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.

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

2 Answers 2


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .