## Is it true for all n, Natural number?

Hello all,

Another quick question for the number theory gurus here:

Let P(n) predicate, n Natural number. Suppose that P(n) satisfies that P(1) is true, and if k in N, P(k) is true, then P(k+2) is true. Is P(n) true for ALL n in N? Why?

-William

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 Recognitions: Gold Member The principal of induction sezs that if its true for N, it's true for N+1, and we begin with a basis of 1 or zero, usually. This means once we have proved it true for 5, we can deduce that it is true for 6, and if true for 6, well..... So if its true for N+2 when its true for N....you can take it from there.
 I would say no. P(1) -> P(3) -> P(5) and so on. In other words, P(n) is true for all odd n.

## Is it true for all n, Natural number?

Interesting e(hoOn3... I was thinking something like that, other opinions?

 No. The answer is no. Well, depending on how N was constructed. You could construct N differently and then be able to say yes.
 If you say no, why no? N = Natural numbers, thanks for your reply!
 Recognitions: Homework Help Science Advisor just construct a counter example. it is trivial. P(1), and hence P(3), P(5),.. are true, but P(2) etc can be completely arbitrary. Are you telling me you can't think of an obvious statement that is true for (something to do with) odd numbers but is false for even numbers?
 Glad there is edit here... its late, im not thinking straight. I didnt catch the counter example, sorry.
 I am not following you at all. Matt Grime pretty much gave you a counter example.
 Wrong. Matt didn't give a counter-example per say. He said that one may construct a statement P(n) that is true for odd n but false for even n.
 Recognitions: Homework Help Science Advisor I think I pretty much did give a counter example. At least if you take second to think what the words ODD and EVEN are doing there. Isn't there a really obvious property that ODD numbers have that EVEN ones don't? And can you show it by induction? Yes. It is a vacuous proof, but still a proof by induction nonetheless.