Isn't 2m+ 2n= 2(m+n) simpler?
Then is even and is even
If you're going to use a 'contradiction' method, I think you have to be a little more careful with your use of notation. The definition of your sets should be
Then your hypothesis would be:
I agree, it does look more or less trivial, but what you should have said is
which is now in the format
So that you can now say:
using the part of the definition that is even
Do you understand this subtle, but important, point?
Yes- thanks again, I realised that I needed to show something analogue to the definition of the even number, but I can see that it is much clearer to substitute of the natural numbers as opposed to . But the most important thing missing from the proof is this substitution of an equivalent natural number p.
This reminds me of another probelm that I'm having proving the distributive laws in set theory- it is somehow simple, but hard to express in the right way!
It follows that,
Unfortunately, it's not very clear to read.
Part 1 of your proof is fine.Part1:
This is not so good. One way to check to see if it makes sense is this: only works if is a prime number, or a product of single powers of prime numbers. (For instance, is OK, but isn't.) But you haven't used the fact anywhere in your proof that 3 is a prime.Part2:
You need to use the Fundamental Theorem of Arithmetic and say that if , where the are primes, then . So , for some , and therefore .
I'm not sure about the Fundamental Theorem of Arithmetic, I had a look at the wikipedia page and it looks interesting, but the course is moving very quickly, unfortunately and although I can't write very good proofs yet, we have already moved onto fields (help!)- anyway, I thought of a good idea for this problem, but I'm not sure how to proove it.
The idea is that
This would mean that I can jump straight into the proof, because can only ever equal one of the 3 possibilities above. But the proof becomes something like:
, so I would like to say something like:
Proof by contraditiction, suppose
[Errr... well, I- this seems to be true, but I don't know how to...?]
[... do the same for the other or]
This is a contradiction, therefore
Well, this is a terrible proof, but on the good side, if done correctly this would actually be an "if and only if" proof! (?)
Well done. I think this is pretty well OK. I should just change the last part a bit (the part where you weren't so sure) so that instead of:
Contradiction, since and both leave remainder when divided by .