Constructing a sequence Given two distinct, positive real numbers, how can I use these two numbers (and their non-zero integer linear combinations) to construct a sequence converges to zero? The sequence can only be of the two original positive numbers, or their non-zero integer linear combinations.
 A: Let's assume $0\lt x=x_0\le y=y_0$ are the two real numbers.  Define
$$x_{n+1}=\min(x_n,y_n-x_n)\quad\text{and}\quad y_{n+1}=\max(x_n,y_n-x_n)$$
The $x_n$'s (for $n\gt0$) is a sequence of nonzero integer linear combinations of $x_0$ and $y_0$.  It's not hard to show that it converges to $0$:  the inequalities
$$0\le x_{n+1}\le x_n\le y_n\le y_{n-1}$$
imply that each sequence, being non-negative and non-increasing, has a limit, say $x_n\to L$ and $y_n\to M$, with $0\le L\le M$, at which point $M=\max(L,M-L)$ implies $L=0$.
Remark (added later):  This construction can be thought of as the euclidean algorithm run in excruciating slow motion -- you gain simplicity of description at the expense of speed of convergence.
A: I misread the question, and thought the asker was just trying to show the sequence exists. Still, I'll leave the answer here since it is not totally trivial to show.

Call the two numbers $a$ and $b$. Let $c = \inf\{r: r > 0, r = ka + lb$ for some integers $k$ and $l\}$. It suffices to show that $c = 0$. Suppose $c$ were not zero; we will arrive at a contradiction. 
If there were distinct $k_n a + l_n b$ decreasing to $c$, then $(k_{n+1} - k_n)a + (l_{n+1} - l_n)b$ would decrease to zero as $n$ goes to infinity, implying $c = 0$, a contradiction. So we can assume there are not distinct $k_n a + l_n b$ decreasing to $c$. In other words there are some $k$ and $l$ such that $ka + lb = c$.
Next, observe that if there were $k'$ and $l'$ such that $k'a + l'b$ were not an integer multiple of $c$, then $mc < k'a + l'b < (m+1)c$ for some integer $m$, so that $0 < (k' - mk)a + (l' - ml)b < c$, contradicting minimality of $c$. 
So all $k'a + l'b$ are integer multiples of $c$. In particular $a$ and $b$ are integer multiples of $c$, meaning $a$ and $b$ are rational multiples of each other. Writing $a = {m \over n} b$ for integers $m$ and $n$ then $na - mb = 0$. This implies $c =0$, a contradiction and we are done.
A: ok, suppose we can use the floor function and we can compare numbers. what we do is pretty much the euclidean algorithm. let's start with $a > b$ and let $x_1 := a$, $x_2 := b$. assume $x_1, \ldots, x_n$ are already defined and the sequence is strictly decreasing and positive. if at any point we get $x_n = 0$ we stop (that's the case if $a$ and $b$ are proportional), so lets assume this doesn't happen. let now
$$x_{n+1} := x_{n-1} - [\frac{x_{n-1}}{x_n}]x_n$$
$x_{n+1}$ is an integer combination of $a$ and $b$ if all the $x_n$'s are. moreover $x_{n+1} < x_n$ because 
$$x_{n+1} - x_n = x_{n-1} - ([\frac{x_{n-1}}{x_n}] + 1 )x_n  < 0$$ since $([\frac{x_{n-1}}{x_n}] + 1 )> \frac{x_{n-1}}{x_n}$. this is a decreasing and bounded sequence hence it has a limit point $g$. if it's $0$ then we're done, so let's assume $g>0$. the floor function is right-continuous  (continuous from the right, i.e. it's continuous at $y$ when considered on an interval $[y, y+\epsilon)$. since our sequence is decreasing that's all we need - we can plug $g$ into the recurence definition which gives
$$g = g - [\frac{g}{g}]g = 0$$ and so we're done.
edit:
oh, to see that the combinations are never $0$ observe that if $a$ and $b$ are not proportional over $\mathbb{Q}$ (the only case we need to consider) this can never happen since if $x_{n-1}$ and $x_n$ are integer combinations of $a$ and $b$ and $x_{n+1} = 0$ then this means $[ \frac{x_{n-1}}{x_n}] = \frac{x_{n-1}}{x_n}$ so $\frac{x_{n-1}}{x_n} \in \mathbb{Z}$. let $x_{n-1} = ka + mb$, $x_n = ja + nb$, then $ka + mb = l(ja + nb)$ where all $k, m, j, n, l$ are integers. this means $a$ and $b$ are proportional over $\mathbb{Q}$, so we're done
