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Lgbariddle!! is \[\huge a^{\frac mn} = \sqrt[n]{a^m} = (\sqrt[n] a)^m\]
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yup
Can be proved
so tell me...what happens when a is negative?
for example \[\huge (-1)^{2/4}\]
\[\huge \sqrt[4]{-1} \implies \text{imaginary}\] \[\huge \sqrt[4] {(-1)^2} \implies 1\]
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you're right, again
that's where your algebra is now shaken :troll:
this is the part where i advertise this: http://openstudy.com/updates/4fdaecf3e4b0f2662fd13fd6
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