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

Question

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.

Answer 1

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

Answer 2

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 $e$.