How do you simplify #1/2^0#?

1 Answer
Mar 27, 2016

#1/2^0 = color(green)(1)#

Explanation:

Any (non-zero) number to the power of zero equals #1#
So #1/color(red)(2^0) = 1/color(red)(1) = 1#

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What is the logic behind "any number to the power of zero equals #1#"?

#a^n=a^(n-1)xxa#
#color(white)("XXX")#that is #a# multiplied by itself #n# times
#color(white)("XXX")#is the same as
#color(white)("XXXXXX")a# multiplied together #(n-1)# times
#color(white)("XXXXXX")#and then multiplied by #a# one more time.

This could be re-written as
#a^(n-1)=a^n div a# (provided #a!=0#)
So
#color(white)("XXX")a^2=a^3div a#
#color(white)("XXX")a^1=a^2diva#
#color(white)("XXX")a^0=a^1diva#
but
#color(white)("XXX")a^1=a#
and therefore
#color(white)("XXX")a^0=adiv a=1# (again, provided #a!=0#)