# When dividing by a fraction, why can you not take the reciprocal of term involving addition/subtraction?

Given something like:

$$\frac{a}{\frac{a}{b}}$$ You would multiply the numerator $a$ by the reciprocal of the denominator, $\tfrac ba$ to get: $$a\cdot\frac{b}{a}= \frac{ab}{a}=b$$ Given $$\frac{1}{\frac 1a + \frac 1b}$$ By taking the LCM and adding the denominators you get: $$\frac{1}{\left(\frac{a+b}{ab}\right)}$$ Given the reciprocal division rule in example one: $$\frac{ab}{a+b}$$

Why can you not take the reciprocal of $\frac 1a + \frac 1b$ to begin with? I did this and ended up with: $$1\cdot\left(\frac a1 + \frac b1\right) =a + b$$

However $\frac{1}{1/a + 1/b}$ is not the same as $a+b$ so this is incorrect. I was trying to find a similar example online but I could not, why is this incorrect? Does the rule only work with one fraction as the denominator and not terms linked by addition and/or subtraction?

• This is exactly the reason you should be using Mathjax! – Bram28 Nov 11 '17 at 21:54
• Sorry! I am reading the page now! – Sphygmomanometer Nov 11 '17 at 21:56
• Cool! I did the first edit for you ... – Bram28 Nov 11 '17 at 21:59
• Thanks, I am working on the rest! – Sphygmomanometer Nov 11 '17 at 22:03
• Your edit made the result of the first multiplication $a$ instead of $b$. – Ross Millikan Nov 11 '17 at 22:09

## 2 Answers

When you took the reciprocal of $\frac 1a+\frac 1b$ to get $a+b$, you implicitly assumed that the reciprocal of a sum is the sum of the reciprocals. This is not true.

Note that $\frac 1a$ is $a^{-1}$. Just as the square of a sum is not the sum of the squares (in general), that is, $(x+y)^2\neq x^2+y^2$, so also the reciprocal of a sum is not in general the sum of the reciprocals: $(x+y)^{-1}\neq x^{-1}+y^{-1}$.

• Nope, not me. My name is pretty common. – Michael Weiss Nov 11 '17 at 23:00

The reciprocal of $\frac{1}{a} + \frac{1}{b}$, is by definition $$\frac{1}{\frac{1}{a} + \frac{1}{b}} = \frac{1}{\frac{a+b}{ab}} = \frac{ab}{a+b}.$$

So your reasoning fails because $\frac{a}{1} + \frac{b}{1}$ is not the reciprocal of $\frac{1}{a} + \frac{1}{b}$.