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I was communicated the following 1994 Miklos Schweitzer problem:

Is there an ordering of the real numbers such that whenever $x<y<z$ (in this ordering), we have $y \neq (x+z)/2$?

I really have no idea how to approach this problem. It is not really on my top list of priorities right now. I'm just curious about the answer.

1) It is obvious that such an ordering should exist? Why? If such an ordering does exist, is it helpful somehow?

2) If we want to prove that such an ordering does not exist, then what is the right path to the proof?

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Any problem that appears in Miklos Schweitzer is going to very hard (much harder than IMO). I would not even mention the word obvious. – Aryabhata Feb 12 '12 at 8:32
The wording of the problem just states 'ordering'. If this is the case (rather than 'total ordering', as you say in the title) then a trivial example is the discrete ordering, where $x \le x \Leftrightarrow x=x$. Then there do not exist $x,y,z$ such that $x < y < z$, so the statement is vacuously true. (I'm quite sure the problem $does$ want a total order though.) – Clive Newstead Feb 12 '12 at 9:42
@Clive: In many places the term "ordering" is used only in the context of linear orders, much like well ordering is always linear, despite having a very good notion for well orders which are non-linear. – Asaf Karagila Feb 12 '12 at 10:32
My intuition gives me the positive answer. I would try to define such an order using a transfinite induction method, on each step "killing" triples $(a,(a+b)/2,b)$. EDIT: Oh, I've just noticed the answer of Aryabhata. But looks like the authors of the article also use AC and similar arguments. – Damian Sobota Feb 12 '12 at 13:39
up vote 3 down vote accepted

Looks like the answer is yes, there is such an ordering. An article titled "Chaotic ordering of rationals and reals" appeared in December 2011 issue of the American Mathematical Monthly.

The first page of the article can be seen here:

Apparently the question is due to Erdos and Graham.

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