## Fractions are Hard!

### 3.8 A common confusion in algebra class

 Why is writing $$\dfrac{x+y}{a+b}=\dfrac{x}{a+b}+\dfrac{y}{a+b}$$ valid but writing $$\dfrac{x+y}{a+b}=\dfrac{x+y}{a}+\dfrac{x+y}{b}$$ is not?

This is a question about when it is valid to tease apart an addition.

The only belief of arithmetic we have that attends to teasing apart an addition is the Distributive Rule: $$a\left(b+c\right) = ab+ac$$.

Applying this rule to $$a=\dfrac{1}{n}$$ and we see $$\dfrac{1}{n}\times \left(b+c\right) = \dfrac{1}{n}\times b + \dfrac{1}{n}\times c$$. That is,

$$\dfrac{b+c}{n}=\dfrac{b}{n}+\dfrac{c}{n}$$.

(We’re using here the result that $$\dfrac{1}{n}\times a=\dfrac{a}{n}$$  from lesson 2.8.)

So we certainly have, in arithmetic, the means to tease apart additions in a numerator.

Comment: For $$\dfrac{x+y}{a+b}$$, write $$n$$ for $$a+b$$ so that this reads $$\dfrac{x+y}{n}$$. This equals $$\dfrac{x}{n}+\dfrac{y}{n}$$, which is $$\dfrac{x}{a+b}+\dfrac{y}{a+b}$$. So we really can tease apart of an addition in a numerator.

 Question: Can we tease about a subtraction in a numerator? Is $$\dfrac{x-y}{n}$$ sure to equal $$\dfrac{x}{n}-\dfrac{y}{n}$$? (HINT: $$x-y$$ can be rewritten $$x + \left(-y\right)$$.)

Nothing we have discussed has dealt with teasing apart additions in a denominator. Maybe it simply cannot be done?

 Question: Choose some numbers for $$x$$, $$y$$, $$a$$, and $$b$$. Does $$\dfrac{x+y}{a+b}$$ equal $$\dfrac{x+y}{a}+\dfrac{x+y}{b}$$ for your numbers? (Try it again just to make double sure. It really doesn’t work in general.)   Are there any values $$a$$ and $$b$$ for which, by pure coincidence, $$\dfrac{1}{a+b}$$ happens to equal $$\dfrac{1}{a}+\dfrac{1}{b}$$?

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