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?
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$\begingroup$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 $$\frac{1}{2^2} = \frac{2^0}{2^2} = 2^{-2}.$$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$$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 $1$.
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