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Consider a sum $\frac{a}{b}+\frac{c}{d} = \frac{x}{y}$ where each fraction is reduced. Alternatively using the familiar process of lowest common denominators, we have $\frac{a}{b}+\frac{c}{d} = \frac{a\cdot\frac{d}{(b,d)}+c\cdot\frac{b}{(b,d)}}{[b,d]}$ where $(b,d)$ denotes the gcd and $[b,d]$ denotes the lcm. My question is, when is it true that $y = [b,d]$? For example, this does not hold for $\frac{5}{6}+\frac{1}{14} = \frac{38}{42} = \frac{19}{21}$ where $21\neq [14, 6]$.

Are there any simple necessary and sufficient conditions for $y = [b,d]$?

Edit It has been suggested that

$y = [b,d] \iff (b, d) = 1$

is a necessary and sufficient condition. I'm interested in either a proof or a counterexample if possible.

Edit 2 After a bit of searching, I've found $\frac{1}{24} + \frac{1}{16} = \frac{5}{48}$ as a counterexample.

I am still looking for nice conditions for this to be satisfied and I feel that I should give an explanation of exactly what type of condition I am seeking. Angela has provided a necessary and sufficient condition, but it does not seem to be "simpler" than simply adding the fraction and seeing if it reduces. This is perhaps ambitious, but I am looking for a condition which is simple enough to use by inspection for simple fractions.

  • 2
    The suggested condition is not necessary: $\frac{1}{12}+\frac{1}{18} = \frac{5}{36},$ where $b=12$, $d=18$, $[b,d]=36$, $(b,d) = 6$. The suggested condition is sufficient, since $y\neq[b,d]$ if and only if $1 \lt \gcd((b,d),a\delta+c\beta)$, where $b=(b,d)\beta$, $d=(b,d)\delta$.2012-01-14

1 Answers 1

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let $x=\gcd(b,d)$

$b=xe$

$d=xf$

$\frac{a}{b}+\frac{c}{d}=\frac{af+ce}{efx}$

$efx=\operatorname{lcm}(b,d)$

$e,f$ are relative primes

The fraction simplifies when $\gcd(af+ce,x)\not=1$, which is when $\gcd(ad+bc,bd)\not=\gcd(b,d)$.