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

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