#
Solve this:

Q.71. Find (- 1)$\times $(- 1)$\times $...... 2 m times, m is a natural number?

(A) - 1

(B) 1

(C) 2

(D) - 2

The solution is as under:

$\left(-1\right)\times \left(-1\right)\times \left(-1\right)\times ......2mtimes\phantom{\rule{0ex}{0ex}}or,={\left(-1\right)}^{2m}\left\{inexponents,whenbaseissame,powerisadded,{a}^{m}\times {a}^{n}={a}^{m+n}\right\}\phantom{\rule{0ex}{0ex}}Now,asmisanaturalnumber,andnaturalnumbersareintegerswhichstartsfrom1andgoesuptoinfinity,\phantom{\rule{0ex}{0ex}}hence2mwillbeaneveninteger\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}or,={\left(-1\right)}^{2m}=1\left\{as{\left(-1\right)}^{evenint}=1\right\}$

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards

**
**