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.

  • $\begingroup$ As discussed in math.stackexchange.com/questions/98620/…, $y = [b,d]$ iff $(b,d)=1$. $\endgroup$
    – lhf
    Jan 13, 2012 at 13:29
  • $\begingroup$ @lhf I don't think that's true. $\frac{5}{6} + \frac{3}{14} = \frac{22}{21}$ so $(6,14) \neq 1$ but $21 \neq [14, 6]$. Even in the example I gave in the question, this does not hold. $\endgroup$
    – EuYu
    Jan 13, 2012 at 14:49
  • $\begingroup$ I don't see a contradiction. $\endgroup$
    – lhf
    Jan 13, 2012 at 16:08
  • $\begingroup$ @lhf My mistake. I got it mixed up. $\endgroup$
    – EuYu
    Jan 13, 2012 at 16:25
  • 2
    $\begingroup$ 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 $b=(b,d)\beta$, $d=(b,d)\delta$. $\endgroup$ Jan 14, 2012 at 4:47

1 Answer 1


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





$e,f$ are relative primes

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


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