The other day, I was solving problems regarding rational expressions. I noticed that every single time, after simplifying the numerator, the fraction is already in lowest terms. So I wondered if this is always true.

After a bit of experimentation, I was able to construct this counterexample:

\[ \frac{x+4}{(x+1)(x+2)} - \frac{x}{(x+2)(x+3)} \]

\[= \frac{(x+4)(x+3)-x(x+1)}{(x+1)(x+2)(x+3)}\]

\[= \frac{x^2+7x+12-x^2-x}{(x+1)(x+2)(x+3)}\]

\[= \frac{6x+12}{(x+1)(x+2)(x+3)}\]

\[= \frac{6(x+2)}{(x+1)(x+2)(x+3)}\]

\[= \frac{6}{(x+1)(x+3)}\].

Can you think of a way to generate infinitely-many such problems?

Here are some assumptions:

All the coefficients are integers.

The addends are already in lowest terms.

After simplifying the numerator, the numerator and the denominator stilll has a common factor.

## Comments

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TopNewestHey Mark, I am wondering what you mean by "such problems" here. Do you mean to find pairs of fractions that when added/subtracted, result in a numerator that has a common factor with the denominator? Also, the starting fractions must be linear functions of \(x\) in the numerator? Please write back. – Josh Silverman Staff · 2 years, 5 months ago

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Also, the starting fractions must be linear functions of \( x \) in the numerator? - Not necessarily. :-) – Mark Lao · 2 years, 5 months ago

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