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.