Page 277 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proof by Mathematical Induction 263
As we saw in the last example, the hardest part of a proof by mathematical
induction is usually the induction step, in which you must prove the statement
∀n ∈ N(P(n) → P(n + 1)). It is usually best to do this by letting n be an
arbitrary natural number, assuming P(n) is true, and then proving that P(n + 1)
is true. The assumption that P(n) is true is sometimes called the inductive
hypothesis, and the key to the proof is usually to work out some relationship
between the inductive hypothesis P(n) and the goal P(n + 1).
Here’s another example of a proof by mathematical induction.

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Example 6.1.2. Prove that ∀n ∈ N(3 | (n − n)).
Scratch work

As usual, the base case is easy to check. The details are given in the following
proof. For the induction step, we let n be an arbitrary natural number and assume
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that 3 | (n − n), and we must prove that 3 | ((n + 1) − (n + 1)). Filling in the
definition of divides, we can sum up our situation as follows:
Givens Goal
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n ∈ N ∃ j ∈ Z(3 j = (n + 1) − (n + 1))
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∃k ∈ Z(3k = n − n)
The second given is the inductive hypothesis, and we need to figure out how it
can be used to establish the goal.
According to our techniques for dealing with existential quantifiers in proofs,
the best thing to do first is to use the second given and let k stand for a par-
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ticular integer such that 3k = n − n. To complete the proof we’ll need to
find an integer j (which will probably be related to k in some way) such that
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3 j = (n + 1) − (n + 1). We expand the right side of this equation, looking
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for some way to relate it to the given equation 3k = n − n:
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2
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(n + 1) − (n + 1) = n + 3n + 3n + 1 − n − 1
2
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= (n − n) + 3n + 3n
2
= 3k + 3n + 3n
2
= 3(k + n + n).
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Itshouldnowbeclearthatwecancompletetheproofbyletting j = k + n + n.
As in similar earlier proofs, we don’t bother to mention j in the proof.
Solution

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Theorem. For every natural number n, 3 | (n − n).
Proof. We use mathematical induction.