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...

  • $\begingroup$ How far did you get with your contradiction? Have you tried a direct approach? $\endgroup$ – abiessu Mar 18 at 5:03
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    $\begingroup$ $\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? $\endgroup$ – fleablood Mar 18 at 5:13
  • $\begingroup$ I did establish that: ad + bc = kbd then ad = b (kd - c) so b| ad $\endgroup$ – 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.


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