# If a, b are elements of a ring and m, n ∈ Z, show that (na) (mb) = (mn) (ab)

If a, b are elements of a ring and m, $$n \in Z$$, show that $$(na) (mb) = (mn) (ab)$$

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unett

We have to show that if a, $$\displaystyle{b}∈{R}$$ and $$\displaystyle{m},{n}∈{Z}$$, then $$(na)(mb)=(nm)(ab).$$
Notice that

$$(na)(mb)={(a+....+a)(b+....+b)}n \times m \times =a{(b+...+b)}+...+a{(b+...+b)}ms =$$

$${{(ab+...+ab)}+...+{(ab+...+ab)}}m \times n \times ={m(ab)+...+m(ab)}n \times =(nm)(ab)$$

Hence the proof.