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?

Solution 1:

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 $(x^n)^m = x^{nm}$.

Solution 2:

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 $5^{-1}=\frac{1}{5}$.

Solution 3:

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 $5^{-2} = \frac{1}{5^2}$.