Bell’s Inequality simplified

[Lazarus]

Bell’s Inequality Simplified

 

P(a,b) -P(a,b’)+P(a’,b)+P(a’,b’) <= 2

 

a+b-a-b’+a’+b+a’+b’

------------------------------ <=2

a+b+a’+b’

 

b+a’+b+a’

--------------- <=2

a+b+a’+b’

 

2(b+a’)

-------------------- <=2

(b+a’)+(a+b’)

 

2

----------------- <=2

1 + (a+b)

-------

(b+a’)

 

1

----------------- <=1

1 + (a+b)

-------

(b+a’)

 

and

 

(a+b)

0 <= ---------- <= infinity

(b+a’)

 

Therefore: 1 divided by 1 or greater is less than or equal to 1.

[TakenItSeriously]

So is that directed at assuming that is sometthing I'm failing to understand within my thread. I assume it is since your posting an accepted arguement in the speculations forum. I dont mind responding in this thread since you did me the kindness of not derailing my arguements.

 

Its just that my premise precludes the set theory he used wich assumes a non biased even distribution of all possible combinations, as well as precluding the impact of entanglement itself.

 

it could be possible that just be one of those points is true, btw.(edited to claiify)

 

I can actually give an explanation of how this could work in a simple explanation if your willing to wait and see How it works in my thread. which Im working on even now. But this gives you at least an idea of where I'm heading.

 

My premise is that not all combinations are possible and many are biased in their distribution.

 

Furthermore the rest of the set theory arguement is offset by the entanglement itself where information is shared in hindsite between Alice and Bob, despite the fact that they are unaware of it in realtime or proper time, if you like which plays havoc with set theory.

 

Thats because the results are recorded in the hind site of knowing all results. and thats where the information of entanglement was actually shared.

 

I hope that makes sense.

 

If I was wrong about my assumption I oppologize but this is still my reply as its directly directed at the postulate you gave.

Edited March 7, 2017 by TakenItSeriously

[imatfaal]

Just no. You can only divide through by a factor for instance (a+b) and cancel when the factors are multiplied

 

[(a+b)(c+d)]/(a+b) = 1*(c+d)

 

but

 

[(a+b)+(c+d)]/(a+b) Does not equal 1+(c+d)

 

Just sub in numbers for a,b,c,d and you will see that it cannot be the case

 

and - by the by - technically that formula is closer to CHSH than Bells original which relies on correlation values.

 

Please get some basic education in before trying to overturn what is commonly regarded as one of the best pieces of modern science

[TakenItSeriously]

Just no. You can only divide through by a factor for instance (a+b) and cancel when the factors are multiplied

 

[(a+b)(c+d)]/(a+b) = 1*(c+d)

 

but

 

[(a+b)+(c+d)]/(a+b) Does not equal 1+(c+d)

 

Just sub in numbers for a,b,c,d and you will see that it cannot be the case

 

and - by the by - technically that formula is closer to CHSH than Bells original which relies on correlation values.

 

Please get some basic education in before trying to overturn what is commonly regarded as one of the best pieces of modern science

You do realize that I am not argueing against entanglement, but that my agruement is based on the corelations that Bell was basing his theory on.

[imatfaal]

You do realize that I am not argueing against entanglement, but that my agruement is based on the corelations that Bell was basing his theory on.

My post was directed to the OP - not to you. I have replied to you in your thread

[TakenItSeriously]

oops sorry

[Lazarus]

Just no. You can only divide through by a factor for instance (a+b) and cancel when the factors are multiplied

 

[(a+b)(c+d)]/(a+b) = 1*(c+d)

 

but

 

[(a+b)+(c+d)]/(a+b) Does not equal 1+(c+d)

 

Just sub in numbers for a,b,c,d and you will see that it cannot be the case

 

and - by the by - technically that formula is closer to CHSH than Bells original which relies on correlation values.

 

Please get some basic education in before trying to overturn what is commonly regarded as one of the best pieces of modern science

 

Lazarus:

Sorry my notation confused you. Here is another try to make it clear.

 

2(b+a’) / [(b+a’) + (a+b’)] <= 2

 

2 / [(b+a’) / (b+a’) + (a+b’) / (b+a’)] <= 2

 

2 / [1 + (a+b’) / (b+a’)] <= 2

 

I have to fess up that I got sloppy and dropped some primes but that should have been pretty obvious if your eye sight is better than mine.

[imatfaal]

OK - that makes sense if you are dividing both denominator and numerator of the LHS by (b+a') . Which is allowed - but was not at all clear

 

Can you elaborate on the last line?

[Lazarus]

OK - that makes sense if you are dividing both denominator and numerator of the LHS by (b+a') . Which is allowed - but was not at all clear

 

Can you elaborate on the last line?

 

 

This is where I failed to include the primes.

 

2

----------------- <=2

1 + (a+b) Correction 1 + (a+b')

-------

(b+a’)

 

1

----------------- <=1

1 + (a+b) Correction ! + (a+b')

-------

(b+a’)

 

and

 

(a+b) Correction (a+b')

0 <= ---------- <= infinity

(b+a’)

[imatfaal]

Sorry but I am pretty sure you have still messed up your primed and non-primed. Why not use a,b,c=a' and d=b' ? Much easier to follow

[Lazarus]

Sorry but I am pretty sure you have still messed up your primed and non-primed. Why not use a,b,c=a' and d=b' ? Much easier to follow

 

 

Lazarus

The primes make it difficult. You are right, some of them were in the wrong place. Here is the abc version with both kinds of notation.

 

P(a,b) -P(a,b’)+P(a’,b)+P(a’,b’) <= 2

P(a,b) -P(a,d)+P(c,b)+P(c,d) <= 2

 

a+b-a-d+c+b+c+d

------------------------ <=2 a+b-a-d+c+b+c+d / a+b+c+d <=2

a+b+c+d

 

b+c+b+c

----------- <=2 b+c+b+c / a+b+c+d <=2

a+b+c+d

 

2(b+c)

------------ <=2 2(b+c) / (b+c)+(a+d) <=2

(b+c)+(a+d)

 

2

------------ <=2 2 / [1 + (a+d) / (b+c)] <=2

1 + (a+d)

-------

(b+c)

 

1

------------ <=1 1 / [1 + (a+d) / (b+c)] <=1

1 + (a+d)

-------

(b+c)

 

and

 

(a+d)

0 <= ------- <= infinity 0 <= (a+d) / (b+c) <= infinity

(b+c)

[imatfaal]

So you have gone from one mathematical truism to another* (presuming you have the algebra correct) - the question is "and...?"

 

*the sum of two positive numbers (they are counts remember) divided by the sum of two other positive numbers is always gonna be between 0 and infinity unless denominator or numerator is zero in which case you either get zero or undefined

[Lazarus]

So you have gone from one mathematical truism to another* (presuming you have the algebra correct) - the question is "and...?"

 

*the sum of two positive numbers (they are counts remember) divided by the sum of two other positive numbers is always gonna be between 0 and infinity unless denominator or numerator is zero in which case you either get zero or undefined

 

Lazarus:

The significance is that it shows that at least one of the 4 values has to be negative to violate Bell’s inequality. With probabilities and counts that cannot happen but by using Correlation Coefficients instead, negative numbers are possible since the formula to go from Probability to Correlation Coefficients is P*2 -1 which CHSH uses. That may be acceptable. On this forum I am forbidden to say that the only reason CHSH violates Bell’s Inequality is that 3 of the 4 tests are really the same test and that is because the results depend on the angle between the polarizers, not the raw settings of them, so I won’t mention it.

[swansont]

Yes, circular logic that has been offered up repeatedly without justification is problematic. Basically it's an admission that QM is correct, so why bother with analyzing the experiment? (IOW, the result from QM is not the result you get if there were hidden variables, which is the whole point of looking at the inequality)