# Prove the sum of two rational number is equal to $\frac{e}{lcm(b,d)}$ for some integer $e$.

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.

• 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. Sep 12, 2019 at 6:05
• What does k(b, d) mean? gcd(b,d) was a scalar k? But my text book doesn't state in this way...
– Will
Sep 12, 2019 at 6:14
• 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 Sep 12, 2019 at 6:48
• 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. Sep 13, 2019 at 4:17

## 2 Answers

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$$e=a\frac d{\gcd(b,d)}+c\frac b{\gcd(b,d)}\in\mathbb Z.$$

• Some of your fractions are upside down. Sep 12, 2019 at 6:17
• 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?
– Will
Sep 12, 2019 at 6:18
• @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? Sep 12, 2019 at 6:22
• @JiayeWang I've edited my answer. What do you think now? Sep 12, 2019 at 6:23
• 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. Sep 12, 2019 at 6:24

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 $$e$$.

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