# Showing that there's a Cauchy sequence where Xn<X for all n?

## Homework Statement

"If x is a real number, show that there exists a Cauchy sequence of rationals Xl, X2,... representing X such that X n < x for all n."

## Homework Equations

- All Cauchy sequences are convergent
- All Cauchy sequences are bounded.

## The Attempt at a Solution

These proofs that involve Cauchy sequences have been rough on me, and I'm trying to start working through them rather than just hunting for solutions.

But I just don't know quite where I should start. What should I be assuming or trying to show? Do I start with a series where Xn<X for all n that converges to x and show it's Cauchy? Just kinda stumped so far :/

Use the density of rationals and then find a convergent sequence that converges to x from "below".

So should I start by assuming something is Cauchy, or by taking a sequence and then proving it to be Cauchy?

No, that is not what I said. :-)

Pick a rational number in (x-1,x) call it the first element.
Pick the next in (x-(1/2),x) call it the second. Keep doing this ...
Pick the nth term from (x- (1/n) ,x).

Show the above sequence is convergent and it converges to x.

Since x_n is convergent is it cauchy ?

HallsofIvy
Homework Helper
If x is any real number, then it can be written in a decimal expansion. let xn be the number formed from the first n decimal places of x.

For example, if $x= \pi$, then x= 3.1415926... so the "Cauchy sequence of rational numbers" is 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, 3.1515926, ...

No, that is not what I said. :-)

Pick a rational number in (x-1,x) call it the first element.
Pick the next in (x-(1/2),x) call it the second. Keep doing this ...
Pick the nth term from (x- (1/n) ,x).

Show the above sequence is convergent and it converges to x.

Since x_n is convergent is it cauchy ?
Awesome, thanks! I was able to figure it out with this. Much appreciated.

HallsofIvy
Any real number has an expression as a decimal expansion: $a.b_1b_2b_3\cdot\cdot\cdot$ or, equivalently, $a+ b_1/10+ b_2/100+ b_3/1000+ \cdot\cdot\cdot$, where a is an integer and every b is an integer from 0 to 9.
Look at the sequence $a$, $a.b_1$, $a.b_1b_2$, $a.b_1b_2b_3$, ...
Or, equivalently, $a$, $a+ b_1/10$, $a+ b_1/10+ b_2/100$, $a+ b_1/10+ b_2/100+ b_3/1000$, ...