Question: As an exercise I try to prove this inequality principle : the sense of an inequality is reversed if each side is multiplied by the same negative number. 

In steps $(2)$ and $(4)$ of the proof ( below) , I need to use the fact that : 

> 
> $$(a) (-1) = -a$$
> 
> 
> 

So I've asked myself the question : how to prove this fact ? 

The number : $-a$ is defined as the opposite (or additive inverse) of the number $a$ . Opposites ( additive inverses) are not defined in terms of multiplication. So, though basic, this fact is not absolutely trivial. 

It seems to me that I cannot use here the " rule of signs" since the sign of the number a is undetermined. 

Which *axioms* or *basic principles* could be used in an algebraic system to prove this elementary fact? 

Below, my attempt of proof regarding inequalities : 

[![enter image description here](https://i.stack.imgur.com/wbSoj.jpg)](https://i.stack.imgur.com/wbSoj.jpg)

Answer: 
Let $R$ be a commutative ring and let $a,b\in R$. Let $0$ be the zero element of $R$ and $-a$ be the additive inverse of $a$. Here are some basic facts:

(1) $0a = 0$.

Proof: $0+0=0$. Multiplying with $a$ gives $a0 = a(0+0) =a0+a0$. Adding $-a0$ on both sides gives $0 = a0$.

(2) $-(-a) = a$.

Proof: $-(-a)$ and $a$ are both additive inverses of $-a$. By the uniqueness of add. inverses, $-(-a)=a$.

(3) $(-a)b = a(-b) = -ab$.

Proof: $ab + (-a)b=(a+(-a))b= 0b=0$ and so $(-a)b$ is inverse to $ab$. By the uniqueness of add. inverses, $(-a)b=-ab$.

(4) $(-a)(-b) = ab$.

Proof: $(-a)(-b) = -((-a)b) = -(-(ab))$ by (3) and $-(-ab) = ab$ by (2).

In particular, by (3), $a(-1) = -(a1) = -a$.
Three key points:

* Multiplication gives the same answer as repeated addition.
* Subtraction is addition of negatives
* Addition starts from 0.

These transform it as follows:$$a(-1)=0+\underbrace{(-1)+\cdots+(-1)}\_{\text{a times}}=0-a=-a$$