The Secret to Mathematics?
Date: 09/27/2004 at 19:11:12 From: Peggy Subject: Dividing fractions with variable dividend or divisor I understand how to divide fractions--invert and multiply--but what if the dividend or divisor is a variable like 5/6 5 ----- = - n 4 The relationship of 5 to 5 is 1, and the relationship of 6 to 4 is 1/3. So I guess the answer would be one divided by one third, but then you have to invert it, right? So would that be three?
Date: 09/28/2004 at 02:46:09 From: Doctor Ian Subject: Re: Dividing fractions with variable dividend or divisor Hi Peggy, You're making this more complicated than it has to be. Think back to when you first learned about division, i.e., that 12 = 3 * 4 12 / 3 = 4 12 / 4 = 3 are all just different ways of expressing the same fact. Does this look familiar? Okay, so you have 5/6 ----- = 5/4 n What are the corresponding facts? They are 5/6 ----- = n 5/6 = n * 5/4 5/4 Does either of those get you closer to an answer? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Date: 09/28/2004 at 11:35:51 From: Peggy Subject: Dividing fractions with variable dividend or divisor No, I'm afraid not. I get the "reverse of division is multiplication" thing, but I'm not sure at what point I have to use a reciprocal.
Date: 09/28/2004 at 12:03:16 From: Doctor Ian Subject: Re: Dividing fractions with variable dividend or divisor Hi Peggy, If you rewrite 5/6 ----- = 5/4 n in the equivalent form 5/6 ----- = n 5/4 now you just divide 5/6 by 5/4... which you do by multiplying by the reciprocal of 5/4: 5 4 - * - = n 6 5 But you don't really need to use reciprocals at all, if you don't want to. For example, starting from 5/6 ----- = 5/4 n you could multiply both sides of the equation by n, to get 5 5 - = - * n 6 4 Now you can multiply both sides by 4, to kill the fraction on the right: 5*4 --- = 5 * n 6 Now multiply by 6 to kill the fraction on the left: 5*4 = 5*6*n The 5's cancel: 4 = 6*n Dividing by 6 on both sides solves for n: 4/6 = n Now, if you think about it, you'll see that I _did_ sort of multiply by the reciprocal of 5/4, but in several steps; and I could have done that in one step: 5/6 4 5 4 ----- * - = - * - n 5 4 5 5/6 * 4/5 ----------- = 1 n 4/6 ----------- = 1 n And now I can just multiply both sides by n. Or, if I'm comfortable with cross-multiplying, I could start by transforming the original equation into a proportion: 5/6 5/4 ----- = ----- n 1 Of course, I can always flip a proportion upside-down: n 1 ----- = ----- 5/6 5/4 And now I can multiply both sides by 5/6 to get 5/6 n = ----- 5/4 The point is to avoid getting locked into thinking that there is one "right" way to approach a problem. There are always lot of possible ways to go, and the habit you want to get into is stepping back and thinking about what the _easiest_ one will be. It's important to keep in mind that there is really only one rule in mathematics, which is that you're allowed to write down anything that's true. In algebra, this means you start with an equation (which you assume to be true, for purposes of finding out what follows from it), and so long as you do the same thing to both sides of the equation, what follows from it will be true as well (provided you don't divide by zero!). If you want to multiply both sides by something, go ahead. The worst that will happen is that it won't lead you towards a solution. Or it may take you on a longer path to the solution. But mainly, when you look at something like 5/6 ----- = 5/4 n you want to get into the habit of thinking something like, "Boy, I wish that n wasn't in the denominator", which suggests an immediate way to get it out of the denominator: multiply both sides by n. Or thinking, "Boy, I'd be happier if this looked like a proportion, 'cause those are easy", because you can always set up a proportion by dividing by 1. In either case, what you're looking for is a way to make the problem look more like something you know how to solve already. And if there is a secret to mathematics, this comes close to being it: Beyond simple arithmetic, you're almost never looking for an answer. What you're looking for is a way to transform your current problem into a simpler problem, which has the same solution. If you keep doing that, eventually you'll end up with a problem that's so simple you can see the solution just by looking at it. Does this make sense? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Date: 09/29/2004 at 11:13:53 From: Peggy Subject: Thank you (Dividing fractions with variable dividend or divisor) Wow, thanks! That makes sense! I wish someone would tell my TEACHER that there is more than one way to solve these things!
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