Can someone explain how 2^-1=0.5? I know it's a stupid question, but my head is stuck.
I have this :
Binary System Problem
-3
$\begingroup$
binary
binary-programming
-
0$2^{-1}=\frac1{2^1}$. The general rule is $a^{-b}=\frac{1}{a^b}$ – 2017-01-18
2 Answers
1
Hint: $2^1\cdot 2^{-1}=2^{1-1}=2^0=1$. Since $2^1=2$, $2^{-1}$ be the number that gives $1$ when multiplied by $2$, i.e., $\frac12$.
2
I'm not sure if my explanation will answer your question but you could think of it like this:
$2 = 2^1$ then to get $2^0$ you need to divide $2^1 = 2$ by $2$ (by itself). So $2^{1-1} = 2^0 = 2/2 = 1$.
The same reasoning, to achieve $2^{-1}$ = $2^{0-1} = 2 ^ {-1} = 2^{0} / 2 = 1/2$
-
0Congrats on your first answer. You could perhaps make this even more apparent by starting at (say) $2^3=8$ and working downwards from there, since some people find $2^0=1$ a bit mysterious also. – 2017-01-18