# 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.

• Why should Cauchy sequences be convergent? – Asaf Karagila Sep 11 '13 at 22:41
• As Michael mentioned, you can have a cauchy sequence of rational numbers whose limit is not itself a rational number. It's these limits which complete the rationals into the reals. – Sai Sep 11 '13 at 22:46

This is a simple misunderstanding of definitions. A sequence is $\mathbb{Q}$-convergent if it converges to a rational number L as you stated. However a sequence can be $\mathbb{Q}$-cauchy without being $\mathbb{Q}$-convergent. A real number r is then defined as equivalence classes of $\mathbb{Q}$-cauchy sequences which converge to r in the traditional sense.

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.

I struggled with this exact idea when I first learned the reals-from-rationals construction. Here's the example that made me understand.

Every irrational number has an (infinite) decimal representation. Construct that particular irrational by enumerating the decimal representation, one digit at a time. For example, the sequence that converges to $\pi$ is:

$3$

$3.1$

$3.14$

$3.141$

$3.1415$

$3.14159$

and so on. Each of these terms is clearly rational ($3.14159 = \frac{314159}{100000}$), but the limit is clearly $\pi$.