Life In 19x19

DrStraw wrote:......

Prove that below is not true.

given,
{a, b, c, d, X} ⊂ {x|x is a real number}
a<X<b <=> c<X<d

then,
a=c and b=d

MJK wrote:

DrStraw wrote:......

Code: Select all

Prove that below is not true.

given,
{a, b, c, d, X} ⊂ {x|x is a real number}
a<X<b <=> c<X<d

then,
a=c and b=d

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)

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<b <=> 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.

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

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.

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

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.

Bill Spight wrote:
Well, if a<x<b => 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.

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.

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')

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: Select all

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')

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

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