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,




(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?



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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