unsolvable (math) problem for a whole year, help me!

[uPWarrior]

You are right, I was focusing on the equivalence transformation and totally messed up the negation part.

[lemmata]

Counterexamples are nice because they make short proofs. You should love them!

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What is the following set X?

X =
{(a1, a2, b1, b2, x)|
      ({a1, a2, b1, b2, x} ⊂ {x|x is a real number})
  and (a1<x<a2 ⇔ b1<x<b2)
  and ~(a1=b1 and a2=b2)
}

('~' means 'not')

When you ask "what is", you must mean something more specific that you are forgetting to tell us. You have already defined X by describing the conditions that characterize its membership. That already says what set X "is". Let us also shorten the definition of X by letting R denote the real line. The parts in "" can be omitted.

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X = { "set of all" (a1, a2, b1, b2, x) in R^5
   | "such that" (a1<x<a2 ⇔ b1<x<b2) and ~(a1=b1 and a2=b2) }

MJK, can you use the following sets to make X? There's a long way and a short way. The short way takes advantage of the relationship between E4 and F.

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A={(a1,a2,b1,b2,x) in R^5| a1<a2 and b1<b2 and max(a1,b1) < x < min(a2,b2)}
B={(a1,a2,b1,b2,x) in R^5| a1<a2 and b1<b2 and (x <= min(a1,b1) OR x >= max(a2,b2)}
C={(a1,a2,b1,b2,x) in R^5| a1<a2 and b1 >= b2 and (x <= a1 OR x >= a2)}
D={(a1,a2,b1,b2,x) in R^5| a1 >= a2 and b1<b2 and (x <= b1 OR x >= b2)}

E1={(a1,a2,b1,b2,x) in R^5| a1 > a2 and b1 > b2}
E2={(a1,a2,b1,b2,x) in R^5| a1 > a2 and b1 = b2}
E3={(a1,a2,b1,b2,x) in R^5| a1 = a2 and b1 > b2}
E4={(a1,a2,b1,b2,x) in R^5| a1 = a2 and b1 = b2}

F={(a1,a2,b1,b2,x) in R^5| ~(a1=b1 and a2=b2)}

If you stumble, try verifying that (a1<x<a2 ⇔ b1<x<b2) holds for every member of the sets A, B, C, D, En.

Does this help visualize cross-sections of the 5-dimensional set?

Long way: X= (A u B u C u D u E1 u E2 u E3 u E4) INTERSECT F
Short way: X= (A u B u C u D u E1 u E2 u E3)

Note that A, B, C, D, E1, E2, E3, E4 have no overlap.

[Bill Spight]

Yes I am trying to clearly define a set of a five-dimensional space.

But you only have 2 unbound variables in your original problem.

Edit:

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-b/2 < a+b < b/3
a/3 < a+b < -a/2

Now let's introduce another binding:

a = 1-b

which gives us

-b/2 < 1 < b/3
(1-b)/3 < 1 < (b-1)/2

Now let b = 3.5

which gives us

-3.5/2 < 1 < 3.5/3
-2.5/3 < 1 < 2.5/2

which is true.

What is not true is this:

-b/2 < x < b/3 
<=>
(1-b)/3 < x < (b-1)/2

when x and b are unbound.

In that case the second inequality must hold for [i]all[/i] x in the range of the first inequality. It is the constraint upon x that allows both inequalities to hold.

[iazzi]

The new statement is not equivalent to the original one. The point there is that in the transformed inequalities the bounds depend on the term between them. After you say x = alpha+beta, you should only leave one other independent variable, traditionally y=alpha-beta, but any other combination (e.g. 2alpha+beta) would be ok.

For example
-b/2 < a+b < b/3
a/3 < a+b < -a/2
can become, leaving only a as an independent variable
-x/2 + a/2 < x < x/3 - a/3
a/3 < x < -a/2
which are trivially equivalent, but do not have the same bounds.

now the question is, given four functions of x: a(x), b(x), c(x), d(x) and knowing that
a(x)<x<b(x) iff c(x)<x<d(x)
can we say that a=c, b=d?

The answer is, clearly, no, as the original problem shows.

Of course, if a, b, c, and d are constants, then the answer is yes, and the original problem is not a counterexample.

[lemmata]

The new statement is not equivalent to the original one.

He is no longer interested in the original problem. It is all crossed out. The "new problem" seems to be a principle that he previously believed to be true, which he erroneously applied to the original problem.

EDIT:
The original problem that is now crossed out (the equation image) seems to be the following.

Let a,b be reals such that -(3/2)b < a < -(2/3)b. Is it the case that a=b=0?

The short answer is that this cannot be true because it immediately leads to the contradiction that 0<0. In fact, the premise implies that b>0 and a<0.

The longer answer probably has to do with where the OP's logical error was. I believe Dr. Straw covered that and the OP agreed.