# Is $a-a=0$ defined or can it be proved by using any axioms?

Following is a partial proof for the trichotomy of integers from Terence Tao's book Real Analysis:

Lemma 4.1.5 (Trichotomy of integers).

Let $$x$$ be an integer. Then exactly one of the following three statements is true:

(a) $$x$$ is zero;

(b) $$x$$ is equal to a positive natural number n; or

(c) $$x$$ is the negation -n of a positive natural number n.

Proof. We first show that at least one of (a), (b), (c) is true. By definition, $$x$$ = $$a-b$$ for some natural numbers $$a, b$$. We have three cases: $$a > b, a = b, or a < b$$.

If $$a > b$$ then $$a = b + c$$ for some positive natural number $$c$$, which means that $$a-b = c-0 = c$$, which is (b).

If $$a= b$$, then $$a-b =a-a= 0-0 = 0$$ which is (a).

If $$a < b$$, then $$b > a$$, so that $$b-a = n$$ for some natural number $$n$$ by the previous reasoning, and thus $$a-b = -n$$, which is (c).

Can anyone explain the below statement

If $$a= b$$, then $$a-b =a-a= 0-0 = 0$$ which is (a).

How is $$a-a=0-0$$?I understand this might be a trivial question but i am also new to real analysis.Any help would be greatly appreciated.

• Hint: what do the axioms say about $0$? How is $-$ defined?
– J.G.
Commented Jun 16, 2020 at 15:09
• If $a=b$, then $a-b= a-a= 0$. That $a-a = 0 = 0-0 = 0+ 0$ does not mean $a=0$. Commented Jun 16, 2020 at 15:18

Answering this question calls for a careful look at Tao's text, in particular Definition 4.1.1:

An integer is an expression of the form $$a-b$$, where $$a$$ and $$b$$ are natural numbers. Two integers are considered to be equal, $$a-b = c-d$$, if and only if $$a + d = c + b$$.

(There is a footnote attached to "expression" elaborating on the notion of equivalence relation on ordered pairs of natural numbers.) That's how one gets from $$a-a$$ to $$0-0$$.

This answers the question in the title:

Is $$a-a=0$$ defined or can it be proved by using any axioms?

It boils down to the definition of $$x-y$$.

One common definition is $$x-y=x+(-y)$$, where by definition $$-y$$ is such that $$y+(-y)=0$$.

In this sense, $$a-a=0$$ is by definition.