$a,b \in Z$, $a>b$ if and only if $a − b$ is a positive natural number (Terence Tao - Anlysis $1$ ex $4.1.7$). Note: Integer $x = n-m$ for some $n,m \in N$. Substraction was introduced as $x-y = x + (-y)$ where $x,y \in Z$. That is $x = n-m, y = d-k$ for some $n,m,d,k \in N$ and $-y = k-d$
take $a = x-y$, $b = c-d$ for any $x,y,c,d \in N$.
$a>b$ then $\exists m \in N, a = b+n$ => $x-y=c-d+n$
$a-b = (x-y)-(c-d) = (c-d+n)-(c-d) = (c-d+n)+(-(c-d))=(c-d+n)+(d-c)=(c-d+n)+(d-c - 0) = c-d+d-c-(n-0)=c-d-c-d-(n-0) = -(n-0)=-n$ so I get that $a-b$ is equal to negative number. Can you point out where I am wrong and how to solve the problem?
I forgot that on one of the previous exercises I prooved quite a handy law: $x+(-x) = (-x)+x = 0, x \in Z$ (the proof was by contradiction).
Using that:
$(c-d) + n = (c-d)+(n-0) = (c+n)-(d+0) = (c+n)-d$ so far so good
$a-b = ((c+n)-d) - (c-d) = ((c+n)-d) + (-c - (-d))=(c+n+(-c)) - (d+(-d)) = (n+c+(-c)) - (d+(-d))$ since $n, c \in N$
$a-b=n+0-0=n-0=n$
The only part at this point in which I am not entirely sure is the $-(c-d) = +(-c-(-d))$ by the common sense it is true. However, I guess, this must be proved. The way the $-(c-d)$ is defined in book is $+(d-c)$. Hence,
Suppose, they are not equal, then:
$(d-c) \not= (0-c)-(0-d)$, but $(0-c)-(0-d) = (0-c)+(d-0) = d-c$, so they is a contradiction. They are indeed equal.
Here I use that $-x = (0-x), x \in Z$