1. inequalities

Given

$a\not=b\not=c\not=0$

$a>0,\ \ \ b>0,\ \ \ c>0$

and the following inequalities

$4a > 2c > b$

$a+b > c$

$a+c > b$

$2a+b > 2c$

Is it possible to deduce the numerical order of a, b and c ?

$i.e.\ \ \ a > b > c\ \ \ or\ whatever\ is\ correct.$

Thanks

Pro

2. Re: inequalities

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.

3. Re: inequalities

Thanks Ackbeet,

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

Pro

4. Re: inequalities

Originally Posted by Ackbeet
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 have one question to ask that really bothers me :

How do we get those six inequalities that you wrote above?

i know that if you position those variables on the line of real Nos then Geometry shows the possible order .

But mathematically (algebraically) how??

5. Re: inequalities

Hi psolaki,

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 $\ \ a\not=b\not=c$
in relation to my o.p.

To put it another way (same result) just take all combinations of a b & c

abc
acb
bac
bca
cab
cba

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.

hth

pro

6. Re: inequalities

Originally Posted by procyon
Hi psolaki,

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 $\ \ a\not=b\not=c$
in relation to my o.p.

To put it another way (same result) just take all combinations of a b & c

abc
acb
bac
bca
cab
cba

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.

hth

pro
Thanks ,man ,but my problem is that i tried hard to find a theorem or an axiom or a definition to base those orderings .

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??

7. Re: inequalities

Hi,

It's not a 'rule', it's pure logic. If $\ \ a\not=b\not=c$
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.

Pro

8. Re: inequalities

Why, pure logic has no rules?? ,anyway thanks for the explanation