First, we can cancel common terms in the numerator and denominator:
#1/color(red)(2) xx color(red)(2)/8 = 1/cancel(color(red)(2)) xx cancel(color(red)(2))/8 = 1/1 xx 1/8#
#1/1 =1# therefore we can rewrite the problem as:
#color(blue)(1) xx 1/8#
Then, knowing 1 times "anything" is the "anything" or in math terminology #1 xx x = x#:
#1 xx 1/8 = 1/8#
Another way to solve this problem is:
To multiply fractions you multiply the numerators over the denominators multiplied:
#color(red)(a)/color(blue)(b) xx color(red)(c)/color(blue)(d) = color(red)(a xx c)/color(blue)(b xx d)#
Applying this rule to our problem gives:
#color(red)(1)/color(blue)(2) xx color(red)(2)/color(blue)(8) = color(red)(1 xx 2)/color(blue)(2 xx 8) = color(red)(2)/color(blue)(2 xx 8) = (cancel(color(red)(2))1)/color(blue)(cancel(2)1 xx 8) = 1/8#
