# Prove that the sum is not an integer

Prove that if a / b and c / d are two irreducible rational numbers such that gcd (b, d) = 1 then the sum (a/b + c/d) is not an integer.

I was thinking about the proof by contradiction, but then I haven't find the correct answer yet...

• How far did you get with your contradiction? Have you tried a direct approach? – abiessu Mar 18 at 5:03
• $\frac ab+\frac cd=\frac {ad+bc}{bd}$. The $bd|ad+bc$ so $b|ad+bc$ so $b|ad$. Do you see why that's not possible? – fleablood Mar 18 at 5:13
• I did establish that: ad + bc = kbd then ad = b (kd - c) so b| ad – Dot Point Mar 19 at 15:20

Do note that we need at least one of $$b,d$$ to be $$\neq 1$$

Note that $$a/b+c/d=\frac{ad+bc}{bd}$$

For this to be an integer, we must have $$ad+bc$$ divisible by $$bd$$

Since $$\gcd(b,d)=1$$, showing that $$ad+bc$$ is not divisible by any one of them $$\neq 1$$ is sufficient. We show that $$b\neq 1$$ does not divide it.

Suppose it does. Then $$b\mid ad+bc\iff b\mid ad$$. Since $$\gcd(b,d)=1$$, this is equivalent to $$b\mid a$$, a contradiction

We have to assume that either $$b \neq 1$$ or $$d \neq 1$$.

$$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}.$$ If this fraction is an integer, then $$\exists n \in \Bbb Z$$ such that $$nbd=ad+bc$$, so $$b(nd-c)=ad$$. But $$\frac{a}{b}$$ is in lowest terms so $$a$$ and $$b$$ have no common prime factors, and by hypothesis $$(b, d)=1$$ so $$b$$ and $$d$$ have no common prime factors. But $$ad=b(nd-c) \Rightarrow b~|~ad$$, and that can't happen if $$b$$ has any prime factors, so $$b=1$$. But if $$b=1$$, then $$\frac{a}{b}$$ is an integer but $$\frac{c}{d}$$ is not. This contradiction establishes that the fraction $$\frac{ad+bc}{bd}$$ cannot be an integer.