Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

How to shorten this fraction?

$R_1+R_2$ divided by $\frac1{R_1} + \frac1{R_2}$

The answer is $R_1R_2$. I just don't know how to get there.

share|cite|improve this question

First, simplify the denominator:

$$\frac{1}{A}+\frac{1}{B} = \frac{B+A}{AB},$$

and now we can simplify the whole "castle":

$$\frac{A+B}{\frac{1}{A}+\frac{1}{B}} = \frac{A+B}{\frac{A+B}{AB}}= \frac{AB}{A+B}\cdot \frac{A+B}{1} = AB.$$

share|cite|improve this answer
Welcome to the site! – Zev Chonoles Aug 17 '11 at 19:07
Thank you, Zev! – Álvaro Lozano-Robledo Aug 17 '11 at 19:14

We can use a common trick for simplifying a fraction, multiplying by an expression equal to 1: $$\frac{R_1+R_2}{\frac{1}{R_1}+\frac{1}{R_2}}=\frac{R_1+R_2}{\frac{1}{R_1}+\frac{1}{R_2}}\cdot\left(\frac{R_1R_2}{R_1R_2}\right)=\frac{(R_1+R_2)(R_1R_2)}{\left(\frac{1}{R_1}+\frac{1}{R_2}\right)(R_1R_2)}=\frac{(R_1+R_2)(R_1R_2)}{R_2+R_1}=R_1R_2$$

share|cite|improve this answer

Multiply through by $R_1R_2$:


share|cite|improve this answer

Write $$ \frac{R_1+R_2}{\frac{1}{R_1}+\frac{1}{R_2}} $$ Multiply both numerator and denominator by $R_1R_2$ and cancel the $R_1+R_2$ in the numerator and denominator.

share|cite|improve this answer

$R_1 + R_2$ divided by $\frac{1}{R_1} + \frac{1}{R_2}$

= $R_1 + R_2$ divided by $\frac{R_2}{R_1 R_2} + \frac{R_1}{R_1 R_2}$

= $R_1 + R_2$ divided by $\frac{R_2 + R_1}{R_1 R_2}$

= $R_1 + R_2$ times $\frac{R_1 R_2}{R_2 + R_1}$

= $\frac{R_1 + R_2}{1} \cdot \frac{R_1 R_2}{R_2 + R_1} = \frac{1}{1} \cdot \frac{R_1 R_2}{1}$

$= R_1R_2$

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.