10
$\begingroup$

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?

$\endgroup$
1
  • 10
    $\begingroup$ Because we would like the rule $x^nx^m = x^{n+m}$ to be preserved when extending exponentiation to negative exponents. This forces you to define $x^0 = 1$, and that in turn enforces $x^{-n} = 1/x^n$. $\endgroup$
    – Alex B.
    Commented Oct 12, 2011 at 4:12

5 Answers 5

23
$\begingroup$

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

$\endgroup$
8
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
2
  • 4
    $\begingroup$ No need to "accept" that $5^0=1$. This follows by the same reasoning. $\endgroup$
    – Alex B.
    Commented Oct 12, 2011 at 4:17
  • 2
    $\begingroup$ Alex, you offered an excellent answer which has been accepted for good reason. Without taking anything away from that, my experience has been that some people understand (or at least, claim to understand) that $5^0 = 1$ while still being confused about negative exponents. So I started there. I suppose I could have used "know" instead of accept. In any case, I never wrote that anyone needs to accept anything. $\endgroup$
    – Adam Saltz
    Commented Oct 12, 2011 at 17:48
0
$\begingroup$

$$ \begin{align} a\cdot 3^{10} & = a\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3 \\ \\ a\cdot 3^6 & = a\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3 \end{align} $$ 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.

$10 = 6 + 4$

$6 = 10 + (-4)$

$$ \begin{align} a\cdot 3^{10} & = a\cdot 3^{6+4} \\ \\ a\cdot 3^6 & = a\cdot 3^{10 + (-4)} \end{align} $$

$\endgroup$
0
$\begingroup$

There are plenty of examples. It's just logical. $$ \begin{align} 10^{5} &= 100000\\ 10^{4} &= 10000\\ 10^{3} &= 1000\\ 10^{2} &= 100\\ 10^{1} &= 10\\ 10^{0} &= 1\\ 10^{-1} &= .1 = 1/10\\ 10^{-2} &= .01 = 1/100\\ 10^{-3} &= .001 = 1/1000\\ 10^{-4} &= .0001 = 1/10000\\ 10^{-5} &= .00001 = 1/100000 \end{align} $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .