# Analysis Proof

1. Sep 5, 2011

### Zhalfirin88

1. The problem statement, all variables and given/known data
Prove that if x in R and epsilon > 0 are arbitrary, then there exist r in Q such that |x - r | < epsilon

2. Relevant equations

3. The attempt at a solution

I'm stumped on this one. I tried using the reverse triangle inequality, but I seemingly hit dead ends with it.

2. Sep 5, 2011

### tylerc1991

Can you use the fact that the rationals are dense in the reals? This means that any real number can be approximated by a rational number.

3. Sep 5, 2011

### Zhalfirin88

I know what you are saying, but how do I write that (this is my first proofs class).

By the denseness of Q, you can say that epsilon < p < q < ... < r < x

But I'm not sure how you get the inequality in there. Feel like I'm missing something

4. Sep 5, 2011

### tylerc1991

The assumption is that you are given an $x \in \mathbb{R}$ and an $\varepsilon > 0$. Using the fact that the rationals are dense in the reals, any real number can be approximated by a rational number. In other words, there exists a $r \in \mathbb{Q}$ such that $r$ is arbitrarily close to $x$, or $|x - r| < \varepsilon$.