Cauchy sequence of increasing rational numbers Trying to work my way through this problem ... is this in the right direction?
Let $x$ be a real number. Show that there exists a Cauchy sequence of rationals $x_1,x_2,\cdots$ representing $x$ such that $x_n \leq x_{n+1}$ for every $n$.
We define $x_1,x_2,\cdots$ as follows. Let $y$ be the largest integer less than or equal to $x$, and let $y = x_1$. We define $x_2$ as the average of $x_1$ and $x$; put generally, we define $x_i$ as the average of $x_{i-1}$ and $x$. We know that $x_n \leq x_{n+1}$ by induction. For $P(1)$, we have $x_1 \leq x_2$ because $y = y/2 + y/2 \leq y/2 + x/2$ because $x > y$. For $P(n)$, we want to show $x_n \leq x_{n+1}$. We know $x_n = \frac{x_{n-1}+x}{2}$ and $x_{n+1} = \frac{x_{n}+x}{2}$. Since $x_{n-1} \leq x_n$ by the inductive hypothesis, we have $x_n \leq x_{n+1}$.
Now, we want to show that such a sequence is a Cauchy sequence representing $x$. To do this, we need to show that for every $n$, there exists an $m$ such that for every $j,k \geq m$, $|x_j-x_k| \leq 1/n$, and given a Cauchy sequence $(y_i)$ that represents $x$, we need to show that $(x_i) \sim (y_i)$.
I feel like this is so off; are there general tips for constructing a Cauchy sequence? Also, how do I find $m$?
 A: Your argument fails when you define $x_2$ as the average of $x_1$ and $x$, because $x_2$ is irrational if $x$ is irrational.
Let $a_0$ be the largest integer not exceeding $x$. Let $a_1$ be the largest nonnegative integer such that $a_0 + a_1/10 \le x$. Having chosen nonnegative integers $a_1,\ldots, a_n$, let $a_{n+1}$ be the largest nonnegative integer such that $$a_0 + \frac{a_1}{10} + \cdots + \frac{a_{n+1}}{10^{n+1}} \le x$$ This defines a sequence $(a_n)_{n=1}^\infty$ of integers such that $a_n \ge 0$ whenever $n \ge 1$. Now set $$x_n = a_0 + \frac{a_1}{10} + \cdots + \frac{a_n}{10^{n}}\quad (n = 0,1,2,\ldots)$$ Then $x_n \le x_{n+1}$ and $x - 10^{-n} < x_n \le x$ for all $n \ge 0$. By the squeeze theorem, $(x_n)_{n=1}^\infty$ converges to $x$. In particular, the $x$-sequence is Cauchy.
A: Basically what @kobe said, but I would phrase it differently:
Hint: Think about what decimals do. If you want to write pi as a decimal, you might write 3.14... or 3.14159... or 3.141592... etc. Of course (once you've had a few math classes) you probably round to the nearest decimal instead of truncating and leaving dots, but it's the same idea.
So, in response to your original question of "Am I approaching this in the right way?" Well, yes, in the sense that you're starting at a rational y and bumping it up to approach x from below, but also no, in the sense that you've missed the true problem with x, which is that it might not be rational but you only want to use rational numbers. Based on this, you could come up with all sorts of ways to approach x from below using rationals; you have to use rationals, which decimals certainly are, and you have to get closer and closer, and the decimal setup (aka using negative powers of 10) gives a beautifully intuitive way to justify that you've done that, but you can come up with all sorts of ways to vary the decimal example to get a reasonable sequence.
