Question

Let R be a ring with identity of characteristic $\mathbf{n}\mathbf{\ge }\mathbf{0}$. Prove that $\mathbf{na}\mathbf{=}{\mathbf{0}}_{\mathbf{R}}$for every $\mathbf{a}\mathbf{\in }\mathbf{R}$.

Short answer

It is proved that na=0_R .

Step-by-step solution

Definition of ring

A ring is a fundamental algebraic structure. It consists of a set equipped of a set with two binary operations that generate the arithmetic operation of addition and multiplications.

Showing that na=0R 

Let R be a ring with identity and characteristic $n\ge 0$ and 1_R be the identity of R. Therefore n1_R=0 . Now to show for every $a\in R$, $na=0_R$.

Consider left-hand side and simplify as:

$\begin{array}{rcl}\mathrm{na}& =& \mathrm{a}+\mathrm{a}+\mathrm{a}+....+\mathrm{a}\\ & =& 1_{\mathrm{R}}\mathrm{a}+1_{\mathrm{R}}\mathrm{a}+...+1_{\mathrm{R}}\mathrm{a}\\ & =& \left(\mathrm{n}{1}_{\mathrm{R}}\right)\mathrm{a}\\ & =& 0_{\mathrm{R}}\mathrm{a}\end{array}$

This gives na=0_R .

It is proved that na=0_R.