Given two real numbers $a$ and $b$, define an $a$-$b$-sum as a finite sum of $a$'s and $b$'s, i.e. a sum:
$$m\cdot a + n\cdot b$$ where $m,n$ are non-negative integers.
Is there a pair of numbers $a<0$ and $b>0$, such that for every $\epsilon>0$ there is an $a$-$b$-sum $S \in (0,\epsilon)$?
The claim is obviously untrue if $a$ and $b$ are integers, since in that case the sum is also integer so there is no $S \in (0,1)$.
It is also untrue if $a$ and $b$ are rationals, since in that case the sum is always an integer multiple of $\frac{1}{pq}$ (where $p,q$ are the denominators of $a,b$ respectively) so there is no $S\in (0,\frac{1}{pq})$.
Is there a pair of irrational numbers that makes the claim true?
Take $a=\sqrt{2}$ and $b=-1$ .
From Kronecker's Theorem there is a positive integer $c$ such that : $$\{c\sqrt{2}\} < \epsilon $$ (This case of the theorem can be proved by a simple pigeon-hole argument )
Now simply take the following $(a,b)$-sum :
$$c \cdot a + \lfloor c\sqrt{2} \rfloor \cdot b=c\sqrt{2}-\lfloor c\sqrt{2} \rfloor=\{c\sqrt{2} \}<\epsilon$$ as wanted .