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I have a general question about strong induction:

Assuming that the base case is 0, if I let my inductive hypothesis be that for all 0 <= k < n some statement is true, and if I prove that that statement is true for n, does this mean that I've proved it for all natural numbers?

Some examples do something a little differently: Assuming that for all 0 <= k <= n is true which includes n, they prove it's true for n + 1..are these two equivalent? Or is only the second method valid? Thanks.

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They are equivalent. Note that both methods show that assuming the statement is true for $0\le k\le 16$, it is also true for $17$. The first method does this when $n=17$, the other when $n=16$. Since with both methods one shows the induction step for all $n$, the difeerence in notation does not matter.

Note however that the first variant (correctly) allows one to drop the base case: The induction step for $n=0$ shows that, assuming the statement is true for all $k$ with $0\le k<0$ it also holds for $0$; but as $0\le k<0$ is impossible, the assumption is vacuously true, hence the case $n=0$ follows without special considereation. This does not work with the second formulation because to do the same trick, we'd have to use $n=-1$, which we don't want to.

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  • $\begingroup$ Thanks for the great explanation! $\endgroup$ – Daniel Cook Oct 25 '13 at 22:31

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