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Why the following induction proof is wrong?

Claim: Natural numbers $0,1,2,3,\dots$ are all even.

Proof by strong induction:

Base case: $n=0$ is an even number, hence the statement is true for $n=0$.

Inductive step: Assume that the statement is true for $n=0,1,2,\dots,k$, and consider $n=k+1$. By assumption, both 1 and $k$ are even numbers, and hence so is their sum $k+1$. It thus follows that the statement holds for all $n=0,1,2,3,\dots$

Omid Sadeghi
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Patrick
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    When did you prove "$1$ is an even number" ? In the heridity step, $k$ is any non negative number. If you consider $k=0$, you see easily that the reasoning does not hold. – Suzet Jul 22 '18 at 03:32
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    Isn't there a proof like this which shows that all horses are the same color? https://en.wikipedia.org/wiki/All_horses_are_the_same_color – Mason Jul 22 '18 at 04:01
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    "Isn't there a proof like this which shows that all horses are the same color?" It's not really the same. In the horse color you have a method (albeit faulty) of getting from $n$ to $n+1$ by subtracting one and adding one (and assuming condition that $n>1$ which was not in your base case). Here you simply assume with no justification at all that $1$ is even. – fleablood Jul 22 '18 at 04:41
  • related: https://math.stackexchange.com/questions/428151/questions-on-all-horse-are-the-same-color-proof-by-complete-induction – Henry Jul 22 '18 at 14:58
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    @fleablood I think the similarity between the two proofs is that, in both cases, the alleged induction step is OK when $k$ is large enough. People are fooled because they think of the induction variable as big, and neglect the first few steps of the inductive process. – Andreas Blass Jul 22 '18 at 15:06
  • That's true. But I can see people making a mistake about large group when it is subtle like the horses (for a large group you can remove one and have some left; not so the base case). Here it just seems to be subtle. Assume true for n=0,1,2...k is nothing more or less than saying assume you've shown it for several base cases. Why the hell would anyone assume that or get fooled by that.... I have to wonder, did the OP actually not see it, or was the op claiming he didn't see why the "rules" of induction didn't allow it. – fleablood Jul 22 '18 at 15:19
  • Possible duplicate of Questions on "All Horse are the Same Color" Proof by Complete Induction – Matthew Towers Jul 22 '18 at 17:15

2 Answers2

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The following statement is not true for $k = 0$

"By assumption, both $1$ and $k$ are even numbers, and hence so is their sum $k + 1$."

For $k = 0$, you only know that $0$ is even not $1$. So you can't make the assertion above for all $k$.

Key Flex
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The problem is that your proof of the induction step should include the step from $0$ to $1$; and that's where your argument fails.

Key Flex
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Martin Argerami
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