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

Bill Spight · **#21**

bayu wrote:

I think the original question is:

Fix 4 reals a, b, c, d.

Now you can ask, whether the following statement is true:

if for all x we have (a<x c<x<d)
then a=c and b=d.

This statement is true if a<b or c<d, and it is false if not (e.g. a=1,b=0,c=3,d=2)

From afar it looks like a confusion of quantifiers.

Well, if a<x c<x<d AND c<x<d => a<x<b, then by symmetry we cannot have a<c because that would mean that c<a. Dr Straw has diagnosed the problem.

MJK · **#22**

lemmata wrote:

You are confusing yourself because you did not word your problem correctly to yourself.

Quote:

Is the following implication true FOR EVERY 5-tuple (a1,a2,b1,b2,x) of real numbers?

If (a1<x<a2 if-and-only-if b1<x<b2) then (a1=b1 AND a2=b2)

To disprove it, you need to find that FOR SOME 5-tuple (a1,a2,b1,b2,x) of real numbers the premise (part after if) is true but the conclusion (part after then) is false. Counterexamples are obvious and plentiful at this stage.

Is it possible for two distinct open intervals to contain the same point? If this is obvious to you, then your mystery might be solved.

Thank You! You have exactly understood and clarified my problem.

However, there are still some mysteries left on me.

As in my previous post, my very original question does contain a counterexample (quoted below), so it might be enough to disprove the fact.

Code:

Let,
a = -B/2
b = B/3
c = A/3
d = -A/2
x = A+B

(A, B are real numbers.)

-B/2<A+B<B/3 <=> A/3<A+B<-A/2
a<X<b <=> c<X<d

however, the two intervals on each side are not the same.
-B/2 != A/3 or B/3 != -A/2

But what I want to know is the reason or the mechanism behind such result of being disprovable which contradicts with my intuition. I need a different kind of proof for example as below in 'Proof 2'.

Code:

Is the following implication true for every 2-tuple (a, b) of real numbers?

if (a^2 = b^2), then (a = b)

No.

Proof 1
counterexample a = -1, b = 1

Proof 2
a^2 = b^2
⇔ a^2-b^2 = 0
⇔ (a+b)(a-b) = 0
⇔ a = ±b

a = ±b <== a = b

Therefore, the implication is not true.

bayu · **#23**

Bill Spight wrote:

Well, if a<x c<x<d AND c<x<d => a<x<b, then by symmetry we cannot have a<c because that would mean that c<a. Dr Straw has diagnosed the problem.

I'm not worrying about the maths. I'm a fellow mathematician

(and on your side concerning infinities, diagonals and all the other highly controversial topics that crop up among go players here:). I believe that there is still a misunderstanding between the OP and the people that are trying to help. And my mother tongue is not english neither, so I might be off, in both understanding and helping.

Your statement is correct except in the kind of degenerate case that both intervalls in question are empty.

I just got sniped by the OP. And in his example the intervall A/3<A+B<-A/2 is empty (if A is positive). Maybe, part of the misunderstanding lies here.

MJK · **#24**

bayu wrote:

And in his example the intervall A/3<A+B<-A/2 is empty (if A is positive). I'm pretty sure, the misunderstanding lies here.

The interval is not empty, and n({(A, B)|A/3 < A+B < -A/2}) = ∞
(n(X) means number of the elements of set X)
I never stated the numbers should be positive but rather {A, B} ⊂ {x|x is a real number}

As lemmenta clarified, there are no more misunderstandings. The existence of counterexamples are obvious.

Again I shall restate my CURRENT problem more clearly.

Code:

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')

RBerenguel · **#25**

MJK wrote:

bayu wrote:

And in his example the intervall A/3<A+B<-A/2 is empty (if A is positive). I'm pretty sure, the misunderstanding lies here.

The interval is not empty, and n({(A, B)|A/3 < A+B < -A/2}) = ∞
(n(X) means number of the elements of set X)
I never stated the numbers should be positive but rather {A, B} ⊂ {x|x is a real number}

As lemmenta clarified, there are no more misunderstandings. The existence of counterexamples are obvious.

Again I shall restate my CURRENT problem more clearly.

Code:

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')

This set is somewhat weirdly stated, since it feels like there is a kind of free variable in there (x) but isn't.

As I told you before, 3<5<7, 1<5<7 satisfies this, kind of. The iff part defining this set feels weird, though, since it seems to state it for all x in the set at the same time. And without much thinking, I think there should be no way this happens for every x in a set which has an x. Also, as before, if you want to study this set:

a1=x &
x=a2 &
b1=x &
b2=x.

This defines the boundary of the set of suitable x values, for any x in \mathbb{R}. And this boundary (which is outside the set, since the inequalities are strict) seems to be just a point, hence X=empty because there can't be any value x satisfying this, given any choice of a1, a2, b1, b2 in \mathbb{R}.

bayu · **#26**

We're getting closer

let's drop the two extra conditions for a second. The hairy one seems so be the first one, which is not stated properly, so that it can lead to misunderstandings.

There are 2 ways to interpret your statement:

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

Are a1, a2, b1 and b2 some fixed real numbers? Then X is a subset of the reals. The subset itself depends on the choice of the a's and b's.

If a1, a2, b1 and b2 are variables too than your set X is a subset of a 5-dimensional space.

I suspect the former.

MJK · **#27**

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

As RBerenguel pointed out, it is currently not well defined.

I will need to think more to do this.

Perhaps some clues of the properties of set X will occur after being clearly defined.

RBerenguel · **#28**

MJK wrote:

Perhaps some clues of the properties of set X will occur after being clearly defined.

Very likely!

uPWarrior · **#29**

Have you tried describing the equivalence operator differently so that your intuition improves?

a <=> b is the same as (a AND B) OR (!a AND !b),

therefore

a<X c<X<d is the same as (a<X=X>=b AND c>=X>=d)

Is the set clearer to you if describe it like this? It basically means that the value X is always between a and b and between c and d, as long as a-b and c-d have the same sign.

Tryss · **#30**

uPWarrior wrote:

Have you tried describing the equivalence operator differently so that your intuition improves?

a <=> b is the same as (a AND B) OR (!a AND !b),

therefore

a<X c<X<d is the same as (a<X=X>=b AND c>=X>=d)

Is the set clearer to you if describe it like this? It basically means that the value X is always between a and b and between c and d, as long as a-b and c-d have the same sign.

Be careful, the negation of "a<X<b" is not "a>=X>=b", but "X<a OR b<X"

uPWarrior · **#31**

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

lemmata · **#32**

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

MJK wrote:

Code:

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.

Code:

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.

Code:

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?

Bill Spight · **#33**

MJK wrote:

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:

Code:

-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 · **#34**

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 · **#35**

iazzi wrote:

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.

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

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