) + 1
  4. Also one more than a multiple of 3, so every case is covered.(3k+2)2(3k + 2)^2=9k2+12k+4{}= 9k^2 + 12k + 4=3(3k2+4k+1)+1{}= 3(3k^2 + 4k + 1) + 1

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Proof

Not needed for Year 12Hard

Infinitely many primes · Year 13 · 5 marks

  1. Prove by contradiction that there are infinitely many prime numbers.
    (5)
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  1. Assume the opposite.Assume there are finitely many primes, p1,p2,…,pnp_1, p_2, \dots, p_n
  2. Build a number from every prime on the list.Let N=p1p2⋯pn+1N = p_1 p_2 \cdots p_n + 1
  3. So none of the listed primes divides NN.Dividing NN by any pip_i leaves remainder 1
  4. Every integer greater than 1 has a prime factor.So NN is prime, or has a prime factor not on the list
  5. Either way there is a prime missing from the list: a contradiction, so there are infinitely many primes

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