# Why does 2^-2 equal 1/2^2?

My intuitive way of thinking about it is that it is $2/2/2$ or $2/2^2$, So why then is it $1/2^2$? what is the flaw in my thinking?

• look at DanielV's answer math.stackexchange.com/questions/819218/negative-exponents-why – Jack Oct 24 '14 at 5:11
• Does $2^0=2$ in your world? $2^0=1$ in mine which makes sense as deducting one in the exponent is the same as dividing by 2 if one looks at some low powers of 2: 2,4,8,16, etc. – JB King Oct 24 '14 at 5:17
• Are you average? because you are mean. – Ray Kay Oct 24 '14 at 20:23

Repeated multiplication can be seen as $$\overset{\text{m terms}}{a \cdots a} = a^m.$$

Dividing this term repeatedly can then be seen as subtracting from $a^m$, because $$\frac{a^m}{a} = \frac{\overset{\text{m terms}}{a \cdots a}}{a} = \overset{\text{m - 1 terms}}{a \cdots a} = a^{m-1}.$$

So $$\frac{1}{2^2} = \frac{2^0}{2^2} = 2^{-2}.$$

Well, $2^{-n} := {1 \over 2^n} (n \geq 0)$ is a definition.

Why this definition?

Because the usual rules (e.g. $2^{n+m}=2^n2^m$) which were established for ($n,m\in \mathbb{N}$) are now true for all $n,m \in \mathbb{Z}$. Which is nice, and shortens a lot of proofs.

$2^2 = 2\cdot 2 = 4$, $2^1 = 2$, $2^0 = 1$, $2^{-1} = 1/2$, $2^{-2} = 1/4$.

I guess the problem in your way of thinking is that when you're thinking about multiplication, you should "start" from $1$, whereas you're "starting" from $2$. The first $2$ in your expression $2/2/2$ is playing a different role than the other two $2$s - it should really be a $1$.