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9231 · 1.7

Proof by induction — common mistakes

Common exam mistakes on 9231 Proof by induction. Learn what loses marks, then practise the topic with Examiner’s Ink.

Exam tip 1

In the inductive step, you MUST explicitly use your assumption for n=kn=k. A good habit is to underline or highlight where you substitute it in your own working to ensure you haven't missed it. Also, never skip the final concluding sentence; it's usually worth a specific mark and ties the whole proof together.

What happens if I can't prove the inductive step?

First, double-check your algebra. It's the most common source of error. Second, ensure you have correctly used the inductive hypothesis (P(k)P(k)). A proof of P(k+1)P(k+1) that doesn't rely on P(k)P(k) is not a proof by induction. If you're still stuck, reconsider your approach. For divisibility, is f(k+1)f(k)f(k+1)-f(k) helpful, or maybe f(k+1)cf(k)f(k+1) - c \cdot f(k) for some constant cc?

Can I work backwards from the $n=k+1$ case?

You must be very careful. It is not valid to assume the n=k+1n=k+1 result is true and work backwards to the n=kn=k assumption. However, you can use the target expression for n=k+1n=k+1 as a 'goal' on your scrap paper to guide the manipulation of your starting expression. Your final written proof must flow logically from the LHS to the RHS of the P(k+1)P(k+1) statement.

Do I have to write the full concluding sentence every time?

Yes, absolutely. Examiners look for it. A statement like 'Since the statement is true for n=1n=1 and if it's true for n=kn=k it is also true for n=k+1n=k+1, it is true for all positive integers nn by the principle of mathematical induction' is essential for full marks. It shows you understand the logical principle you are using.

Is proof by induction only for positive integers?

No. The statement might be for all integers nn0n \ge n_0, where n0n_0 could be any integer, such as 0, 2, or even -5. Your base case would then be to prove the statement for n=n0n=n_0, and your assumption would be for kn0k \ge n_0.