Reducing Fractions with Large Numbers

Date: 03/18/2004 at 15:46:58
From: Sonya
Subject: What is the answer to this Fraction

If you reduced 32/48 to lowest term, what would the answer be?  Those
are big numbers, which makes it hard for me.

Date: 03/18/2004 at 16:23:16
From: Doctor Peterson
Subject: Re: What is the answer to this Fraction

Hi, Sonya.

Yes, they're big numbers; but you can reduce the fraction bit by bit 
rather than having to do it all in one step--it's a lot like dieting!  
That makes the problem a lot less scary.

Let's take a worse example: 280/588

What we want to do is to divide the numerator and denominator by a 
common factor.  Each time we do that, we are reducing the fraction; 
eventually we will reach a point where we can't do it any more, and 
then we're done.

So we can start with any obvious common factors (such as 10, if they 
both ended with 0); otherwise, just start with the smallest prime 
number, 2, and see if it goes into both numbers.  In this case, it 
does; they're both even.  So we divide 280 by 2 and get 140; we divide 
588 by 2 and get 294.  Now we've reduced the fraction to 140/294.

Then we do the same thing again!  Both numbers are still even, so we 
can divide them again by 2.  (That's important: after you've divided 
by one prime number, you have to check the same prime again, because 
you may not be done with it.)  So we get 70/147.

Now only one number is even, so we move on to the next prime number, 
3.  Neither number is divisible by 3, so we move on to 5, which only 
divides 70; and then to 7, which does divide both numbers.  Dividing 
by 7, we get 10/21.

Now we don't need to try any more primes, because the only factors of 
10 are 2 and 5, and we've already tried them.  We know we're done when 
we can try all the prime factors of one number, and they aren't 
factors of the other.

So we've reduced a really big, ugly fraction, 280/588, to 10/21, 
without ever having to divide by a big number.  The weight came off in 
manageable bits.

If you'd like more examples, see this page:

  Reducing Fractions
    http://mathforum.org/library/drmath/view/58914.html 

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/