Equality on integers is reflexive and symmetric (Exercise 4.1.1, Tao Analysis I)

Question

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.

Answer 1

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:

An integer is (represented as) a formal difference of two natural numbers.

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 $\Bbb Z$: $$\Bbb Z:=(\Bbb N\times\Bbb N)/\!\!=_{\Bbb Z}$$