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There is a real number between any two given real numbers.

Prove that if x,y \in \mathbb{R} with x < y, then there exists z \in \mathbb{R} such that x < z < y.


Proof. (This is about the point in the blog at which I’m just going to use the basic properties of \mathbb{R} without much comment. As usual, if something is unclear, leave a comment, and I’ll clarify.)

    \begin{align*}   x < y & \implies (y-x) > 0 \\   &\implies 2(y-x) > (y-x) \\   &\implies (y-x) > \frac{y-x}{2}. \end{align*}

Then, \frac{y-x}{2} > 0 since y-x > 0 and \frac{1}{2} > 0 (since 2 > 0 and using I.3.5, Exercise #4) and so their product is also greater than 0.
Then, by adding x,

    \[ 0 < \frac{y-x}{2} < y-x \implies x < \frac{y+x}{2} < y.  \]

So, letting z = \frac{y+x}{2}, we have z \in \mathbb{R} with

    \[ x < z < y, \]

as requested. \qquad \blacksquare

7 comments

  1. abomb says:

    My version of anonymous’s solution which I like a lot more than the (x+y)/2 solution because we guarantee there exists a real instead of explicitly constructing one.

    x, y ∈ ℝ such that x < y
    S = { z y – h by T 1.32, that is z > x.
    Hence there is a real z such that x < z < y.

    • abomb says:

      Yah, something went wrong too with my reply. It’s just basically anon’s solution but I was using fancy unicode, which I guess this site doesn’t have an encoding for. Oh well. Great site!!

    • Anonymous says:

      Sorry but something went wrong as I posted the comment.
      S is the set of all the real numbers that are smaller than y.
      Apply theorem I.32 with h = y-x.
      For some real number in S, say z, we have
      z>y-(y-x)=x
      Hence, x<z<y.

      • Anonymous says:

        the only thing left is you need to prove sup S = y before you use z>y-(y-x). Otherwise, only “THERE EXISTS z>Sup S – (y-x)” is true

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