breaking up fractions

Question

I have these two fractions

${11 \over 31 }+{-11 \over 61}$

Adding them gives

$330 \over 1891$

But how do I go back to the two fractions, once I've added them?

I can get the denominators just by factoring 1891, but how do I find the numerators without brute force?

score 4 · Accepted Answer · answered Jun 27 '13 at 20:51

4

Factor the denominator: $${330 \over 1891} = \frac {330}{31\cdot 61} = \frac A{31} + \frac B{61}$$

Then we need to find $$A\cdot 61 + B\cdot 31 = 330\tag{$\dagger$}$$ The problem then becomes finding integers $A, B$ that satisfy this equation $(\dagger)$. To do this, try using Euclid's Algorithm, or the \Extended Euclidean Algorithm.

answered Jun 27 '13 at 20:51

amWhy

210,739

I find the Extended Euclidean Algorithm a nice bit faster than brute force. – Lord_Farin Jun 27 '13 at 20:53
I meant the Extended algorithm... – amWhy Jun 27 '13 at 20:54
How do I generalize this to 3 or more fractions? – Nick ODell Jun 27 '13 at 22:39
@amWhy: Already looks like an excellent day to me! ;-) +1 – Amzoti Jun 28 '13 at 02:35
enjoy the uni, @Babak! – amWhy Jul 06 '13 at 02:20

score 1 · Answer 2 · answered Jun 27 '13 at 20:50

1

Every fraction can be decomposed into infinitely many other fractions. For instance, $1/2=1/12321+12319/24642$. So there's no hope.

answered Jun 27 '13 at 20:50

Brian Rushton

13,375

True. Let me add a restriction, to make the output more useful: all of the denominators must be prime numbers. – Nick ODell Jun 27 '13 at 21:20

score 1 · Answer 3 · answered Jun 27 '13 at 20:50

1

You need to find integers $m$ and $n$ such that $$61m + 31n = 330.$$

For this you use Euclid's algorithm.

answered Jun 27 '13 at 20:50

MJD

67,568
43
308
617

score 0 · Answer 4 · answered Aug 10 '14 at 23:21

0

I eventually got around to writing a library to do this.

Link to github.

answered Aug 10 '14 at 23:21

Nick ODell

231

breaking up fractions

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