How is multiplying by a fraction the same as dividing by the inverse of that fraction

1. Aug 9, 2011

jimgavagan

Supposedly this proof answers my question.

8 / 16 = .5
8 / 8 = 1
8 / 4 = 2
8 / 2 = 4
8 / 1 = 8
8 / .5 = 16
8 * 2/1 = 8 / .5

I'm just wondering how this proof answers my question?

2. Aug 10, 2011

Dr. Seafood

Re: How is multiplying by a fraction the same as dividing by the inverse of that frac

It doesn't ... and that's not a proof o__O

It follows kind of from "definition" of division: $x \over y$ actually means $x \cdot y^{-1}$ where $y^{-1}$ is the number such that $y \cdot y^{-1} = 1$, called the "multiplicative inverse" or "reciprocal" of y. This number is unique. Obviously $({a \over b})^{-1} = {b \over a}$ (since ${a \over b} \cdot {b \over a} = 1$) as long as we have $a, b \ne 0$. (Also of note: $0^{-1}$ does not exist!

So it turns out that since ${x \over y} = x \cdot y^{-1}$, setting $y = {a \over b}$ we get

"${{x} \over {a \over b}}$" $= {x} \cdot ({a \over b})^{-1} = {x} \cdot {b \over a}$ (again provided $a, b \ne 0$). I hope this explanation helps.

Last edited: Aug 10, 2011
3. Aug 10, 2011

jimgavagan

Re: How is multiplying by a fraction the same as dividing by the inverse of that frac

ooooooooooooooooooooooooooh awesome! :D

Very much appreciated!