Dividing by a Fraction
Date: 7 Jun 1995 13:15:08 -0400 From: Cad09 Subject: Dividing by a Fraction Hello, my name is Janet. I am a sixth-grade student at Cadwallader Elementary school in San Jose, California. I have a question: Why is it when dividing fractions, you have to multiply by the reciprocal? Thank you.
Date: 7 Jun 1995 13:54:24 -0400 From: Dr. Ken Subject: Re: Dividing by a Fraction Hello Janet. Well, the answer has to do with what division IS. I'll bet that before you learned division, you learned multiplication. You learned that 6 x 7 = 42, and then a while later you learned that 42 % 6 = 7. In this sense, multiplication and division do opposite jobs; the technical term for this is that multiplication and division are "inverse" operations. (This computer doesn't have the normal division symbol that you may be used to, so I'm going to use the % symbol when I mean divide) You may also have learned that whenever you divide by a number, that's really the same thing as multiplying by the "inverse" of that number. When I say inverse here, I mean "one over that number." Like 42 % 6 is the same as 42 x 1/6. Technically, what the inverse of a number means is that it's a number you can multiply your first number by to get 1, for instance the inverse of 6 is 1/6 since 6 x 1/6 = 1, and the inverse of -3/4 is -4/3, since -3/4 x -4/3 = 1. Now look at this. One way we can write division is to write it as a fraction: the number 42 % 6 is the same as the number 42 --- 6 . So let's say we have the division problem 42 7 --- % --- 6 3 . Instead of writing that as a division problem, we can write it as a multiplication problem: dividing by a number is the same as multiplying by its inverse. So what's the inverse of 7/3? It's 3/7, since 7/3 x 3/7 = 1. Now we can rewrite the problem as 42 3 --- x --- 6 7 , and then you probably know how to do it from there: 42 3 42 x 3 6 x 1 6 --- x --- = ------- = ------- = --- = 3. 6 7 6 x 7 2 x 1 2 ^ | | |In this step, I cancelled the 42 with the 7, and the 3 with the 6. Thanks for the question! -Ken "Dr." Math
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum