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Checking Proportions using Square Roots

Date: 04/11/2002 at 09:15:00
From: Erica Lasko
Subject: True proportions using square roots?

I am an 8th grade math teacher and our class is studying proportions.  
While we were discussing different ways to decide if a specific 
proportion is a true proportion (by cross-products or reducing the 
fractions or decimal equivalents), a student suggested the following 

"Multiply the two fractions, and then take their square root. If it 
equals one of the fractions you began with, it is a true proportion."

We tried his method with several examples and while it seems to work, 
I was wary of saying it will work all of the time. What do you think?  
My class awaits your answer.

Date: 04/11/2002 at 11:10:32
From: Doctor Ian
Subject: Re: True proportions using square roots?

Hi Erica,

That's a pretty clever observation!

Let's see what happens when we follow his advice. If we have a true
proportion, then it will look like this:

  a   ka
  - = --
  b   kb

where k is the constant of proportionality. If we multiply the two
fractions together, we get

  a   ka   a * ka
  - * -- = ------
  b   kb   b * kb

           k   a * a
         = - * -----
           k   b * b

           a * a
         = -----
           b * b

And the square root of this will, in fact, be one of the fractions, 
assuming that at least one of the fractions was in reduced form to 
begin with. 

But suppose this isn't true, i.e., suppose we start with 

  ma   kma
  -- = ---
  mb   kmb

then the product is

  ma  kma   ma * kma
  - * --- = --------
  mb  kmb   mb * kmb

           km   a * a
         = -- * -----
           km   b * b

           a * a
         = -----
           b * b

And the square root of this won't be either of the fractions.  So the
trick works for, say,

  2    8             16           4     2
  - = --   =>  sqrt( -- ) = sqrt( - ) = -
  3   12             36           9     3

but it doesn't work for, say,

  4    8             32           4     2
  - = --   =>  sqrt( -- ) = sqrt( - ) = - 
  6   12             72           9     3

So it depends in part on how you're using the word 'equal'. If you say 
that the trick works in the latter case because 2/3 is 'equal' to 4/6, 
then you're right, the trick works, but in order to establish that it 
works, you have to solve the same kind of problem that you started out 

I hope this helps. Write back if you'd like to talk more about this, 
or anything else.

- Doctor Ian, The Math Forum 

Date: 04/11/2002 at 22:07:29
From: Erica Lasko
Subject: True proportions using square roots?

Dear Dr. Ian,

I had a question about your reply.  If I use two fractions that are 
not reduced, such as 

  ma   kma
  -- = ---
  mb   kmb

when I multiply them, won't they equal

  ma   kma      k m m a a
  -- = ---   =  ---------   ?
  mb   kmb      k m m b b

Then if I take the square root of the result, it will be

      k m m a a     ma      k       ma
 sqrt ---------- =  -- sqrt -    =  --   
      k m m b b     mb      k       mb

so then that would be one of the fractions we started with.

But I understand what you said, that if we change to decimal 
equivalents then the idea works. I will present the solution to the 
class tomorrow.  I would appreciate it if you can comment on the 
argument above.  

Thanks so much for your help.  My class and I appreciate it.

Erica Lasko
Wayne Highlands Middle School, PA

Date: 04/12/2002 at 08:27:47
From: Doctor Ian
Subject: Re: True proportions using square roots?

Hi Erica,

You're right - I left out an m in both the numerator and denominator 
of the product.  But the idea is still the same (which is why the 
second example in my previous message ended up with a fraction other 
than the two I started with). 

That is, 
    ________     ______     ____
   | km^2 a^2    | km^2    | a^2
   | -------- =  | ---- *  | --- = a/b
  \| km^2 b^2   \| km^2   \| b^2

which is not one of the original fractions.  It doesn't matter whether 
you're canceling m/m or (m^2)/(m^2).  

To get back to the original fraction, you'd have to _partially_ reduce 
the fraction inside the square root, which would be sort of an odd 
thing to do. (If you know enough to cancel out a factor of k, but not 
a factor of m^2, that would indicate that you already _know_ that 
there is a constant of proportionality, which would raise the question 
of why you're doing the trick at all.) 

You're absolutely correct that changing everything to decimal 
equivalents would make the trick work, but changing the original 
fractions to decimal equivalents would make the trick unnecessary in 
the first place. That is, after you've done that, you can just compare 
the two decimal numbers directly:

      3   12
      - = --
      8   32

  0.375 = 0.375

Now the 'trick' reduces to taking the square root of a number 
multiplied by itself:

  0.375 = \| 0.375 * 0.375

In the end, the great value of thinking about this is that it's one of 
those situations that shows why working with variables can give you 
more insight than looking at examples using numbers.  

Does this help? 

- Doctor Ian, The Math Forum 
Associated Topics:
Middle School Fractions
Middle School Ratio and Proportion

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