# Simple algebra question…exponents?

I know this is very basic, but when evaluating

$\frac{b^{-4}}{b^{-4}}$

Apparently I cannot write it as $\frac{b^4}{b^4}$ by moving the expressions to make them positive. Why can't I do that? The answer is supposed to be $b^8$ not 1, like I originally thought.

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The answer is $1$. –  André Nicolas Nov 6 '12 at 1:44
I think you understanding is incorrect. That is clearly 1, ($b \ne 0$). The only way that is $b^8$ is the question is $\frac{b^4}{b^{-4}}$ –  Peter Grill Nov 6 '12 at 1:44
I was looking at my sister's math HW, and I told her it's 1, but apparently that was wrong. That's why I was confused. –  Newbie Nov 6 '12 at 1:45
Well whoever told you it was not 1 was wrong. –  Peter Grill Nov 6 '12 at 1:45
Then it is $b^{-8}$, or $\frac{1}{b^8}$. –  André Nicolas Nov 6 '12 at 1:53

Your answer is flawed. The answer is one because any number $x: x \in \mathbb{R}$ \ $0$ divided by itself equals one.

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Is 0/0 = 1?.... –  Ram Apr 5 '13 at 16:46
@recursiverecursion: it is usually best not to edit another's answer if only to correct a serious mistake. Such a correction is the answerer's responsibility (and choice). –  amWhy Mar 29 '14 at 14:44
Sorry about that. –  recursive recursion Mar 29 '14 at 15:23

Simply because $x/x=1$ for all non zero real numbers. $\frac {b^{-4}}{b^{-4}}$ is an expression of the form $x/x$ (assuming $b\neq0$) and so it equals $1$. One can show this using this other method: $$\frac{b^{-4}}{b^{-4}}={b^{-4}}\frac1{b^{-4}}={b^{-4}}{b^{4}}=b^{-4+4}=b^0=1$$

It is however $b^{-8}$ if $\dfrac{b^{-4}}{b^4}$ which can be shown using the method above.

I hope this helps.
Best wishes, $\mathcal H$akim.

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Thanks @amWhy ! Corrected. $\overset{\cdot\cdot}\smile$ –  Hakim Mar 29 '14 at 14:51