Trying to prove mathematical induction. Going in circles?

[daon]

The question is:
Let P(n) be a statement depending on a variable n\(\displaystyle \in\mathbb{N}\). In order to prove that “P(n) is true for all n” it is sufficient to prove:
Initial case: P(1) is true, and
Inductive case: P(n) implies P(n+1)


My teacher writes the comment "This is what you are trying to prove!" about the bolded text below in my proof. So I guess I am going in circles? I am totally lost for ideas and we have a quiz tomorrow on this. She has given me a "suggestion" to use the following proposition as a given truth:

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Let \(\displaystyle B\subseteq\) \(\displaystyle \mathbb{N}\) suct that:

1. \(\displaystyle 1 \in B\)
2. whenever n \(\displaystyle \in B\) then (n+1) \(\displaystyle \in B\)
Then B = \(\displaystyle \mathbb{N}\)
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I am not sure how to apply this proposition. Here is my incorrect proof:
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Let n\(\displaystyle \in\mathbb{N}\),
Assuming P(1) is true by the hypothesis, we have covered the first number in \(\displaystyle \mathbb{N}\). Also, because n \(\displaystyle \in\mathbb{N}\), by axiom 6.2, (n+1)\(\displaystyle \in\mathbb{N}\). So, if we assume P(n) is true and prove that P(n+1) is true with that assumption (that is, prove that P(n) implies P(n+1)), then P(n) can be said to be true for both n and the successor of n. Since n is representative of all elements in the natural numbers, P(n) is true for all n \(\displaystyle \in\mathbb{N}\);.
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Any help would be great!
Daon

 

[pka]

Here is an axiom. Every non-empty subset of positive integers has a first term.
Now suppose that the proposition, P(n), on the positive integers has these two properties: 1) P(1) is true and 2) if P(n) is true then P(n+1) is also true.

So suppose that P(K) is not true for some positive integer K.
Consider the set S={n: P(n) is not true for n a positive integer}.
Well S is not empty because S contains K.
Thus S has a first element, call it L.
L is not 1 because P(1) is true, 1 is not in S.
Because L is the first term in S, L−1 is not in S.
That means that P(L−1) is true.
But if P(L−1) is true then P(L−1+1) is also true.
Do you see the contradiction?

 

[daon]

Yes, I see the contradiction pka. Thanks for that! I did try something similar but got lost in the terminology and felt I was introducing too many new things. Proofs are still new to me, but I am trying to adapt. I may swing in that direction to correct my proof, but is there any way besides contradiction?

My professor probably wouldn't care which way we proved this, but she thinks contradiction should always be a last resort.

Thanks again.
Daon