Generating problems about rational expressions

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:

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

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

=x2+7x+12x2x(x+1)(x+2)(x+3)= \frac{x^2+7x+12-x^2-x}{(x+1)(x+2)(x+3)}

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

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

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

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

Here are some assumptions:

  1. All the coefficients are integers.

  2. The addends are already in lowest terms.

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

Note by Mark Lao
5 years, 1 month ago

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Hey 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 xx in the numerator? Please write back.

Josh Silverman Staff - 5 years, 1 month ago

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Do you mean to find pairs of fractions that when added/subtracted, result in a numerator that has a common factor with the denominator? - Yes.

Also, the starting fractions must be linear functions of x x in the numerator? - Not necessarily. :-)

Mark Lao - 5 years ago

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