# Proof by induction [closed]

I was just wondering how to prove generally that a statement is true

i think this is correct, but i am not sure:

1) prove that the statement $P$ is true for $1$

2) and then prove that the statement is true for $P(n+1)$, by taking for granted that $P(n)$ is true.

2.5) we can prove that $P(n+1)$ is true by showing that it is true when $n = 1$.

i am not sure about the last part though.

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## closed as not a real question by Andres Caicedo, Austin Mohr, TMM, Norbert, Ｊ. Ｍ.Nov 17 '12 at 12:46

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Part 2.5 is not true: Take $P(n)$ to mean "$n=1\text{ or }n=2$". Now $P(1)$ is true. $P(n+1)$ is also true when $n=1$. However, it does not follow that $P(n)$ is true for all $n$ since $P(3)$ is not true.

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This is axiom 9 of Peano arithmetic. It is the hard one. Is that the system you are working in?

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no, i am taking an introductory course –  George Nov 16 '12 at 5:52
seriously, how is that an answer to my question? –  George Nov 16 '12 at 6:04
My point is that it is an axiom. You don't have to prove it is true, you use it to prove other things are true. Intuitively, it says if you know $P(1),$ and $P(1) \implies P(2)$, you know $P(2)$. Then if $P(2) \implies P(3)$ you know $P(3)$ and so on. If you can prove that in general $P(n) \implies P(n+1)$ you know $P(n)$ for all $n$. –  Ross Millikan Nov 16 '12 at 14:11
i know that, but how do i prove that p(n+1) is true? i can't do any algebraic manipulation. –  George Nov 16 '12 at 16:03