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

by jimgavagan
Tags: dividing, fraction, inverse, multiplying
jimgavagan is offline
Aug9-11, 11:44 PM
P: 24
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?
Phys.Org News Partner Mathematics news on
Hyperbolic homogeneous polynomials, oh my!
Researchers help Boston Marathon organizers plan for 2014 race
'Math detective' analyzes odds for suspicious lottery wins
Dr. Seafood
Dr. Seafood is offline
Aug10-11, 12:42 AM
P: 120
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 is offline
Aug10-11, 01:37 AM
P: 24
ooooooooooooooooooooooooooh awesome! :D

Very much appreciated!

Register to reply

Related Discussions
This is going to be a really difficult question to ask. Multiplying and dividing by Calculus & Beyond Homework 22
Find closest fraction to another fraction Precalculus Mathematics Homework 3
Dividing/Multiplying Imaginary Numbers General Math 6
Evaluate fraction x fraction Precalculus Mathematics Homework 4
Expressing this algebraic fraction as partial fraction Precalculus Mathematics Homework 6