proof by induction using +2

Question

im wondering if this variant of proof by induction is correct

the standard proof by induction states that if an equation/algorithm works for n and you can prove that it works for n+1 then you can assume it works for every integer bigger or equal to n.

Now, if you had 2 base case, (ex: 2 and 3) and you were to prove it works for n+2, can you say that it works for every integer bigger then 2 ?

because suppose you can prove that its correct for n+2,

2+2=4
3+2=5
4+2=6

etc, so you cover every integer bigger then 2

thanks for you help ^^

(also if the +2 version is correct that implies that if you have m consecutive base case and a proof that it works for n+m then it will work for every integer bigger then n)

Answer 1

It depends on what you mean by "it works for n+2". If you mean that there is some statement S(n), and you can prove

If S(n) is True then S(n+2) is True

And if you know S(0) is True, then by induction, it follows that S(2), S(4), S(6), ..., S(n) for all even n is True.

And if also you know S(1) is True, then by a second application of induction, it follows that S(3), S(5), ..,. S(n) for all odd n is True.

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Or, if you can prove

If S(2n-1) and S(2n-2) are True, then S(2n) is True

and also that S(0) and S(1) are True, then by the inductive step it follows that S(2) is True. And since S(1) and S(2) are True, then again by the inductive step S(3) is True. And by successive applications of the inductive step it follows that S(n) is true for all n > 0.

(This is easily adaptable to induction steps where m prior statements S(n-m), ..., S(n-1) are used to prove S(n)...)

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If on the other hand you can only prove

If S(n-1) and S(n) are True, then S(n+2) is True

then even if S(0) and S(1) are True, you are in trouble since your inductive step merely gives you that S(3) is True. It does not prove S(2) is True. Thus the inductive step can not be applied over and over again and thus it fails to achieve lift off...