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I am getting two different answers depending oh how I cancel out the numbers in the numerator and denominator.

When I take $(\sqrt{2}/2)/(\sqrt{2}/2)^2$, I would normally cancel out $(\sqrt{2}/2)$ in both the numerator and denominator so I am left with $1/(\sqrt{2}/2)$. I would then multiply the numerator by the reciprocal to come to $2/\sqrt{2}$.

HOWEVER, when I take $(\sqrt{2}/2)/(\sqrt{2}/2)^2$ and multiply out the denominator, I get $(\sqrt{2}/2)/(2/4)$, I then multiply the numerator by the recipcrocal to get an answer of $\sqrt{2}$....

I should get the same thing shouldn't I, where am I going wrong? Thanks for your help.

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    $\begingroup$ They are the same. $\endgroup$
    – Randall
    Commented Aug 17, 2017 at 1:17
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    $\begingroup$ as is $2\sqrt 2\over2$ there are a lot of equivalent forms. $\endgroup$
    – user451844
    Commented Aug 17, 2017 at 1:22
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    $\begingroup$ At any rate, congrats. You did the problem correctly, twice. $\endgroup$
    – Randall
    Commented Aug 17, 2017 at 1:27

2 Answers 2

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In case you care to see it worked out...

$$ \frac{2}{\sqrt{2}} = \frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} =\sqrt{2} $$

I guess teachers call this "rationalizing the denominator" or something.

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Using the fact that $a^na^m=a^{n+m}$ or $ b^n/b^m=b^nb^{-m}=b^{n-m} $:

$$ \frac{2}{\sqrt{2}}=\frac{2^1}{2^{1/2}} = 2^1\,2^{-1/2} = 2^{1-1/2}=2^{1/2} = \sqrt{2} $$

i.e. the algebra of exponents.

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  • $\begingroup$ Of course! Ah I wasn't thinking! Thanks so much. $\endgroup$
    – Shezmula
    Commented Aug 17, 2017 at 2:17

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