# Thread: Z+Z is Z too

1. ## Z+Z is Z too (Same for Z.Z)

Need verification for 3 proofs.
1-)So now I am doing the integers and there is this statement saying an integer + an integer is integer.
The lecturer did a proof that I didn't understand so I wrote my own, but I don't know if it's a proof, appreciate if you guys can give me a feedback.

So we defined an integer as z=m-n when n,m are natural numbers

so said for ever integer z=m-n and z*=m*-n* when n,n*,m,m* are natural numbers
so z+z*=m-n+m*-n*=m+m*-(n+n*)
So according to the statements that I proved before a natural plus natural is a natural
and this is actually the definition of the integers since m+m* and n+n* are natural numbers.

2-)The second proof is Z.Z is an integer too, so I took the same integers that I defined and multiplied them, so I got
(m-n)(m*-n*)=(mm*-mn*-nm*+nn*)=(mm*+nn*)-(mn*+nm*)
So I proved that natural number + natural number is natural number and natural number times natural numbers is natural numbers, so we know that n,m,n*,m* are naturals so we get that mm*+nn* is natural and mn*+m*n is natural and if we write instead of them a and b then we get z.z*=a-b when a and b are naturals which is the definition of integers.

3-)For every z integer there can't exist w an integer that z<w<z+1
Proof:
z=m-n, w=p-r when m,n,p,r are natural numbers so I subtracted z from the inequality and got
0<p-r+n-m<1 and we got 0<(p+n)-(r+m)<1 and I defined p+n and r+m as k and l respectively
and we got 0<k-l<1 and I added l in the inequality and got l<k<l+1 and I proved in the natural numbers that this situation can't happen because from one side k-l is natural number and from the other side we get k-l<1 which is clearly a contradiction therefore the statement.

Happy to get feedback about these. And if there is a simpler proof happy to hear it, because it looks like I do things in a very long way.. Thanks.

2. ## Re: Z+Z is Z too

Hey davidciprut.

The proofs look good: the only suggestion I have for you is to make sure you refer to elements as members of a set (like a E A) in the formal proof.

Also as some food for thought, one of the ways that you prove stuff with natural numbers is to use the well ordering theorems where the natural numbers has a least element (usually 1 but sometimes for some definitions it can be 0). This technique is used a lot in number theory proofs and also within a lot of pure mathematics proofs that range across different areas (not just number theory).

3. ## Re: Z+Z is Z too

For the second one, did you prove the distributive property? Or is that given axiomatically? It is not difficult to prove, but if it is not given, you should probably include its proof.