# Elementary Properties of Ring

Some basic elementary properties of a ring can be illustrated with help of following theorems and these properties are used in developing further concepts in rings and these properties are building of rings.

Theorem:

If $R$ is a ring, then for all $a,b$are in $R$.

(a) $a \cdot 0 = 0 \cdot a = a$
(b) $a\left( { - b} \right) = \left( { - a} \right)b = - \left( {ab} \right)$
(c) $\left( { - a} \right)\left( { - b} \right) = ab$

Proof:

(a) We know that
$a0 = a\left( {0 + 0} \right) = a0 + a0\,\,\,\forall a \in R\,\,\,\,\,\,\,\left[ {{\text{using}}\,{\text{distributive law}}} \right]$

Since $R$ is a group under addition, applying right cancellation law,
$a0 = a0 + a0 \Rightarrow a + a0 = a0 + a0 \Rightarrow a0 = 0$

Similarly, $0a = \left( {0 + 0} \right)a = 0a + 0a\,\,\,\forall a \in R\,\,\,\,\,\,\,\left[ {{\text{using}}\,{\text{distributive law}}} \right]$
$\therefore \,\,\,0 + 0a = 0a + 0a\,\,\,\,\,\,\,\left[ {{\text{because}}\,0 = 0a + 0a} \right]$

Applying right cancellation law for addition, we get $0 = 0a$ i.e. $0a = 0$

Thus $a0 = 0a = 0$

(b) To prove that $a\left( { - b} \right) = - ab$ we should show that $ab = a\left( { - b} \right) = 0$

We know that $a\left[ {b + \left( b \right)} \right] = a0 = 0$ because $b + \left( { - b} \right) = 0$ with the above result (a)
$ab + a\left( { - b} \right) = 0\,\,\,\,\,\,\,\left[ {{\text{by}}\,{\text{distributive}}\,{\text{law}}} \right]$
$\therefore \,\,\,a\left( { - b} \right) = - \left( {ab} \right)$

Similarly, to show $\left( { - a} \right)b = - ab$, we must show that $ab + \left( { - a} \right)b = 0$

But $ab + \left( { - a} \right)b = \left[ {a + \left( { - a} \right)} \right]b = 0b = 0$
$\therefore \,\,\, - \left( a \right)b = - \left( {ab} \right)$ hence the result

(c) Actually to prove $\left( { - a} \right)\left( { - b} \right) = ab$ is a special case of forgoing article. However its proof is given as under:
$\left( { - a} \right)\left( { - b} \right) = - \left[ {a\left( { - b} \right)} \right] = \left[ { - \left( {ab} \right)} \right] = ab$

Because $- \left( { - x} \right) = x$ is a consequence of the fact that in a group inverse of the inverse of an element is element itself.