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As title state: $\frac{a}{b} + \frac{c}{d}=\frac{e}{lcm(b,d)}$ for some integer $e$. Here is what I tried:

$\frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd}$ Since $gcd(b,d)lcm(b,d)=bd$, so I got $\frac{ad+bc}{gcd(b,d)lcm(b,d)}$. Since these are two rational numbers, so $a, b, c, d \in \mathbb{Z}$, then by Bezout's identity: $ad+bc=gcd(b,c)$, so $\frac{gcd(b,d)}{gcd(b,d)lcm(b,d)} = \frac{1}{lcm(b,d)}$, therefore there exist an integer $e$ such that $e=1$.

That's my proof, I think there is an error within it, because I can't find an example support this proof... Can someone help me to point out where my errors are? Appreciate all helps.

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    $\begingroup$ It's not $ad+bc=(b,d)$, but rather $ad+bc=k(b,d)$. I'm fairly certain that Bezout's identity has different conditions. $\endgroup$
    – abiessu
    Sep 12, 2019 at 6:05
  • $\begingroup$ What does k(b, d) mean? gcd(b,d) was a scalar k? But my text book doesn't state in this way... $\endgroup$
    – Will
    Sep 12, 2019 at 6:14
  • $\begingroup$ A counterexample for your proof: Let $A=1/4$ and $B=1/6$ be two rational numbers and we have $A+B= 1/4+1/6=5/12=5/lcm(4,6)$. Also, there is a similar type of question where e=1 is valid, just for extra knowledge check this $\endgroup$
    – IamKnull
    Sep 12, 2019 at 6:48
  • $\begingroup$ It's not that $k$ is a function, but rather $(a,b)$ is shorthand for $\text{gcd}(a,b)$ and after that $k$ is just a scalar value. $\endgroup$
    – abiessu
    Sep 13, 2019 at 4:17

2 Answers 2

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That is not correct, because Bézout's identity doesn't say that.

However, it can be corrected. It follows from what you did that$$\frac ab+\frac cd=\frac{a\frac d{\gcd(b,d)}+c\frac b{\gcd(b,d)}}{\operatorname{lcm}(b,d)}$$and, since $a$, $\frac d{\gcd(b,d)}$, $c$, and $\frac b{\gcd(b,d)}$ are all integers, you can just take$$e=a\frac d{\gcd(b,d)}+c\frac b{\gcd(b,d)}\in\mathbb Z.$$

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  • $\begingroup$ Some of your fractions are upside down. $\endgroup$
    – darij grinberg
    Sep 12, 2019 at 6:17
  • $\begingroup$ Thank you for your answer, but I am afraid my understanding of this subject is not that advanced yet... I can't see how you get the RHS in the first step... can you show more details? $\endgroup$
    – Will
    Sep 12, 2019 at 6:18
  • $\begingroup$ @darijgrinberg I don't think so, but I wrote several times $\operatorname{lcm}(b,d)$ where I should have written $\gcd(b,d)$. I've edited the answer. What do you think now? $\endgroup$
    – José Carlos Santos
    Sep 12, 2019 at 6:22
  • $\begingroup$ @JiayeWang I've edited my answer. What do you think now? $\endgroup$
    – José Carlos Santos
    Sep 12, 2019 at 6:23
  • $\begingroup$ Ah, that works too :) But it's overcomplicating things. If you replace $d/\gcd(b,d)$ by $\lcm(b,d)/b$ etc., then you don't need the gcd-lcm relation, just the definition of the lcm. $\endgroup$
    – darij grinberg
    Sep 12, 2019 at 6:24
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We know that $\operatorname{lcm}(b,d)\gcd(b,d)=bd$, and so $\gcd(b,d)=bd/\operatorname{lcm}(b,d)$.

Then

$$\frac ab+\frac cd=\frac{ad+bc}{bd}=\frac{ad+bc}{\operatorname{lcd}(b,d)\gcd(b,d)}.$$

Of course, $\gcd(b,d)$ divides $d$ and $b$, and so $$\frac{ad+bc}{\gcd(b,d)}=a\frac{d}{\gcd(b,d)}+c\frac{b}{\gcd(b,d)}$$ is an integer $e$.

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  • $\begingroup$ $\frac ab+\frac cd=\frac{ad}{bd}+\frac{bc}{bd}$ $\endgroup$
    – pancini
    Sep 12, 2019 at 6:36
  • $\begingroup$ But why you change $bd$ to $\gcd(b,d)$? $\endgroup$
    – Will
    Sep 12, 2019 at 6:37
  • $\begingroup$ As you said in your post, $bd=\operatorname{lcm}(b,d)\gcd(b,d)$. $\endgroup$
    – pancini
    Sep 12, 2019 at 6:38
  • $\begingroup$ But where is the $\text{lcm}(b,d)$...? $\endgroup$
    – Will
    Sep 12, 2019 at 6:39
  • $\begingroup$ The last line is just the integer $e$. Then $\frac ac+\frac cd=\frac{e}{\operatorname{lcm}(b,d)}$ $\endgroup$
    – pancini
    Sep 12, 2019 at 6:40

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