Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
As discussed in math.stackexchange.com/questions/98620/…, $y = [b,d]$ iff $(b,d)=1$. –  lhf Jan 13 '12 at 13:29
    
@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. –  EuYu Jan 13 '12 at 14:49
    
I don't see a contradiction. –  lhf Jan 13 '12 at 16:08
    
@lhf My mistake. I got it mixed up. –  EuYu Jan 13 '12 at 16:25
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 $b=(b,d)\beta$, $d=(b,d)\delta$. –  Arturo Magidin Jan 14 '12 at 4:47
show 2 more comments

1 Answer

up vote 1 down vote accepted

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 $\gcd(ad+bc,bd)\not=\gcd(b,d)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.