Page 278 - HOW TO PROVE IT: A Structured Approach, Second Edition

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264 Mathematical Induction
3
3
Base case: If n = 0, then n − n = 0 = 3 · 0, so 3 | (n − n).
3
Induction step: Let n be an arbitrary natural number and suppose 3 | (n − n).
3
Then we can choose an integer k such that 3k = n − n. Thus,
3
2
3
(n + 1) − (n + 1) = n + 3n + 3n + 1 − n − 1
3 2
= (n − n) + 3n + 3n
2
= 3k + 3n + 3n
2
= 3(k + n + n).
3
Therefore 3 | ((n + 1) − (n + 1)), as required.
Once you understand why mathematical induction works, you should be able
to understand proofs that involve small variations on the method of induction.
The next example illustrates such a variation. In this example we’ll try to figure
n
2
out which is larger, n or 2 . Let’s try out a few values of n:
n n 2 2 n Which is larger?

0 0 1 2 n
1 1 2 2 n
2 4 4 tie
3 9 8 n 2
4 16 16 tie
5 25 32 2 n
6 36 64 2 n

n
It’s a close race at first, but starting with n = 5, it looks like 2 is taking a de-
2
cisive lead over n . Can we prove that it will stay ahead for larger values
of n?

2
n
Example 6.1.3. Prove that ∀n ≥ 5(2 > n ).
Scratch work
2
n
We are only interested in proving the inequality 2 > n for n ≥ 5. Thus, it
would make no sense to use n = 0 in the base case of our induction proof. We’ll
take n = 5 as the base case for our induction rather than n = 0. Once we’ve
checked that the inequality holds when n = 5, the induction step will show that
the inequality must continue to hold if we repeatedly add 1 to n. Thus, it must
also hold for n = 6, 7, 8,.... In other words, we’ll be able to conclude that the
inequality holds for all n ≥ 5.