# Homework Help: Evaluate an expression with a “new” operation

Problem Statement:

Imagine that a new operational symbol for mathematics has have been developed. This symbol is $\sim$ and is represented by $$a\sim b=\frac{ab}{a-b}$$ Using this new symbol, find the value of $(2\sim3)\sim4$

Any help?

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Welcome to Math.SE! Since you appear to be new here, I have a couple of things I'd like to share. First, we like problems that show attempts at solutions; however, if you don't even know where to start, that's fine--but please just say so. :) Second, we like descriptive titles that can help others know exactly what the question is about. Third, the "homework" tag is something we call a "meta" tag. This basically means it should be used with some other tag, and not by itself. –  anorton Jan 30 '13 at 2:59
Question: Do you really mean to write $(2-4)-4$? Or do you mean to actually use the new operational symbol somewhere? –  anorton Jan 30 '13 at 3:00
The answer is $-2.4$. Can you replicate this? –  gnometorule Jan 30 '13 at 3:01
It's obviously a typo. It's the new symbol. –  gnometorule Jan 30 '13 at 3:02
Additions to the "welcome:" Fourth Some people around here find questions phrased in the imperative (e.g. "Find..." or "Solve...") to be rude, and can trigger downvoting. Fifth We also use a tool called LaTeX to render math. I've suggested an edit that adds this-- please make sure I haven't changed the problem by accident. :) I'm writing up an answer now, btw. –  anorton Jan 30 '13 at 3:06

We want to evaluate: $$(2\sim3)\sim4$$

First, simplify the parenthesis. Note that, for the part in the parenthesis, $a=2, b=3$. $$a\sim b = \frac{ab}{a-b}$$ $$2\sim3 = \frac{(2)(3)}{2-3}$$ $$2\sim3 = \frac{6}{-1} = -6$$

Now, your expression is: $$(-6)\sim4$$

I'll leave the rest for you to do. If you need more help (or need more of the problem worked out), let me know! :)

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Assuming that you meant a~b=ab/(a-b) then 2~3 = 2*3/(2-3) = -6 and (2~3)~4 = (-6)~4 which I will let you work out for yourself.

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Below is a way of viewing these operations that will greatly simplify computing big $\sim$ products.

$$\begin{eqnarray}\rm a\sim b\ &=&\rm\ \frac{ab}{a-b}\, =\, \frac{1}{b^{-1}-a^{-1}}\, =\, \frac{1}{b'-a'}\ \ where\ \ x' = x^{-1}\\ \\ \rm\Rightarrow\ \ (a\sim b)' &=&\ \rm b'-a'\\ \\ \rm\Rightarrow\ \ ((a\sim b)\sim c)' &=&\ \rm c'-(a\sim b)'\\ \\ &=&\ \rm c'-(b'-a')\\ \\ &=&\,\rm \frac{1}4-\left(\frac{1}3-\frac{1}2\right)\\ \\ \\ &=&\, \frac{3-(4-6)}{12}\, =\, \frac{5}{12} \\ \\ \rm\Rightarrow\ \ (a\sim b)\sim c &=&\rm\ \left(\frac{5}{12}\right)' = \frac{12}5\quad \text{by taking prime of prior} \end{eqnarray}$$

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