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.

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
