Proving that between any two real numbers there exist a real number Formally, I want to prove that if $x$ and $y$ are real numbers such that $x \lt y$ then there exists a real number $z$ such that $x \lt z \lt y$.
I want to know whether, in constructing the proof, I should think in terms of continuity of the real line or not. In other words, is the proof of the existence of a real between any two reals equivalent to proving the real line is continuous?
This is a question in a calculus textbook I am going through as part of a self study project to brush up on my calculus. The text is called Calculus Volume 1 written by Tom M. Apostol page 19 exercise 1.14*.
 A: First, let me say a bit about the end of your question. I'm not sure what you mean by "the real line is continuous," but I suspect you mean that the real line is connected or complete. If so, then no, that's a different thing; for example, $\mathbb{Q}$ has the "density" property here (between any two rationals, there's another rational) but is neither connected (consider $(-\infty,\pi)$ versus $(\pi,\infty)$, for example) or complete (consider the sequence $3,3.1,3.14,3.141,...$, for example).
Since connectedness/completeness can be a bit technical at first, let me say the following: intuitively, a space is connected if it doesn't "break into a bunch of separate pieces," and is complete if it "doesn't have any gaps" - the set $\mathbb{Q}$ of rationals, even though it's dense, has lots of gaps and breaks apart really easily.

As to proving the claim you're looking at, I suspect you're thinking too hard about it. First, try to guess what a good formula for such a $z$ might be, and then try to prove (using whatever formal system you've been given) that it actually works. And this formula will be quite simple. 

HINT: if on a quiz Sam scored $10/10$ and Alex scored $6/10$, then together they got an ---- score of $8/10$. Now, how do you calculate the "---" of two numbers?

