Simple proof? Natural Numbers...

[daon]

I proved this last semester, but our axiom system has changed and I can't use it. Can you check my proof and let me know if anything is wrong with it? I saw a well thought-out WOP proof but it seemed much more difficult than this:

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Prop: There exists no natural number m, s.t. for all n \(\displaystyle \in \L \mathbb{N}\), n < m < n+1.

Pf: I will assume there exists such an m and come to a contradiction.

We know m<n+1 so there exists a natural number b such that n+1=m+b.

Also, n<m implies n+1<m+1. And, n+1=m+b for the above b, so, m+b<m+1. This imples that b<1. This contradicts that b is a natural number. Thus there is no natural number m such that n<m<n+1.
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I'm still not very confident with my proofs, so I will be asking a lot of these kind of questions

Thanks,
Daon

 

[pka]

Can you check my proof and let me know if anything is wrong with it? Prop: There exists no natural number m, s.t. for all n \(\displaystyle \in \L \mathbb{N}\), n < m < n+1.

First, it is next to impossible to check a proof if we do not know the definitions and axioms. For one thing, some consider zero to be a natural number. That does not appear to be the case in your text?

Secondly, I think that you have misstated the problem.
Surely it says “For each natural number, n, there is no other natural number, m, such that n<m<n+1”. That is the usual problem: Prove that there is no natural number between a natural number and its successor.

 

[daon]

Yes pka, you are correct on both accounts. That is, zero is not a natural number and I am to prove: If n is a natural number, then there is no natural number m such that n<m<n+1. Have I proven that?

My axioms are that: addition and multiplication have commutativity and associativity, as well as distributivity, and we have a multiplicitive identity (1). We also know that it is well ordered.

Thanks,
Daon

 

[pka]

What axiom justifies "We know m<n+1 so there exists a natural number b such that n+1=m+b."? If there is one then I think your proof is will work. I am accustomed to working with a much weaker system. One in which you could not justify that statement.

 

[daon]

What axiom justifies "We know m<n+1 so there exists a natural number b such that n+1=m+b."? If there is one then I think your proof is will work. I am accustomed to working with a much weaker system. One in which you could not justify that statement.

That is how my book defines greater-than and less-than. i.e. "a>b means b = a + c for some natural number c." This is a Real Analysis course, I'm not sure why the axioms differ from what I (and it seems you, too) are used to. Thanks again.