Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
The answer is $-2.4$. Can you replicate this? – gnometorule Jan 30 '13 at 3:01

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! :)

share|cite|improve this answer

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.

share|cite|improve this answer

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.