real analysis homework 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
 A: 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 $ax - by < \epsilon$.
A: The thing about real numbers is the reals have the least upper bound property (which could just as legitimately be called the greatest lower bound property).
The statement that for any $\epsilon > 0$ there are integers $a,b$ so that $0 < ax + bx  <\epsilon$ is nothing more or less than a claim that $S = \{ax+by| a,b \in \mathbb Z; ax+by>0\}$ is non-empty and $\inf S = 0$.
(To see the equivalence of the claims, note there existing an $0<ax + by < \epsilon$ is to claim there exist $ax + by >0$ so $S = \{ax+by| a,b \in \mathbb Z; ax+by>0\}$ is non-empty; that by definition all $ax+by \in S$ are so that $ax+by > 0$ so $0$ is a lower bound, but $ax+by < \epsilon$ means any $\epsilon>0$ is not a lower bound while $0$ is, and that is the definition of $\inf$.)
So let's prove $\inf S = 0$.
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(Out line:  If we assume $d= \inf S\ne 0$ then as $0$ is a lower bound, $d > 0$.  Now we can find two distinct $ax + by; ex + fy$ so that $d \le ax + by < ex + fy < 2d$, we'd have $0 < (a-e)x + (b-f)y < d$ which would be a contradiction.)
(But although $2d > d$ so $2d$ is not lower bound so we *must have an $d \le ax + by < 2d$ is is possible that $ax + by$ is the only possible value.  That would mean $ax + by =\min S$ and $\min S = \inf S = d = ax+by$.  We avoid this by proving $S$ has no minimum value.)
Let's go:
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First to show $d=\inf S\ge 0$ exists we must show $S$ is non-empty and $0$ is lower bound of $S$.  $S = \{ax+by| a,b \in \mathbb Z; ax+by>0\}$ is obviously non-empty.  Just consider $|x| + |y| =\pm 1\cdot x + (\pm 1)\cdot y > 0$.  And by definition, if $ax+by \in S$ then $ax+ by > 0$ so $0$ is a lower bound.
(Oh, I guess we need to point out that neither $x$ nor $y$ may be $0$... That's clear as $\frac xy$ is defined and irrational).
To show there is no minimum element of $S$:
We know $|x| \ne |y|$ as that would make $\frac xy$ rational. Wolog we can assume $0<|y| < |x|$.  If $a'x + b'y = c> 0$ then $a|x| + b|y|=c$ for appropriate values of $a = \pm a'; b  =\pm b'$.
Suppose $a|x| + b|y| = c > 0$.   If $c \ge |y|$ we can have $c>a|x| + (b-1)|y|> c-|y| >0$.
And if $c< |y|< |x|$.  By archimedian principal there is a natural $n$ so that $nc < |y|\le (n+1)c$.
So $an|x| + bn|y| = nc <|y|$ and $an|x| + (bn-1)|y|=nc - |y| < 0$ and $-an|x| + (1-bn)|y|= |y|- nc > 0$.  But $|y|-nc \le (n+1)c - nc = c$.  So $an|x| + (bn-1)|y| \le c$, so we have  $-an|x| + (1-bn)|y|\in S$ and  $-an|x| + (1-bn)|y| \le c$.
(Hmmm.... I need to show that  that $-an|x| + (1-bn)|y|=|y|-nc \ne c$ or that $|y|\ne (n+1)c$.)
( But $|y|=(n+1)c$  would mean $a|x| + b|y| = c = \frac {|y|}{n+1}$ and that $a|x|= |y|(\frac 1{n+1} -b)$. $a = \frac {|y|}{|x|}(\frac 1{n+1} -b)$. But as $\frac {|y|}{|x|}$ is irrational this means $\frac 1{n+1} - b=0$. But $b$ is an integer and $n$ is a natural number so that's impossible.)
So $-an|x| + (1-bn)|y| = nc-|y| < c$.
So!  $c$ is not the least possible element of $S$ and $\min S$ does not exist.
Now..... Suppose $d=\inf S \ne 0$.  Then $\inf S > 0$.   That mean $\frac 32 d > d$ so $2d> d$ is not a lower bound of $S$.  We also know $d \in S$ is impossible as that would mean $d = \min S$ and we know $\min S$ doesn't exist.
So there exists $ax + by$ so that $d < ax+by < 2 d$.  ANd as $ax+by>d$, $ax+by$  can't be a lower bound so there must exist $ex + fy$ so that $d <  ex+fy <ax + by < 2 d$.
So $0 < (a-e)x +(b-f)y < d=\inf S$ which is impossible as $0 < (a-e)x +(b-f)y\implies (a-e)x +(b-f)y\in S$.
So $d=\inf S = 0$.
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Which means for any $\epsilon > 0$, then $\epsilon$ is not a lower bound of $S$ so there is an $ax + by \in S$ so that $0 < ax+ by < \epsilon$.
