None of that matters at all. The fact that you're trying to use induction does not mean you can ignore basic logic. This "proof" introduces the assumption that 1 and then proceeds toward a conclusion, conveniently forgetting that the conclusion applies only to cases where 1 is even. For the conclusion to apply to the general case, it must be proven either that the assumption is true for all cases or that the conclusion is also valid where the assumption does not hold. That an inductive technique happens to be used somewhere has absolutely no bearing.

What's been said above pretty much gives the answer to the problem already, but I thought I'd add a different viewpoint to it.

Basically, the proof in the problem might look plausible because of slightly confusing notation "n = 0, 1, .., k". Instead, the principle of induction may be stated in terms of sections which are defined as Sn = {i is natural | i < n} (here natural numbers include 0). So a set like {0, 1, .., k} would become Sk+1.

Now, proof by induction is used to prove statements of the kind "proposition P holds for all natural numbers". For convenience let P(Sn) mean "proposition P holds for all natural numbers less than n". With this, a general proof by induction may look like this:

1) P holds for 0
2) for all n P(Sn+1) implies that P holds for n + 1
3) Hence P holds for all natural numbers.

Finally, using all of the above, the claim in the original proof is equivalent to following statements:

1) 0 is even
2) for all n Even(Sn+1) => 1 is even and n is even => n + 1 is even
3) Hence all natural numbers are even

By the usual definition of evenness 0 is indeed even. Next, in the second part there's need to prove two implications. Proposition "1 is even and n is even => n + 1 is even" is again true because of the properties of even numbers. On the other hand the first one is a bit more complex. Decomposing it into "Even(Sn+1) => 1 is even" and "Even(Sn+1) => n is even" it's easy to see that second one is always true (Sn+1 contains n). So the validity of the whole proof rests on the remaining proposition:

Even(Sn+1) => 1 is even

and to bring down the whole proof it's sufficient to find a single counterexample to it. Now if n = 0 then Sn+1 = {0}, and there's no way to prove "1 is even" from "Even({0})" (which is just "0 is even" of course").

So again, it's the same thing that's been said in this thread, just a bit more rigorous. Hopefully I didn't confuse things further ^_^.

This will fail for any attempted proof where it is not practical to examine the antecedent in the step. Xiongmao gave a rigorous explanation above; I'll confine myself to an English one.

In induction, you must prove two propositions. If you prove both, there is no call to examine any other.

All inductions amount to:

Given an ordered set with an open upper bound, such as the whole numbers (the natural numbers and zero), prove:

* The proposition is true for the first member of the set.

* If the proposition is true for any member of the set, then it is true for that member's successor.

You cannot invalidate an induction by attacking the antecedent: saying there is a member that must not be in the set. You must find a member for which the consequent is false. Anything less is no disproof at all.

In this case the proof fails, precisely and only because the statement "if 0 is even, then 1 is even" is false.