# How to write inverse of integer as sum of fractions

I was reading this article about partial fractions and at the bottom of the article there was a paragraph about integers. However, I cannot seem to get it right each time.

For example:

1/8 = 1/2^3 = 1/2 - 1/4 - 1/8


but

1/9 = 1/3^2 ≠ 1/3 - 1/9

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sorry which part is not clear for you? – dato datuashvili Apr 13 '12 at 10:53
why it doesn't work for 1/9. 1/3 - 1/9 = 2/9. – SemT Apr 13 '12 at 10:58

$\frac{1}{9} = \frac{a}{3} + \frac{b}{9}$

That implies, $3a + b =1$.

put a and b values and satisfies the above equation.

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The summation formula for a finite geometric series is $$\sum_{k=0}^{n-1}a^k=\frac{a^n-1}{a-1}=\frac{1-a^n}{1-a}\tag{1}$$ or $$\sum_{k=0}^{n-1}a^{-k}=\frac{1-a^{-n}}{1-a^{-1}}=\frac{a^n-1}{(a-1)\,a^{n-1}}$$ In particular, when $a=2$ in the top formula, we can express $2^n-1$ as a sum of lower powers of $2$. But for integers $a>2$ (like $a=3$), the denominator $a-1$ is not $1$.

$$\frac{x}{a}+\frac{y}{b}=\frac{c}{d}\tag{2}$$ is solvable in integers (for $x,y$ given $a,b,c,d$) iff $d$ divides $mc$ for $m=\href{http://en.wikipedia.org/wiki/Least_common_multiple}{\operatorname{lcm}}(a,b)$, for then the equation becomes $$\frac{m}{a}x+\frac{m}{b}y=\frac{mc}{d}$$ or, with $g=\href{http://en.wikipedia.org/wiki/Greatest_common_divisor}{\operatorname{gcd}}(a,b)=\frac{ab}{m}$, $a'=\frac{m}{a}=\frac{b}{g}$, $b'=\frac{m}{b}=\frac{a}{g}$ and $c'=\frac{mc}{d}=\frac{abc}{dg}$, the famous linear diophantine equation $$a'x+b'y=c'$$ which is solvable (with infinitely many solutions) since $\operatorname{gcd}(a',b')=\operatorname{gcd}(\frac{b}{g},\frac{a}{g})=\frac{\operatorname{gcd}(a,b)}{g}=\frac{g}{g}=1$.

As we increase the number of terms on the left in $(2)$, we can of course represent more numbers $\frac{c}{d}$, but we will always hit a brick wall when a prime $p$ (or prime power $p^k$) dividing $d$ does not also divide the product of $c$ with the LCM of the denominators on the LHS.

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