# Proofs of elementary ring properties

The following proofs of elementary ring properties use only the axioms that define a mathematical ring:

## Basics

### Multiplication by zero

Theorem: ${\displaystyle 0a=a0=0}$

(Click "show" at right to see the proof of this theorem or "hide" to hide it.)

${\displaystyle 0a=(0+0)a=(0a)+(0a)}$

By subtracting (i.e. adding the additive inverse of) ${\displaystyle 0a}$  on both sides of the equation, we get the desired result. The proof that ${\displaystyle a0=0}$  is similar.

### Zero ring

Theorem: A ring ${\displaystyle (R,+,\cdot )}$  is the zero ring (that is, consists of precisely one element) if and only if ${\displaystyle 0=1}$ .

(Click "show" at right to see the proof of this theorem or "hide" to hide it.)
Suppose that ${\displaystyle 1=0}$ . Let ${\displaystyle a}$  be any element in ${\displaystyle R}$ ; then ${\displaystyle a=a\cdot 1=a\cdot 0=0}$ . Therefore, ${\displaystyle (R,+,\cdot )}$  is the zero ring. Conversely, if ${\displaystyle (R,+,\cdot )}$  is the zero ring, it must contain precisely one element. Therefore, ${\displaystyle 0}$  and ${\displaystyle 1}$  is the same element, i.e. ${\displaystyle 0=1}$ .

### Multiplication by negative one

Theorem: ${\displaystyle (-1)a=-a}$

(Click "show" at right to see the proof of this theorem or "hide" to hide it.)

${\displaystyle (-1)\cdot a+a=(-1)\cdot a+1\cdot a=((-1)+1)\cdot a=0\cdot a=0}$

Therefore ${\displaystyle (-1)\cdot a=(-1)\cdot a+0=(-1)\cdot a+(a+(-a))=((-1)\cdot a+a)+(-a)=0+(-a)=(-a)}$ .

Theorem: ${\displaystyle (-a)\cdot b=a\cdot (-b)=-(ab)}$
To prove that the first expression equals the second one, ${\displaystyle (-a)\cdot b=((-1)\cdot a)\cdot b=(a\cdot (-1))\cdot b=a\cdot ((-1)\cdot b)=a(-b).}$
To prove that the first expression equals the third one, ${\displaystyle (-a)\cdot b=((-1)\cdot a)\cdot b=(-1)\cdot (a\cdot b).}$
A pseudo-ring does not necessarily have a multiplicative identity element. To prove that the first expression equals the third one without assuming the existence of a multiplicative identity, we show that ${\displaystyle (-a)\cdot b}$  is indeed the inverse of ${\displaystyle (a\cdot b)}$  by showing that adding them up results in the additive identity element,
${\displaystyle (a\cdot b)+(-a)\cdot b=(a-a)\cdot b=0\cdot b=0}$ .