# 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 ∈ Z, show that (na) (mb) = (mn) (ab)

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

$$=a{(b+...+b)}+...+a{(b+...+b)}m$$

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

$$={m(ab)+...+m(ab)}n$$

$$=(nm)(ab)$$
Hence the proof.