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

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?

Dr. Seafood

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

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

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

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

jimgavagan

ooooooooooooooooooooooooooh awesome! :D

Very much appreciated!