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

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.

• Can you provide more context? How is $\Bbb Z$ defined? Sep 1, 2020 at 21:08
• Tao defines the integers $\mathbb{Z}$ as formal differences of natural numbers : An integer $n$ is any expression of the form $x - y$. Two integers $a - b$ and $c - d$ are equal if and only if $a + d = b + c$. Sep 1, 2020 at 21:15

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