How do you find the set in which the real number #-sqrt(0.0625)# belongs?

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Answer 1 · 2017-03-09T06:18:00

#-sqrt(0.0625)# is a rational number.

Note that:

#0.0625 = 0.125/2 = 0.25/4 = 0.5/8 = 1/16 = (1/4)^2#

So:

#sqrt(0.0625) = sqrt((1/4)^2) = 1/4#

So:

#-sqrt(0.0625) = -1/4#

This is a rational number, since it is expressible in the form #p/q# for integers #p, q# with #q != 0#. For example: #p=-1# and #q=4#.