Question

If R is a ring with identity and $\mathbf{m}\mathbf{,}\mathbf{n}\mathbf{\in }\mathbf{Z}$prove that $\left(m{1}_R\right)\left(n{1}_R\right)\mathbf{=}\left(\mathrm{mn}\right){\mathbf{1}}_{\mathbf{R}}$.

Short answer

It is proved that $\left(\mathrm{m}{1}_{\mathrm{R}}\right)\left(\mathrm{n}{1}_{\mathrm{R}}\right)=\left(\mathrm{mn}\right)1_{\mathrm{R}}$.

Step-by-step solution

Definition of ring

A ring is 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 (m1R)(n1R)=(mn)1R  in different cases 

Let R be a ring with identity and$\mathrm{m},\mathrm{n}\in \mathrm{Z}$.

Case: 1 when$\mathrm{m}>0,\mathrm{n}>0$

Consider:

$\begin{array}{rcl}\left(\mathrm{m}{1}_{\mathrm{R}}\right)\left(\mathrm{n}{1}_{\mathrm{R}}\right)& =& \left(1_{\mathrm{R}}+1_{\mathrm{R}}+.....1_{\mathrm{R}}\left(\mathrm{m}\mathrm{times}\right)\right)\left(1_{\mathrm{R}}+1_{\mathrm{R}}+.....1_{\mathrm{R}}\left(\mathrm{n}\mathrm{times}\right)\right)\\ & =& \left(1_{\mathrm{R}}+1_{\mathrm{R}}+.....1_{\mathrm{R}}\left(\mathrm{mn}\mathrm{times}\right)\right)\\ & =& \left(\mathrm{mn}\right)1_{\mathrm{R}}\end{array}$

Therefore in this case claims holds$\left(\mathrm{m}{1}_{\mathrm{R}}\right)\left(\mathrm{n}{1}_{\mathrm{R}}\right)=\left(\mathrm{mn}\right)1_{\mathrm{R}}$.

Case: 2 when$\mathrm{m}>0,\mathrm{n}<0$

Let$\mathrm{n}=-\mathrm{n},{\mathrm{n}}_1>0$

Consider:

role="math" localid="1659087930325" $\begin{array}{rcl}\left(m{1}_R\right)\left(n{1}_R\right)& =& \left(m{1}_R\right)\left(-n{1}_R\right)\\ & =& \left(m{1}_R\right)\left(n_1\left(-1_R\right)\right)\\ & =& \left(m{n}_1\right)\left(1_R\left(-1_R\right)\right)\\ & =& \left(mn\right)\left(-1_R\right)\end{array}$

Further simplify this as:

$\begin{array}{rcl}\left(\mathrm{m}{1}_{\mathrm{R}}\right)\left(\mathrm{n}{1}_{\mathrm{R}}\right)& =& \left({\mathrm{mn}}_1\right)\left(-1_{\mathrm{R}}\right)\\ & =& \left(-{\mathrm{mn}}_1\right)\left(1_{\mathrm{R}}\right)\\ & =& \left(\mathrm{mn}\right)\left(1_{\mathrm{R}}\right)\\ & & \end{array}$

Therefore claim hold in this case also.

Case: 3 whendata-custom-editor="chemistry" $\mathrm{m}<0,\mathrm{n}>0$

Let$\mathrm{m}=-{\mathrm{m}}_1,{\mathrm{m}}_1>0$

Consider,

$\begin{array}{rcl}\left(\mathrm{m}{1}_{\mathrm{R}}\right)\left(\mathrm{n}{1}_{\mathrm{R}}\right)& =& \left(-\mathrm{m}{1}_{\mathrm{R}}\right)\left(\mathrm{n}{1}_{\mathrm{R}}\right)\\ & =& \left(\mathrm{m}\left(-1_{\mathrm{R}}\right)\right)\left({\mathrm{n}}_1{1}_{\mathrm{R}}\right)\\ & =& \left({\mathrm{m}}_1\mathrm{n}\right)\left(\left(-1_{\mathrm{R}}\right)1_{\mathrm{R}}\right)\\ & =& \left({\mathrm{m}}_1\mathrm{n}\right)\left(-1_{\mathrm{R}}\right)\end{array}$

Further simplify:

$\begin{array}{rcl}\left(\mathrm{m}{1}_{\mathrm{R}}\right)\left(\mathrm{n}{1}_{\mathrm{R}}\right)& =& \left({\mathrm{m}}_1\mathrm{n}\right)\left(-1_{\mathrm{R}}\right)\\ & =& \left(-{\mathrm{m}}_1\mathrm{n}\right)\left(1_{\mathrm{R}}\right)\\ & =& \left(\mathrm{mn}\right)1_{\mathrm{R}}\end{array}$

Thus the claim also holds for this condition also.

Case: 4 when$\mathrm{m}<0,\mathrm{n}<0$

Let$\mathrm{m}=-{\mathrm{m}}_1,\mathrm{n}=-{\mathrm{n}}_1,{\mathrm{m}}_1>0{\mathrm{n}}_1>0$

Consider:

$\begin{array}{rcl}\left(\mathrm{m}{1}_{\mathrm{R}}\right)\left(\mathrm{n}{1}_{\mathrm{R}}\right)& =& \left(-{\mathrm{m}}_1{1}_{\mathrm{R}}\right)\left(-{\mathrm{n}}_1{1}_{\mathrm{R}}\right)\\ & =& \left({\mathrm{m}}_1\left(-1_{\mathrm{R}}\right)\right)\left({\mathrm{n}}_1\left(-1_{\mathrm{R}}\right)\right)\\ & =& \left({\mathrm{m}}_1{\mathrm{n}}_1\right)1_{\mathrm{R}}\\ & =& \left(\mathrm{mn}\right)1_{\mathrm{R}}\end{array}$

Thus the cliam holds for this case also. Hence, it is proved that $\left(\mathrm{m}{1}_{\mathrm{R}}\right)\left(\mathrm{n}{1}_{\mathrm{R}}\right)=\left(\mathrm{mn}\right)1_{\mathrm{R}}$.