This proof problem haunts me in my dreams.
If $a,b$ are reals such that $a b$, prove that there always exists a rational $z$ such that $a z b$. I got the hint which told me to use the Archimedean Property and the Well-Ordering Theorem.

Question

This proof problem haunts me in my dreams.
If $a,b$ are reals such that $a > b$, prove that there always exists a rational $z$ such that $a > z > b$. I got the hint which told me to use the Archimedean Property and the Well-Ordering Theorem.

experimentx · Answer

3 cases to consider, 
1) both a and b rational
2) a is rational while b is not
3) both a and b are irrational

parthkohli · Answer

@experimentX: Jayesh is too occupied with your proof. Hehe =)

experimentx · Answer

lol yeah :D

shubhamsrg · Answer

well,, we can always say that
a>(a+b)/2 > b

since a and b are rational, so is (a+b)/2 ! ;)

parthkohli · Answer

I can consider the mean of $a$ and $b$ if we talk about (3).

parthkohli · Answer

@shubhamsrg "REALS".
I've received that wrong proof 3 times.

shubhamsrg · Answer

why's that wrong ?

accessdenied · Answer

(a+b)/2 is not rational if a and b are irrationals, which is possible as they can be any real number.  ;)

shubhamsrg · Answer

ohh,,am sorry! never noticed that ! :P

parthkohli · Answer

I got the proof on M.SE chat :) http://chat.stackexchange.com/rooms/36/mathematics

experimentx · Answer

did you download the solution i uploaded on MSE chat?

parthkohli · Answer

@experimentX: Downloading.

parthkohli · Answer

Closing it, thanks guys =)