# Elementary Properties of Rings

Some basic elementary properties of a ring can be illustrated with the help of the following theorem, and these properties are used to further develop and build concepts on 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 the 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)** Proving $$\left( { – a} \right)\left( { – b} \right) = ab$$ is a special case of forgoing the article. However its proof is given as:

$$\left( { – a} \right)\left( { – b} \right) = – \left[ {a\left( { – b} \right)} \right] = \left[ { – \left( {ab} \right)} \right] = ab$$

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