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I am looking for a way to rewrite

$$ \frac{a}{c+1} $$ as

$$ \frac{a}{c} + \cdots $$

in other words, move the $+ 1$ part outside of the denominator.

(a and c are real numbers)

Is this possible? If so, how?

Lee David Chung Lin
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Qqwy
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    This is going to be a smartaleck answer but $\frac{a}{c+1}=\frac{a}{c}+(\frac{a}{c+1}-\frac{a}{c})$. Now... if you insist on continuing, then try simplifying $\frac{a}{c+1}-\frac{a}{c}$. The end result is that it will not be worth having done and does not simplify matters at all. – JMoravitz Sep 10 '19 at 16:51
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    Whether it "simplifies matters at all" depends on your needs. I have done calculations where the exact manipulation you propose is a useful simplification for what I'm doing. –  Sep 10 '19 at 16:54
  • My main point being, this isn't going to be the same as how $\frac{a+1}{c} = \frac{a}{c}+\frac{1}{c}$... I have seen several people who are not well practiced in math try to make the simplification that $\frac{a}{c+1}=\frac{a}{c}+\frac{a}{1}$ which is incorrect. Performing the simplification I pointed out earlier yields you with $\frac{a}{c+1}=\frac{a}{c}-\frac{a}{c^2+c}$ which would appear to be more trouble to calculate by hand than it would have originally been since $\frac{a}{c^2+c}$ is just $\frac{a}{c+1}$ divided by $c$ an additional time. – JMoravitz Sep 10 '19 at 16:59
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    One possibility is the series expansion $\frac{a}{c+1} = \frac{a}{c} - \frac{a}{c^2} + \frac{a}{c^3} - \frac{a}{c^4} + \cdots$ valid if $|c| < 1$. – Daniel Schepler Sep 11 '19 at 00:47
  • Shouldn't that be $|c| > 1$? – Charles Hudgins Sep 11 '19 at 01:33

2 Answers2

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For something slightly different, we have \begin{align} \frac{a}{1 + c} &= \frac{a}{c} \left( \frac{1}{1 + \frac{1}{c}}\right) = \frac{a}{c} \left( \sum_{n = 0}^\infty \frac{(-1)^n}{c^n} \right) \\&= \frac{a}{c}\left(1 - \frac{1}{c} + \frac{1}{c^2} + \cdots \right) = \frac{a}{c} - \frac{a}{c^2} + \frac{a}{c^3} + \cdots \end{align} This series will converge as long as $|c| > 1$.

Charles Hudgins
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We go a bit farther and show that $\tfrac{a}{c+1}$ and $\tfrac{a}{c}$ are algebraically independent whenever $a$ and $c$ are.

Assume $a$ and $c$ are algebraically independent: that is, for all polynomials $f$ with integer coefficients, $$f(a,c)=0\Rightarrow f = 0\text{.}$$ Suppose we had a polynomial relation $$p(\tfrac{a}{c+1},\tfrac{a}{c})=0$$ between $\tfrac{a}{c+1}$ and $\tfrac{a}{c}$. Then since $$p(\tfrac{a}{c+1},\tfrac{a}{c})=(c+1)^{-i}c^{-j}q(a,c)$$ for some natural numbers $i$, $j$ and polynomial $q$, we would have a relation $$q(a,c)=0\text{,}$$ implying that $q=0$. Therefore $p=0$ as well, i.e., $$p(\tfrac{a}{c+1},\tfrac{a}{c})=0\Rightarrow p =0\text{.}$$ QED

K B Dave
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