I would like to rigorously prove the following result for integers $\mathbf{Z}$. This is exercise 4.1.1, page 81, from Analysis-I by Terrence Tao.
Verify that the definition of equality of integers is both reflexive and symmetric.
I would like someone to verify my proof. I just want to make sure, that I am not making implicit assumptions or trivializing my proof. I also want to avoid any circularity.
Proof. (My attempt).
Let $a,b,c,d$ be natural numbers. We define integers $x = a – b$ and $y = c – d$.
The equality relation $=_{\mathbf{Z}}$ on the integers $\mathbf{Z}$ is defined to be the set of all ordered pairs $(x,y)$ given by
$R := \{ (x,y) | (a + d) = (b + c); \text{ such that } x = a – b, y = c – d\}$
(1) Reflexive. Clearly, $a + b = a + b$ for natural numbers $a,b$, so $a – b = a – b$. This implies $x = x$ for all integers $x$.
(2) Symmetric. Moreover, if $x = y$, then $a + d = b + c$, it implies $b + c = a + d$, so $y = x$.
Thank you so much,
Quasar.
Best Answer
As it seems this is part of the process to introduce the integers starting out from the natural numbers, where the idea written in the comments is made rigorous:
This means that we consider the set $\Bbb N\times\Bbb N$ of pairs of naturals, and we introduce an equivalence relation $=_{\Bbb Z}$ on it, which is, despite the notation, not the equality relation on the set of pairs (but it will eventually become equality when taking the quotient set $(\Bbb N\times\Bbb N)/\!\!=_{\Bbb Z}$).
So, the pair (of pairs) $((a,b),\,(c,d))$ is in the relation $=_{\Bbb Z}$ if and only if $a+d=b+c$.
For symmetry, we want to conclude $(c,d)=_{\Bbb Z}(a,b)$ which means $c+b=d+a$, then we have to use commutativity of addition on $\Bbb N$.
Try to prove transitivity.
When it's done, finally we are ready to define $\Bbb Z$: $$\Bbb Z:=(\Bbb N\times\Bbb N)/\!\!=_{\Bbb Z}$$