# 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: $0a=a0=0$

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

$0a=(0+0)a=(0a)+(0a)$

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

### Zero ring

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

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

### Multiplication by negative one

Theorem: $(-1)a=-a$

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$(-1)\cdot a+a=(-1)\cdot a+1\cdot a=((-1)+1)\cdot a=0\cdot a=0$

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

Theorem: $(-a)\cdot b=a\cdot (-b)=-(ab)$
To prove that the first expression equals the second one, $(-a)\cdot b=((-1)\cdot a)\cdot b=(a\cdot (-1))\cdot b=a\cdot ((-1)\cdot b)=a(-b).$
To prove that the first expression equals the third one, $(-a)\cdot b=((-1)\cdot a)\cdot b=(-1)\cdot (a\cdot b).$
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 $(-a)\cdot b$  is indeed the inverse of $(a\cdot b)$  by showing that adding them up results in the additive identity element,
$(a\cdot b)+(-a)\cdot b=(a-a)\cdot b=0\cdot b=0$ .