Why is the math for negative exponents so? This is what we are taught:
$$5^{-2} = \left({\frac{1}{5}}\right)^{2}$$
but I don't understand why we take the inverse of the base when we have a negative exponent.  Can anyone explain why?
 A: If you start with $5^3$ and divide by $5^1=5$ you get $5^2$.  Then if you divide by $5$ you get $5^1=5$.  Then if you divide by $5$ you get $5^0=1$.  Then if you divide by $5$ you get $5^{-1}=\frac{1}{5}$.
A: For natural numbers $n$, $m$, we have $x^nx^m=x^{n+m}$. If you want this rule to be preserved when defining exponentiation by all integers, then you must have
$x^0x^n = x^{n+0} = x^n$, so that you must define $x^0 = 1$. And then, arguing similarly, you have $x^nx^{-n} = x^{n-n}=x^0=1$, so that $x^{-n}=1/x^n$.
Now, you can try to work out for yourself what $x^{1/n}$ should be, if we want to preserve the other exponentiation rule $(x^n)^m = x^{nm}$.
A: We know that positive exponents add, e.g. $5^3 \times 5^2 = 5^5$.  If you accept that $5^0 = 1$, then it makes sense that $5^{-2} \times 5^2 = 5^0 = 1$.  That means that $5^{-2} = \frac{1}{5^2}$.
A: $$
\begin{align}
a\cdot 3^{10} & = a\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3 \\  \\
a\cdot 3^6 & = a\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3
\end{align}
$$
To go from the second line to the first, multiply by 3 four times.
To go from the first to the second, multiply by 3 minus four times.
$10 = 6 + 4$
$6 = 10 + (-4)$
$$
\begin{align}
a\cdot 3^{10} & = a\cdot 3^{6+4} \\  \\
a\cdot 3^6 & = a\cdot 3^{10 + (-4)}
\end{align}
$$
A: There are plenty of examples. It's just logical.
$$
\begin{align}
10^{5} &=  100000\\
10^{4} &=  10000\\
10^{3} &=  1000\\
10^{2} &=  100\\
10^{1} &=  10\\
10^{0} &=  1\\
10^{-1} &= .1 = 1/10\\
10^{-2} &= .01 = 1/100\\
10^{-3} &= .001 = 1/1000\\
10^{-4} &= .0001 = 1/10000\\
10^{-5} &= .00001 = 1/100000
\end{align}
$$
