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

MJK · #1

Everything is in the hide tag below.
Attachment:

qqewew.PNG [ 16.79 KiB | Viewed 10864 times ]

This situation occurred while trying to solve a messed up inequality problem. I soon realized it never works this way and somehow found another way, but after a year I still cannot find an error in that definitely non-sense..... expression of thought(?).

After seeing the thread insisting that a right triangle can at the same time be an equilateral triangle on a 2 dimensional Euclidean plain, I again realized that this website has quite a lot of math people by whom I can probably get some help.

[edit] minor fix of notation error caused by a badly coded [s]sequenceformula writer [/edit][/s]

Better presentation of the problem below.

What is the error of generalizing that,

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

then,
a=c and b=d

?

Code:

Prove that below is not true.

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

then,
a=c and b=d

RBerenguel · #2

I can't follow it. What's the assumption beside them being reals? What's P? A clause? And A? And B?

MJK · #3

RBerenguel wrote:

I can't follow it. What's the assumption beside them being reals? What's P? A clause? And A? And B?

Oops, I'm sorry. I just almost copied directly my handnotes... so.

A, B, and P are all conditions of (alpha, beta). They basically mean the same thing. The diagram below shows the points that meet the condition on an R^2 plane. Alpha to the x-axis and beta to the y-axis.

Attachment:

dgdgd.PNG [ 5.16 KiB | Viewed 10846 times ]

Therefore, all the coloured points on the diagram above are the elements of the set {(alpha, beta)|P)}.

I hope my explanation was sufficient.

RBerenguel · #4

MJK wrote:

RBerenguel wrote:

I can't follow it. What's the assumption beside them being reals? What's P? A clause? And A? And B?

Oops, I'm sorry. I just almost copied directly my handnotes... so.

A, B, and P are all conditions of (alpha, beta). They basically mean the same thing. The diagram below shows the points that meet the condition on an R^2 plane. Alpha to the x-axis and beta to the y-axis.

Attachment:

dgdgd.PNG

Therefore, all the coloured points on the diagram above are the elements of the set {(alpha, beta)|P)}.

I hope my explanation was sufficient.

Makes a little more sense now.

A (or P) is redundant, they are the same inequality. Aside from that, what are you trying to prove/see from the inequalities A & B, then?

MJK · #5

RBerenguel wrote:

Makes a little more sense now.

A (or P) is redundant, they are the same inequality. Aside from that, what are you trying to prove/see from the inequalities A & B, then?

In the last few lines. Why do I get the result that alpha and beta equals zero? There must be my fundamental misunderstanding which I cannot find.

DrStraw · #6

How does -b/2 = a/3 and b/3=-a/2 follow from the previous line?

a=-1 and b=1 make the previous line true, but not this line.

MJK · #7

DrStraw wrote:

How does -b/2 = a/3 and b/3=-a/2 follow from the previous line?

a=-1 and b=1 make the previous line true, but not this line.

But isn't it
a<X c<X<d
a=c and b=d?

Because both inequalities have the same range.

RBerenguel · #8

The thing is, inequalities don't work like this. Even though a+b (alpha+beta) satisfy these two inequalities, it doesn't mean the extremes have to coincide. You wouldn't say:

5<d<7
2<d<9

And conclude that 5=2 and 7=9.

From the last line you could deduce:

-0.5b+0.5a<0<1/3b-1/3a

Which is then, just

a<b.

The best way to study 3-way inequalities is do them by pieces, and always study the boundaries. So the important thing is the lines

-3/2b=a
a=-2/3b
b=1/2a
Etc

And their intersections (the first one is b=0 a=0, but from the other pair you get a=-3/4, b=-3/8). You need all four points from here.

Does this help? It's a little tricky typing this now, since I'm on my iPad

MJK · #9

RBerenguel wrote:

The thing is, inequalities don't work like this. Even though a+b (alpha+beta) satisfy these two inequalities, it doesn't mean the extremes have to coincide. You wouldn't say:

5<d<7
2<d<9

And conclude that 5=2 and 7=9.

Thank you for your such long reply with iPad.

But in your example
5<d<7 <!=> 2<d<9 (not necessary and sufficient condition of d)
5<d<7 => 2<d<9 would be true though

if
1<d<3 <=> a<d<b
a=1 and 3=b
this was what I meant.

RBerenguel · **#10**

MJK wrote:

RBerenguel wrote:

The thing is, inequalities don't work like this. Even though a+b (alpha+beta) satisfy these two inequalities, it doesn't mean the extremes have to coincide. You wouldn't say:

5<d<7
2<d<9

And conclude that 5=2 and 7=9.

Thank you for your such long reply with iPad.

But in your example
5<d<7 <!=> 2<d<9 (not necessary and sufficient condition of d)
5<d<7 => 2<d<9 would be true though

if
1<d<3 <=> a<d<b
a=1 and 3=b
this was what I meant.

In this case it is true, but only because this is one inequality with just one variable (the "solution" is an interval) but in the problem above, it is not.

MJK · **#11**

My question can be summarized as,

Code:

What is the error of generalizing that,

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

then,
a=c and b=d

?

DrStraw · **#12**

MJK wrote:

DrStraw wrote:

How does -b/2 = a/3 and b/3=-a/2 follow from the previous line?

a=-1 and b=1 make the previous line true, but not this line.

But isn't it
a<X c<X<d
a=c and b=d?

Because both inequalities have the same range.

You are trying to create an equivalence where there is none.

EdLee · **#13**

MJK, just curious: what's the background of this question ? Are you trying to understand it for fun ?
Or, is it a homework assignment ?

If it's homework, which school level ? Which subject ? (Mathematics, or physics, or something else, like economics ? )

MJK · **#14**

EdLee wrote:

MJK, just curious: what's the background of this question ? Are you trying to understand it for fun ?
Or, is it a homework assignment ?

If it's homework, which school level ? Which subject ? (Mathematics, or physics, or something else, like economics ? )

Background is a situation I faced during my trying to solve a horrible inequality problem. The original problem is quite far away from this and is already solved by a very different way. So everything is for fun, or because it feels very 'caught' due to 'unknowing'

On which level? I can currently deal with some basic analysis (or actually in the process studying it). The level of the answer is not much important. Best if I can understand everything, but even if not so I will essentially have dealt with more math years later, and most importantly I do want to know the answer.

I will re-state my question below. I already changed my original post with the newer version of my problem.

Code:

What is the error of generalizing that,

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

then,
a=c and b=d

?

DrStraw · **#15**

MJK wrote:

Code:

What is the error of generalizing that,

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

then,
a=c and b=d

?

As I stated above, the error is in "a<X c<X<d"

That simply is not true.

MJK · **#16**

DrStraw wrote:

......

Code:

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

DrStraw · **#17**

MJK wrote:

DrStraw wrote:

......

Code:

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

It is true. What is not true is that the <=> is a logical consequence of your original problem.

MJK · **#18**

Let me make things clear...

First, I am still not much familiar with math in English. What I meant by 'given' is actually 'in a situation that', so 'when' I think is a better word there.

So here is (probably) the final version of this problem.

Code:

Prove that below is not true.

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

then,
a=c and b=d

I realized that the proof is given in my very original version. There is a counterexample.

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,
!(a=c and b=d)

What I am really curious is why such counterexample exists and whether there can be a proof of this problem not using counterexamples.

lemmata · **#19**

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.

bayu · **#20**

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.

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

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