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9231 · 1.7

Proof by induction flashcards

Revision flashcards for Cambridge 9231 Proof by induction (syllabus 1.7). Flip, recall, then mark a real past-paper question.

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    What is the principle of mathematical induction?

    A method to prove a statement $P(n)$ is true for all integers $n \ge n_0$ by showing $P(n_0)$ is true (base case) and that for any $k \ge n_0$, if $P(k)$ is true, then $P(k+1)$ is also true (inductive step).

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    What is the 'Base Case' in a proof by induction?

    The first step, where you prove the statement is true for the smallest integer value specified in the question, e.g., $n=1$.

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    What is the 'Inductive Hypothesis'?

    The assumption that the statement is true for an arbitrary integer $k$, i.e., assuming $P(k)$ is true. This is a crucial step before proving for $k+1$.

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    What is the goal of the 'Inductive Step'?

    To show that if the statement is true for $n=k$, it must also be true for $n=k+1$. You must use the inductive hypothesis to achieve this.

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    Why is the concluding statement so important?

    It formally links the base case and the inductive step to the principle of mathematical induction, completing the logical argument. A proof without it is incomplete and will lose marks.

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    Common mistake: Proving $P(k+1)$ without using the assumption $P(k)$. Why is this wrong?

    The logic of induction rests on the 'if $P(k)$ then $P(k+1)$' link. If you don't use the assumption $P(k)$, you haven't shown this connection, and your proof is invalid.

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    How do you start the inductive step for a summation, e.g., $\sum_{r=1}^{n} f(r)$?

    Write out the sum for $n=k+1$: $\sum_{r=1}^{k+1} f(r) = (\sum_{r=1}^{k} f(r)) + f(k+1)$. Then substitute the assumed formula for the sum to $k$.

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    How do you approach a divisibility proof, e.g., 'prove $f(n)$ is divisible by $d$'?

    Assume $f(k) = dm$ for some integer $m$. Then analyse $f(k+1)$ and manipulate it to show it is also a multiple of $d$. A common technique is to consider $f(k+1) - f(k)$.

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    What if the base case is not $n=1$?

    The base case is the first integer for which the statement is claimed to be true. This could be $n=0$, $n=2$, or any other integer $n_0$. The conclusion would then be 'true for all integers $n \ge n_0$'.

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    How do you prove a statement involving matrices, like $\mathbf{A}^n = \mathbf{B}_n$, by induction?

    Base case: Show $\mathbf{A}^1 = \mathbf{B}_1$. Assumption: Assume $\mathbf{A}^k = \mathbf{B}_k$. Inductive step: Prove $\mathbf{A}^{k+1} = \mathbf{B}_{k+1}$ by writing $\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A}$ and substituting the assumption: $\mathbf{B}_k \mathbf{A}$. Then show this product equals $\mathbf{B}_{k+1}$.