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working on proving:

Let $x,y \in \mathbb{R}$ be such that $x < y$. There exists $z \in \mathbb{R}$ such that $x < z < y$

This seems obvious but I'm having trouble using the properties of the real numbers to get this proof started in a way where it will be succinct. Any help is appreciated.

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    $\begingroup$ Hint: try showing the average $\frac{x + y}{2}$ lies between them. $\endgroup$
    – Theo Bendit
    Commented Apr 8, 2019 at 4:28

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As good a point is any as the midpoint between $x,y$, i.e. $(x+y)/2$. Any scalar multiple (between zero and one) of $x+y$ will do but the midpoint is always the easiest to think over.

So, we suppose

$$x < \frac{x+y}{2} < y$$

and want to prove this. To show the first half of the inequality, notice that, since $x<y$, $x/2 < y/2$. As a result,

$$\frac{x+y}{2} = \frac x 2 + \color{blue}{\frac y 2} > \frac x 2 + \color{blue}{\frac x 2} = x$$

Using the same property - $x/2 < y/2$ - you can similarly get $(x+y)/2 < y$. I'll leave that bit to you.

I suppose if you want to rigorously complete the proof you also have to show that $(x+y)/2$ exists and is a real number in the first place, but we immediately get that since $x,y,2\in \Bbb R$ and the real numbers form a field. Not sure if that level of rigor would be expected in your analysis class though.

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  • $\begingroup$ Thanks so much, this makes perfect sense now $\endgroup$
    – Stanley Hudson
    Commented Apr 8, 2019 at 16:13

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