# Why can we treat the numerator as a constant in partial fractions?

I'm learning the method of partial fraction decomposition as a 'useful dodge' (Silvanus Thompson, Calculus Made Easy) for calculus problems, but I'm not quite following the reasoning. According to Thompson:

"Now, since the partial fractions are proper fractions, the numerators are mere numbers without x at all, and we can call them A, B, C ... as we please. So, in this case, we have:

$$\frac{ 3x+1 }{x^2 - 1} = \frac{A}{ x + 1}+\frac {B }{ x − 1}$$

Why does the fact that partial fractions are proper mean that we treat the numerators of both as a constant?

• Do you know what the term "proper fraction" means specifically in the context of fractions of polynomials? Commented May 26, 2022 at 10:30
• According to Thompson, a fraction is a proper fraction when its numerator has a lesser degree than its denominator. Commented May 26, 2022 at 10:31
• @DanÖz and if the degree of the denominator is $1$, then the degree of the numerator must be? Commented May 26, 2022 at 10:39
• Oh OK I've arrived at the party. I didn't pay close enough attention to the specific case we were dealing with – i thought he was talking about partial fractions in general. If we only have terms of x not x^2 etc, then the numerators will always be a constant. Thankyou. Commented May 26, 2022 at 11:04