Prove that if with , then there exists such that .

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

Then, since and (since and using I.3.5, Exercise #4) and so their product is also greater than 0.

Then, by adding ,

So, letting , we have with

as requested

By absurd, if we suppose there is not a z for x<z<y then we imply that x=z and that is absurd since the letter says x<y.

If we form a set K with x as the Inf K and y as the Sup K, then there is a z in the real numbers for which Sup K – Inf K < z(theorem I.32).

How do you develop the intuition to construct such a proof? How do you know where to start?

Thank you.

at this point, how do we know that 2(y-x)>(y-x)?

Because de teorem I.30 but he take n=2

Thanks for the explanation. I was also thinking there is no theorem tells us that (y-x) + (y-x) = 2(y-x) at this point.

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.

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!!

Maybe an alternative solution:

Let S={s|sy – (y-x)=x

Hence we have x<z<y.

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.

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