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Multiplying Fractions Clarification

  1. Oct 20, 2012 #1
    Hi! Im quite new at Maths in general, just recently started get an intrest for it (Aswell as general physics/philosophy) and im trying to learn on my own, so if this question is totally retarded, feel free to let your anger out haha.
    Anyway, I borrowed a book from a friend that looked fun, "Mathemathics for Engineers".
    And after 5 pages i ran into some trouble, with one of the explanations in the book.
    So they show how we multiply fractions, this is the example used:

    [itex]\frac{2}{3}[/itex] * [itex]\frac{5}{12}[/itex] = [itex]\frac{2*5}{3*12}[/itex]=[itex]\frac{1*5}{3*6}[/itex]=[itex]\frac{5}{18}[/itex]

    So what im not really understanding is how we go from [itex]\frac{2*5}{3*12}[/itex] to [itex]\frac{1*5}{3*6}[/itex].

    What it looks like is that they have divided 2 by 2 and 12 by 2. I understand why this is done, to shorten the number down, the results will still be same, 5/18 or 10/36. But why the 2 and 12? Just because they are the easiest ones to shorten? And if so, which are the rules for shortening numbers down?

    Best regards,
  2. jcsd
  3. Oct 20, 2012 #2


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    Dearly Missed

    Well, you could expand your fraction first as:
    (2*5)/(3*12)=(2*5)/(3*2*6) Agreed that this is legal, writing 2*6 instead of 12?
    Now, knowing the multiplication rule, you can do it BACKWARDS as well, writing:

    (2*5)/(3*2*6)=(2/2)*(5/(3*6)) (That is, "demultiplying" the fraction 2/2 from the other fraction!)

    But, a number divided by itself equals 1, so you get:

    Since multplying any number with 1 equals itself, we may collect our results as, precisely,
  4. Oct 20, 2012 #3


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    The details of why this works has to do with prime numbers. Essentially every number can be expressed as a multiple of prime numbers, such as

    6 = 2*3
    12 = 2*2*3
    18 = 2*3*3


    And prime numbers are already in the form we require, so

    7 = 7*1

    So if we took the long approach of changing every number in the fraction into its prime number factors, we would get

    [tex]\frac{2*5}{3*12} = \frac{2*5}{3*(2*2*3)} [/tex]

    Now, all we need to do from here is cancel out any common factors from the numerator and denominator, so we cancel out 2.

    If we were to try and cancel say, the 5 and 3 or the 5 and 12, we will quickly realize there is no way to do it. Why? Well, just as how we could express every number as a multiple of primes, the theorem also states that every number has a unique representation of prime multiples. You can't multiply some prime numbers together to get a number and then find other primes to multiply together to get to that same number.

    But of course we aren't going to take this long approach of breaking each number down into its prime factors, then cancelling. After enough practice, you should just be able to spot small common factors and cancel them instantly. For example, in the problem you presented, the 2 and 12 were both even, so they both have a factor of 2, so you'd start by cancelling that.
  5. Oct 20, 2012 #4
    Thanks alot for the fast replies! <3 Understand it now!
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