# In partial fractions, why must the degree of the numerator be lower than the denominator?

Specifically, it must be one degree lower. But why must it be smaller?

• It's not necessary, but maybe it's more comforting. – Michael Rozenberg Sep 28 '17 at 9:39
• I don't think so. Every textbook and video says it must be smaller. – Jane Doe Sep 28 '17 at 10:45
• This is probably getting close-votes because you didn't include your thoughts about this, but I think the fact that you're asking the question is a good thought already. – Mark S. Sep 28 '17 at 11:14

## 2 Answers

If you ever have a fraction where the degree of the numerator is not lower, then you could use long division to get simpler fractions. For example, $\dfrac{6x^2+3}{2x^2-x+7}=3+\dfrac{3x-18}{2x^2-x+7}$.

This is analogous to the "improper" fractions of positive integers being those where the numerator is not smaller than the numerator. $\dfrac{11}{3}=3+\dfrac{2}{3}$, etc.

• Thanks for that. – Jane Doe Sep 28 '17 at 19:58

When you make the usual Ansatz for the decomposition, you get something on the right-hand side which (if put on a common denominator again) automatically has lower degree in the numerator. So this can't match your original expression on the left-hand side, unless it has that property too, to begin with.