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.

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.
