# Proof for discrete math

I am struggling with understanding this proof. It was on one of my tests and I am lost on how to prove it.

Prove: For any two real numbers that are not equal, you can find a real number between them.

• I don't know if this is too easy of a proof. But you can always consider the average of two real numbers. – randomgirl Mar 30 '15 at 2:44
• how about the midpoint? – abel Mar 30 '15 at 2:44

Hint: if $a<b$, then $$\frac{a+b}{2} = a + \frac{b - a}{2} = b - \frac{b - a}{2}$$

Let $a,b \in \mathbb{R}$ such that $a < b$. The number $c = (a + b)/2$ satisfies $a < c < b$. Indeed: $$c = \dfrac{a + b}{2} < \dfrac{b + b}{2} = b$$ and $$c = \dfrac{a + b}{2} > \dfrac{a + a}{2} = a$$

Let $x,y,z \in R$

Such that

$x<y<z$

Then $y=\frac{x+z}{2}$

Thus $x<\frac{x+z}{2}<z$

• If $x < y < z$, does it really follow that $y = \frac{x+z}{2}$? – pjs36 Mar 30 '15 at 3:00