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 method. "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 with. I hope this helps. Write back if you'd like to talk more about this, or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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 http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum