Let there be two numbers: $\frac{a}{b}+\frac{c}{d}$, s.t. $(a,b)=(c,d)=(b,d)=1$.
The sum is assumed (for indirect, or contradiction approach) to be $k \in \mathbb {Z}$.
$$\frac{a}{b}+\frac{c}{d}=k $$ $$\Rightarrow ad+bc=kbd$$ $$>ad=kbd-cb \Rightarrow ad=b(kd-c), \text{not possible as } b\nmid ad$$ $$>cb=kbd-ad \Rightarrow cb=d(kb-a), \text{not possible as } d\nmid bc$$
Hence proved.
How to extend (hopefully, the only approach would by induction) for sum of more than two fractions?
Also, have another question that concerns with providing a proof using better symbolic logic ( don't know how to state better), the derivation of $(a,c)=1$, by using the three equalities ($(a,b)=(c,d)=(b,d)=1$).
Any way to prove - or is there no way to prove as $(a,c)$ can go both ways; i.e. $=1, \ne 1$, as @MooS's comment to his answer shows.
My incomplete proof attempt is:
Also, stated using Bezout's identity:
$$ax + by =1 , cx + dy =1, bx + dy =1, \exists x,y \in \mathbb{Z} $$(Have taken $x,y$ to mean dummy variables, and may not actually be the same for all three equalities).
1 Answer
$\begingroup$This would be my proof for the case of two summands:
Actually one only needs to assume that $(b,d)=1$ and that one of the fractions is not an integer, w.l.o.g assume $\frac{a}{b} \notin \mathbb Z$.
If $\frac{a}{b}+\frac{c}{d} \in \mathbb Z$, we have $ad+bc = 0 \mod{bd}$.
Reduce this modulo $b$ to get $ad = 0 \mod b$. Since $d$ is co-prime to $b$, we obtain $a = 0 \mod b$, a contradiction to $\frac{a}{b} \notin \mathbb Z$.
For more summands (assuming the denominators are pairwise co-prime and one of the fractions is not an integer), the induction step is actually trivial, because if you take the sum of two fractions, the common denominator will still be co-prime to all other denominators.
$\endgroup$ 12
