Ordering Improper Fractions

Date: 10/12/2002 at 16:39:40
From: Raphie
Subject: Ordering improper fractions

Dr. Math, 

I have tried to figure out how to explain ordering improper fractions. 
But I don't even understand how to do it in the first place, meaning I 
won't be able to explain it! Please tell me how to do this so I can 
finish my math homework.

Sincerely, Raphie

Date: 10/12/2002 at 17:01:32
From: Doctor Sarah
Subject: Re: Ordering improper fractions

Hi Raphie - thanks for writing to Dr. Math.

Putting improper fractions in order is easier than working with mixed 
numbers. Just give them a common denominator. For example, put the 
following improper fractions in order:

   5/2   6/1   8/3   11/6

The common denominator is 6:

   5/2 = 15/6
   6/1 = 36/6
   8/3 = 16/6
  11/6 = 11/6

So in order, least to greatest:

  11/6 = 11/6
   5/2 = 15/6
   8/3 = 16/6
   6/1 = 36/6

Does this help?

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

Date: 10/12/2002 at 17:10:00
From: Doctor Ian
Subject: Re: Ordering improper fractions

Hi Raphie,

Suppose you have to order some integers, like

   5, 2, 1, 9, 3

At some point, you need to compare pairs of numbers, e.g., 

  5 < 9

right?  Now, that's easy for integers, and harder for fractions, but 
there is one technique that works for both. That is this: Divide one 
number by the other; if you end up with something less than 1, the 
first number is smaller; if you end up with something greater than 1, 
the first number is larger. (If you end up with exactly 1, the numbers 
are equal.)

With integers, it looks like

  5/9 is less than 1, so 5 < 9

  7/4 is greater than 1, so 7 > 4

Now, what if we have two fractions, like 1/3 and 2/5?  We can do the 
same thing:

  1/3    1   5   5
 ----- = - * - = -     
  2/5    3   2   6

which is less than 1, so 1/3 is smaller than 2/5.  For fun, let's try 
it the other way:

  2/5    2   3   6
 ----- = - * - = -     
  1/3    5   1   5
  
which is greater than 1, so 2/5 is larger than 1/3.  (That's good, 
because if we'd arrived at a different answer, we'd be in trouble.)

Let's try the fractions 2/3 and 14/21:

   2/3     2   21   42
 ------- = - * -- = --
  14/21    3   14   42

which is equal to 1... so 2/3 and 14/21 have the same value!

Does this make sense?  

Note that this works for ANY numbers - integers, fractions, and 
improper fractions too. So if I have two improper fractions like 21/15 
and 39/24, then

  21/15    21   24   
 ------- = -- * -- = ...
  39/24    15   39

Now, before I multiply all these out, I want to see if I can cancel 
some factors, because that will make my life easier.  

I know that 21 and 39 are both divisible by 3, so I can divide both of 
them by 3 to get

  21/15     7   24   
 ------- = -- * -- = ...
  39/24    15   13

Also, 15 and 24 are both divisible by 3:

  21/15     7    8
 ------- = -- * -- = ...
  39/24     5   13

I don't see any more cancellations I can do, so I'll go ahead and 
multiply:

  21/15     7    8   56
 ------- = -- * -- = --
  39/24     5   13   65

which is less than 1, so I know that 21/15 is less than 39/24.

Now, with improper fractions, there is something else you can try, 
which is changing them into mixed numbers. For 21/15 and 39/24, this 
won't help much, because we get 1 6/15 and 1 15/24, and we still have 
to compare the fractional parts.  

But if you have improper fractions like 39/24 and 105/41, then 
converting to mixed numbers, 

  39/24 = 1 15/24       105/41 = 2 23/41

you can see right away that one plus a proper fraction is smaller than 
two plus a proper fraction, regardless of what the fractions are. Does 
that make sense? 

Finally, sometimes you might be able to get all the improper fractions 
to have the same denominator.  For example, to order

  8/3, 14/12, 21/16

you might notice that 48 is divisible by all these denominators, so 
you can convert to the equivalent fractions

  8/3 * 16/16 = 128/48

  14/12 * 4/4 = 56/48

  21/16 * 3/3 = 63/48

and now you can just forget about the denominators, so it's just 
like trying to order 63, 64 and 128.

Does this help?

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