Date: 10/16/97 at 10:41:16 From: Jimmy De Armas Subject: Pre-Algebra I was reading over the explanation on dividing fractions and why we have to flip the second fraction, and I am still confused. I understand that we have to do it in order to do the problem, but I want to know the reason why we have to flip it. I want a simpler explanation.
Date: 10/16/97 at 12:12:36 From: Doctor Rob Subject: Re: Pre-Algebra In a fraction its value represents the number of things of size measured by the denominator you add up to get one thing of size measured by the numerator. For example, 31/11 is the number of 11-pound objects you put together to to get one 31-pound object. When the denominator is itself a fraction, as in your situation, this does not change. For example, (14/3)/(2/5) is the number of objects, each weighing 2/5 pound, which are put together to get a weight of 14/3 pounds. How many does it take to make one pound? Answer: 5/2, each weighing 2/5 pound, will make one pound. How did we get 5/2? By inverting 2/5, or, in other words, finding its "reciprocal." Why is this the right answer? Because (5/2)*(2/5) = (5*2)/(2*5) = 10/10 = 1. Then to get 14/3 pounds it will take (14/3)*(5/2) objects, each weighing 2/5 pound. (Of course this equals 35/3, so you'll need 11 and 2/3 objects each weighing 2/5 pound to make 14/3 pounds.) There is nothing special about 14, 3, 2, and 5. They could be replaced by any four numbers - except zero: remember, you can't divide by zero! Another way to look at this is to start with your original compound fraction, and multiply the numerator and denominator of the fraction by 5/2. You get: 14 14 5 14 5 -- -- * - -- * - 3 3 2 3 2 14 5 ---- = -------- = -------- = -- * - 2 2 5 1 3 2 - - * - 5 5 2 Why did we pick 5/2? Because 5/2 is the reciprocal of 2/5, the denominator, and when you multiply any number by its reciprocal, you get 1, which is what we want to create in the denominator. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum