How is multiplying by a fraction the same as dividing by the inverse of that fractionby jimgavagan Tags: dividing, fraction, inverse, multiplying 

#1
Aug911, 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? 



#2
Aug1011, 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. 



#3
Aug1011, 01:37 AM

P: 24

ooooooooooooooooooooooooooh awesome! :D
Very much appreciated! 


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