Prove that If rational numbers $y$ and $z$ are $\epsilon$ close to $x$ then so is $w$ which lies between $y$ and $z$

Question

By $\epsilon$ close I mean $|x-y| \leq \epsilon$ for some rational $\epsilon > 0$

I could prove it by representing $w$ as $w = \theta_1 y + (1-\theta_1)z$ where $0\leq\theta_1\leq1$, and then using triangle inequality for

$\theta_1 |x-y| \leq \theta_1 \epsilon $ and

$(1-\theta_1) |x-z| \leq (1-\theta_1) \epsilon $

But I am not content with the proof because I didn't even prove that $w$ can be represented that way.

I am looking for some other proof which just uses the fact that $w$ lies between $y$ and $z$ without using the linear combination part.

I also tried representing $w = y - \delta_1$ and $w=z+\delta_2$ for positive rationals $\delta_1$ and $\delta_2$, but it was not getting me anywhere.

Answer 1

Assume $y\leq z$ without loss of generality. $|x-y|<\epsilon$ and $|x - z|<\epsilon$, which is to say that $x-\epsilon < y \leq z < x+\epsilon$. Then $x-\epsilon<y\leq w$, and $w\leq z < x+\epsilon$, so we have $x-\epsilon < w < x+\epsilon$, i.e. $|x - w| < \epsilon$.

Answer 2

In finding proofs for such intuitively obvious results it is better to visualize the situation on the real number line. The purpose of using the real number line is not to replace the formal proof (as given in the answer by user florence) but rather to aid yourself in obtaining a formal proof with almost no effort. In the absence of such visualizations one is often lost in symbol manipulation game.

The situation of the current question can be depicted on the number line as follows:

--------------$x - \epsilon$----$y$---------$x$------$z$-------$x + \epsilon$-------------------

Thus one observes that if $y, z$ are $\epsilon$-close to $x$ then both $y, z$ lie in the interval $[x - \epsilon, x + \epsilon]$, it follows that any number $w$ between $y, z$ also lies in that interval and hence is also $\epsilon$-close to $x$ (remember an interval is a set $A$ of real numbers with the property that if two numbers lie in $A$ then all the numbers between them also lie in $A$).

The above argument proves the result when all the numbers concerned are real numbers instead of rational numbers, but since rationals are also reals the desired result is achieved.

You were trying to use a formula for any $w$ between $y$ and $z$ and this sort of suggests that you were trying to use formula for $w$ to establish directly $|w - x| \leq \epsilon$ using some basic rules of inequalities. This is what one should avoid because it unnecessarily turns simple proofs of real analysis into a complex symbol manipulation. Most of the proofs in real analysis (especially the introductory ones) are based on the appreciation of order relation of real numbers and don't deal so much with formulas and their algebraic manipulation.