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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.

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    $\begingroup$ I don't know if this is too easy of a proof. But you can always consider the average of two real numbers. $\endgroup$ – randomgirl Mar 30 '15 at 2:44
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    $\begingroup$ how about the midpoint? $\endgroup$ – abel Mar 30 '15 at 2:44
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Hint: if $a<b$, then $$ \frac{a+b}{2} = a + \frac{b - a}{2} = b - \frac{b - a}{2} $$

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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 $$ c = \dfrac{a + b}{2} > \dfrac{a + a}{2} = a $$

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Let $x,y,z \in R$

Such that

$x<y<z$

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

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

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

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