# negative fractions to negative exponents

• Sep 24th 2006, 05:49 PM
shenton
negative fractions to negative exponents
What is the proper method to solve this:

(-½)^-3

Method A:

(-½)^-3

= (-1^-3)/(2^-3)

= (2^3)/(-1^3)

= 8/-1

= -8

Method B:

(-½)^-3

= (1^-3)/(-2^-3)

= (-2^3)/(1^3)

= -8/1

= -8

Should the negative be placed in the "1" as in method A or in the "2" as in method B? This gets confusing as the question is not written in the form of (-1/2) or (1/-2) but as (-½) where the negative appeared to be shared by both 1 and 2. Or it doesn't matter?

• Sep 24th 2006, 05:54 PM
Quick
Both ways work.

Just to clarify: (-1)/(2)=(1)/(-2)

The proof: you know that: (1/2)x(2/3)=(1x2)/(2x3)=2/6=1/3

and you know that: -1x-1=1

and you know that: -1/-1=1

then: (-1)/(2)=(-1)/(2)

multiply both sides by 1: (-1)/(2)x1=(-1)/(2)x(-1)/(-1)

thus: (-1)/(2)=(-1x-1)/(2x-1)

therefore: (-1)/(2)=(1)/(-2)
• Sep 24th 2006, 06:02 PM
shenton
Quote:

Originally Posted by Quick
Both ways work.

Thanks again for confirming. And that was quick too!

Cheers.