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December 6

Confused regarding necessary criterion for the sum of two irreducible fractions to be reducible

While doing some research on a recreational mathematical problem I stumbled on an interesting Wolfram Student Support thread from 2005. Interesting, because applying it will significantly reduce the complexity of the recreational problem.

Here the thread author refers to having recently read an (uncited) paper on the reduction of fractions. A result from this paper is apparently that

If $\frac{a}{b}$ and $\frac{c}{d}$ are both irreducible fractions, i.e., $\gcd (a,b)=\gcd (c,d)=1$, then it is a necessary criterion that in the prime factorizations of the two denominators b and d, at least one prime factor shall occur an equal number of times for the sum

$\frac{a}{b}+\frac{c}{d}=\frac{a\cdot d+b\cdot c}{b\cdot d}$

to be reducible!

It is written with more words in the thread, but that is what I extract out of it.

The thread auhor states that such a prime is called a "balanced" prime between the two denominators, a quite different definition than the usual for a balanced prime.

For instance, if

$b=12=2^2\cdot 3$

and

$d=20=2^2\cdot 5$,

the number 2 is a balanced prime between the two denominators as it occur twice in the prime factorizations of both denominators.

Thus, the sum

$a/12+c/20=(20\cdot a+12\cdot c)/(240)$

may be reducible for some a semiprime with 12 and some c semiprime with 20.

For instance, with

$a=c=1$

the sum is $32/420$, which can be reduced to $8/105$.

The denominator is actually 240, so reduces to $2/15$ 109.151.42.94 (talk) 10:34, 6 December 2013 (UTC)

But here comes my problem, as I appear to be able to find a counter example of the opposite?!?

If

$b=18=2\cdot 3^2$

and

$d=27=3^3$,

no "balanced" prime number exist between the two denominators as the only common prime number in their factorizations (3) does not occur an identical number of times (2 vs 3).

Thus, for any a semiprime with 18 and any c semiprime with 27, the sum $a/18+c/27=(27\cdot a+18\cdot c)/(486)$ will not be a reducible fraction according to this rule. But, that does not appear to be true (or I am making an embarassing mistake)!? For instance, if $a=c=1$, the sum is $45/486$. But $\gcd (45,486)=9$, thus the sum can be reduced to $5/54$, in contradiction with the necessary criterion?!?

So, I am confused. Have I understood the necessary criterion for reducability correctly?

If so, what is wrong with my counterexample?

Is this rule called anything?

Thanks in advance, --Slaunger (talk) 20:10, 3 December 2013 (UTC)

Apparently the thread author is using a modified form of the addition formula. The example 13/36 + 7/45 = 93/180 is given but (ad+bc)/(bd) would give 837/1620. I'm guessing that what the author actually means is the way some of us were taught to add fractions in elementary school, write both fractions with the least common denominator the add the numerators. If so, the formula the author is actually using (though not writing down) is a/b + c/d = ((ad+bc)/gcd(b,d))/(bd/gcd(b,d)). With the example you give the least common denominator is 54, so the result of 5/54 fits. --RDBury (talk) 21:40, 3 December 2013 (UTC)

Ah, yes, of course it has to be the the summed fraction in the form where it is reduced with $\gcd (b,c)$ in the nominator and denomninator as you so clearly state!

Still, I find the result that

$\frac{a}{b}+\frac{c}{d}$

written as the fraction

$\frac{a\cdot [d/\gcd (b,d)]+[b/\gcd (b,d)]\cdot c}{b\cdot d/\gcd (b,d)}$

to be non-reducible if there does not exist a common prime factor in b and d, which occur an equal number of times to be pretty remarkable!

1. Is it correct?
2. Does this rule have a name? (I cannot find anything about it on Wikipedia)

--Slaunger (talk) 15:03, 5 December 2013 (UTC)

Template:Od It's true. To see this, without loss of generality take gcd(b,d) = 1. The only way $ad+bc$ can have a prime factor in common with d is if c has a prime factor in common with d, but this is impossible if the original fraction was written in lowest terms. Likewise, $ad+bc$ cannot have a prime factor in common with b. Sławomir Biały (talk) 16:41, 5 December 2013 (UTC)

Template:Resolved You are right, thanks! And now I solved the problem, which was finding how many ways 1/2 could be written as a sum of distinct inverse square numbers using terms up to 1/80²! --Slaunger (talk) 13:35, 8 December 2013 (UTC)