# Cauchy sequence, help plz

• Oct 8th 2008, 10:12 AM
Cato
Cauchy sequence, help plz
If anyone can help me with this problem, would be greatly appreciated.

Let b be $\mathbb R$ with b > 0 . Show that there exists a Cauchy sequence of Rational numbers { $\ a_n$} with $\ n \geq 1$ such that $\ a_n$ > 0 for all n of N and that $\ lim_{n -> \inf} a_n = b$
• Oct 8th 2008, 10:20 AM
ThePerfectHacker
Quote:

Originally Posted by Cato
If anyone can help me with this problem, would be greatly appreciated.

Let b be $\mathbb R$ with b > 0 . Show that there exists a Cauchy sequence of Rational numbers { $\ a_n$} with $\ n \geq 1$ such that $\ a_n$ > 0 for all n of N and that $\ lim_{n -> \inf} a_n = b$

If $b$ is rational then there is nothing to prove - why?

Thus, it is safe to assume $b$ is irrational. The for any $n\geq 1$ there exists a rational number $a_n$ so that $-\tfrac{1}{n} < b - a_n < \tfrac{1}{n}$. Define a sequence $\{a_n\}$. Then it turns out this sequence is Cauchy with $\lim a_n = b$.
• Oct 8th 2008, 10:26 AM
Cato
But it says b is a real number, not rational.
• Oct 8th 2008, 10:32 AM
ThePerfectHacker
Quote:

Originally Posted by Cato
But it says b is a real number, not rational.

So what? There are two possibilities. Either it is rational or it is irrational. If rational there is nothing to prove. If irrational then we use the fact that between any two irrational numbers there is a rational number - that is the second case.
• Oct 8th 2008, 10:36 AM
Plato
Between any two numbers there is a rational number.
$\left( {\forall n \in \mathbb{Z}^ + } \right)\left( {\exists r_n \in \mathbb{Q}} \right)\left[ {b < r_n < b + \frac{1}{n}} \right]$.
Now it is easy to see that $\left( {r_n } \right) \to b$.
Any convergent sequence is a Cauchy sequence.
• Oct 8th 2008, 10:44 AM
Cato
Thank you both for the quick replies