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

by jimgavagan
Tags: dividing, fraction, inverse, multiplying
 P: 120 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.