and the following inequalities
Is it possible to deduce the numerical order of a, b and c ?
It is not possible. There are six cases:
1. a < b < c
2. a < c < b
3. b < a < c
4. b < c < a
5. c < a < b
6. c < b < a
Now, no expression containing a is ever less than anything, so I zeroed in on cases 4 and 6. The following assignments satisfy all inequalities:
Case 4. a = 5, c = 2, b = 1
Case 6. a = 5, c = 1, b = 1.5
Therefore, you cannot deduce the correct ordering. My observation that no expression containing a is ever less than anything is, I think, part of the reason you can't make the deduction.
I had come to a similar conclusion but wasn't sure if I'd missed something. I could get some partial results which appeared promising, but then they slowly melted away, lol.
Another example (where a is less) is a=2, b=4, c=3
Oh well, back to the asylum
I'll jump in as Ackbeet is offline, he can correct me later if I'm wrong
They are simply all the possible relationships for
in relation to my o.p.
To put it another way (same result) just take all combinations of a b & c
and place a > between them. One of them must be correct (as there is no other way they can be ordered), but in this particular case, it is impossible to find out which is correct with the given information.
I wonder is there an ordering theorem justifying those ordering.
By "what rule" are we allowed to place the ordering in between a combination ,as you showed in your answering post??
It's not a 'rule', it's pure logic. If
then we must have a largest, middle and smallest number.
If we try all possible combinations (only 6 of them as given) and can find examples where each can be bigger than the other, as Ackbeet showed for a > b and a > c then we can say that there is not enough information in the original inequalities to put a, b and c in order of size.