# How to add compound fractions?

How to add two compound fractions with fractions in numerator like this one:

$$\frac{\ \frac{1}{x}\ }{2} + \frac{\ \frac{2}{3x}\ }{x}$$

or fractions with fractions in denominator like this one:

$$\frac{x}{\ \frac{2}{x}\ } + \frac{\ \frac{1}{x}\ }{x}$$

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"fractions in denominator" - for those you reciprocate the denominator. –  Guess who it is. Jul 14 '11 at 10:25
It might be better to use complicated instead of complex. –  Henry Jul 14 '11 at 22:02
Or "compound fractions". –  Américo Tavares Jul 14 '11 at 22:04
I changed "complex fractions" to "compound fractions" in the title. –  Américo Tavares Jul 15 '11 at 7:46

The multiplicative inverse of a fraction a/b is b/a. (Wikipedia)

• Division by a number or fraction is the same as multiplication by its inverse or reciprocal.

Division by $r$ is equal to the multiplication by $\dfrac{1}{r}$: $$\dfrac{\ \dfrac{p}{q}\ }{r}=\dfrac{p}{q}\cdot \dfrac{1}{r}=\dfrac{p\cdot 1}{q\cdot r}=\dfrac{p}{q r}, \quad (1)$$

Division by $\dfrac{t}{u}$ is equal to the multiplication by $\dfrac{u}{t}$:

$$\dfrac{\ s}{\ \dfrac{t}{u}\ }=s\cdot\dfrac{u}{t}=\dfrac{s\cdot u}{t}=\dfrac{su}{t}.\quad (2)$$

• Sum of fractions

$$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}.\quad (3)$$

Apply $(1)$ to

$$\dfrac{\ \dfrac{1}{x}\ }{2}=\dfrac{1}{x}\cdot\dfrac{1}{2}=\dfrac{1\cdot 1}{x\cdot 2}=\dfrac{1}{2x}$$

and $(2)$ to

$$\frac{\ \dfrac{2}{3x}\ }{x}=\dfrac{2}{3x}\cdot\dfrac{1}{x}=\dfrac{2\cdot 1}{3x\cdot x}=\dfrac{2}{3x^2}.$$

So by $(3)$ we have

$$\dfrac{\ \dfrac{1}{x}\ }{2} + \dfrac{\ \dfrac{2}{3x}\ }{x}=\dfrac{1}{2x}+\dfrac{2}{3x^2}=\dfrac{1\cdot 3x^2+2\times 2x}{2x\cdot 3x^2 }=\dfrac{3x^2+4x}{6x^3}=\dfrac{x(3x+4)}{x(6x^2)}=\dfrac{3x+4}{6x^2}.$$

For

$$\dfrac{x}{\; \dfrac{2}{x}\ } + \dfrac{\; \dfrac{1}{x}\; }{x}$$

we have

$$\dfrac{\; x\; }{\dfrac{2}{x}} + \dfrac{\; \dfrac{1}{x}\; }{x}=\dfrac{x\cdot x}{2} + \dfrac{1}{x\cdot x}=\dfrac{x^2}{2}+\cfrac{1}{x^2}=\dfrac{x^2\cdot x^2+2\cdot 1}{2\cdot x^2}=\dfrac{x^4+2}{2x^2}.$$

We can apply the property Division by a fraction is the same as multiplication by its inverse or reciprocal to the following fraction

$$\dfrac{\;\dfrac{a}{b}\;}{\dfrac{c}{d}}=\dfrac{a}{b}\cdot \dfrac{d}{c}=\dfrac{a\cdot d}{b\cdot c}=\dfrac{ad}{bc}\qquad (4).$$

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You can clarify compound fractions in LaTeX by adding dummy space in either denominator or numerator; e.g., \frac{\ \frac{a}{b}\ }{\frac{c}{d}} vs. \frac{\frac{a}{b}}{\frac{c}{d}} to get $$\frac{\ \frac{a}{b}\ }{\frac{c}{d}}\quad\mathrm{vs.}\quad \frac{\frac{a}{b}}{\frac{c}{d}}.$$This might be useful in some of your expressions. –  Arturo Magidin Jul 14 '11 at 21:08
@Arturo: Thanks! I didn't know. –  Américo Tavares Jul 14 '11 at 21:10
I improved formatting of compound fractions, as per Arturo Magini's advice. –  Américo Tavares Jul 14 '11 at 21:32
@Arturo: done. (filling text) –  Américo Tavares Jul 14 '11 at 21:35
The simplification above of $$\dfrac{x(3x+4)}{x(6x^2)}=\dfrac{3x+4}{6x^2}$$ is valid if and only if $x\ne 0$. –  Américo Tavares Jul 14 '11 at 22:02

One easy way to figure this out is that dividing by $x$ is the same as multiplying by $1/x$ (but all bets are off when $x=0$, as division by $0$ is undefined). So

\begin{align*} \frac{ \frac{a}{b} }{c} &= \frac{1}{c} \frac{a}{b} = \frac{a}{bc} \\ \\ \frac{a}{\frac{b}{c}} &= a \frac{1}{\frac{b}{c}} = a \frac{c}{b} = \frac{ac}{b} \end{align*}

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And yet generations of students keep insisting on ambiguously stacking fractions... –  Simon Jul 14 '11 at 10:52
I'd like to point out, that, however, $x= \frac{1}{\frac{1}{x}}$ only when $x\not=0$. –  lentic catachresis Jul 14 '11 at 14:27
@Bruno: Well noted. The solution has been updated to reflect that. –  Michael Chen Jul 14 '11 at 14:38

Here is a start for the first one:

$$\frac{\frac{1}{x}}{2} + \frac{\frac{2}{3x}}{x} = \frac{x\frac{1}{x}}{2x} + \frac{2\frac{2}{3x}}{2x} = \frac{1}{2x} + \frac{\frac{4}{3x}}{2x} = \frac{1}{2x} + \frac{4}{3x}\frac{1}{2x} = \frac{1}{2x} + \frac{4}{6x^2} = \frac{1}{2x} + \frac{2}{3x^2}$$

Now try to derive $\displaystyle\frac{3x + 4}{6x^2}$ from this.

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