If Q-Cauchy sequences must limit to a rational, how can they construct the reals? I'm currently in a graduate Math for Economists course, and we spent last week learning how to construct the reals from the rationals using Cauchy sequences and their equivalence classes. I know that I'm missing something, because from the definitions presented to me, I fail to see how the Q-Cauchy sequences can represent the irrationals.
We define each real number,r, as the equivalence class of Cauchy sequences with r as their limit. Yet, without the reals, we only have Q convergence. So a sequence Q-Converges if for all $n>N, |a_n-L|< E,$ where $L,E$ are rational.
So, how can we define $\pi$, or $\sqrt{2}$? There is no Q-Cauchy sequence which Q-converges to anything other than a rational.

I'm not sure I understand your question properly but here goes..
It is not true that all Cauchy sequences in $\mathbb{Q}$ converge to a limit in $\mathbb Q$. For example consider $a_{n} = F_{n+1} / F_n$, where $F_n$ is the $n$th Fibonacci number. It is well known that this converges to the golden ratio $\phi = \dfrac{1+\sqrt{5}}{2}$ which is irrational. There is an 'elegant' sequence of rationals that converges to $\sqrt{2}$ but I can't think of it off the top of my head.
In other words, the metric space of rational numbers is not complete, however adding irrational numbers to it turns it into a complete metric space.
