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OpenStudy (anonymous):
x
OpenStudy (anonymous):
1-over-anything is the reciprocal of that 'anything.'
OpenStudy (zarkon):
depends if you have
\[\frac{1}{1/x}\]
or \[\frac{1/1}{x}\]
OpenStudy (zarkon):
I assume it is the first not the second
OpenStudy (anonymous):
Explanation... please recall that we define exponent rules so they have the following properties:
\[
(x^a)^b=x^{ab}
\]Thus:
\[
\frac{1}{x}=x^{-1}\implies\frac{1}{\frac{1}{x}}=\frac{1}{x^{-1}}=(x^{-1})^{-1}=x^{(-1)(-1)}=x^1=x
\]
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OpenStudy (zarkon):
if you don't specify then for 1/1/x the division is done from left to right 1/1 first..then divide by x
OpenStudy (anonymous):
Yes, but, the one stated is a trivial question in comparison to the other, so I'd imagine it'd be otherwise.
OpenStudy (zarkon):
then the OP should have typed 1/(1/x) ;)
OpenStudy (anonymous):
Ah, semantics, semantics. But, true enough.
OpenStudy (anonymous):
Parentheses, parentheses, my kingdom for more parentheses!