Let $x$ and $y$ be real numbers. And suppose that $\frac{x}{y}$ is irrational. Prove the following: for each $\varepsilon > 0$, there exist integers $a$ and $b$ such that $0 < ax + by < \varepsilon$.
My approach is: Consider the set $S=\{ax+by>0:a,b\in\mathbb Z\}$. Let $c=\min S$. I claim that $x$ and $y$ are both integer multiples of $c$.
If not then $x=c \times m+r$, where $0\leq r<c,m$ and $m \in \mathbb Z$. From here we have $r=x-mc=x(1-am)+y(-bm)$. thus $r \in S$. Since $r<c$ it is contradiction with $\min S$. So $r=0, \ x=cm$. Similarly we can show $y=ct$, where $t \in\mathbb Z$. Meanwhie $\frac{x}{y}=\frac{cm}{ct}=\frac{m}{t}$ and $\frac{x}{y}$ ratio is irrational but m/t is integer ratio…… my question is it something wrong here if not how continue
Best Answer
Restricting ourselves to $a$ and $b$ as positive integers, if $x = \sqrt{2}$ and $y = 1$, then $ax + by > \sqrt{2} + 1$, so cannot be made arbitrarily small.
However for any irrational of the form $x$/$y$ we have |$x/y - b/a$| can be made arbitrarily small, since rational numbers are dense in irrational numbers. We can then, for any $\epsilon$ always find $a$ and $b$ such that $ax - by < \epsilon$.