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I was told to do a proof by contradiction and I am not sure if what I came up with is valid if you can confirm or assist me I would greatly appreciate it.

Prove If $n\in\mathbb{N}$ is not even then $\exists$ $k\in \mathbb{N}$ s.t. $n=2k-1$

Assume to the contrary that $\exists k\in \mathbb{N}$ s.t. n=2k.

Then by definition of an even number, $n$ is even, which is a contradiction to "$n\in\mathbb{N}$ is not even".

Thus, if $n\in\mathbb{N}$ is not even then there must be a $k\in \mathbb{N}$ s.t. $n=2k-1$.

user348547
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3 Answers3

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You are to prove,

Prove if $n∈\mathbb{N}$ is not even then $∃k∈\mathbb{N}$ s.t. $n=2k−1$.

To prove this, you assumed to the "contrary" that if possible let, $$\exists k\in\mathbb{N}\mid n=2k$$then you proceed to show that (by Definition of even numbers), $n$ is even and this contradicts the hypothesis that "$n$ is not even".

But this only shows that if $n$ is not even then, $$\nexists k\in\mathbb{N}\mid n=2k$$$\color{red}{\text{But how does this show that}\ \exists k\in\mathbb{N}\mid n=2k-1?}$

For a proof of the claim, see here, especially Henning Makholm's answer.

Community
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  • Hmm isn't all these in my answer, even down to the remarks in Henning's answer? Your answer in that thread, on the other hand, is rather distinct (though of course it boils down to the equivalence between induction and well-ordering). – user21820 Jun 23 '16 at 05:48
  • @user21820: To be precise, (considering your comment to form a distinct part from your answer), the answer to your question is "no" because although your answer includes the essential point, it doesn't say explicitly why in regards to this question OP has done a mistake, which I have attempted to do. –  Jun 23 '16 at 05:53
  • True, but I prefer to let the asker to think through and apply the analogy to his own attempt on his own. Anyway doesn't matter. =) – user21820 Jun 23 '16 at 05:55
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You're not proving anything at all. Here's an analogy:

You want to prove "If it rains then the field will be wet." but your argument is "Assume to the contrary that it does not rain."...

What you're asked to prove cannot be done without induction. It would be tricky to do so directly. The better way is to first prove (by induction) that any natural number $n$ is equal to either $2k$ or $2k+1$ for some natural $k$. Then it immediately follows that a natural number that is not even must be equal to $2k+1 = 2(k+1)-1$ for some natural $k$.

user21820
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  • For those who are familiar with first-order PA, there is an interesting way to prove that induction is necessary. The axioms of discrete ordered semirings (https://en.wikipedia.org/wiki/Peano_axioms#Equivalent_axiomatizations) are satisfied by integer polynomials with positive leading coefficient plus the zero polynomial, with the usual polynomial arithmetic and ordering by degree then coefficients. But the polynomial $X+1$ is not even and yet cannot be expressed as one plus twice another polynomial. Therefore the theorem in question requires induction. – user21820 Jun 23 '16 at 05:29
  • Re: prior comment, see also this answer on extending parity to polynomial rings (there germs of polynomials at $+\infty)\ $ – Bill Dubuque Jun 23 '16 at 14:03
  • @BillDubuque: Ah that's interesting; thanks for the link! There's also a related post not long ago about parity and non-standard models of PA but I can't find or recall what it was, nor who wrote it. Do you happen to know of it? – user21820 Jun 23 '16 at 14:17
  • No, I missed that (I have not been on the site for most of the past year). – Bill Dubuque Jun 23 '16 at 14:45
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in your reasoning you admit that for all $n\in \Bbb{N} $ the $n$ must write $n=2k$ or $n=2k + 1$ that is even or odd; in this case your demonstration is correct

m.idaya
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