# Cancelling out problem (2 different answers?)

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.

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

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.

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.

• Of course! Ah I wasn't thinking! Thanks so much. Aug 17, 2017 at 2:17