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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?

Michael Weiss
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Sphygmomanometer
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2 Answers2

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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 $\frac{1}{a} + \frac{1}{b}$.

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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: $(x+y)^{-1}\neq x^{-1}+y^{-1}$.

Michael Weiss
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