1/sign(a.h) is indeterminate when "a" is orthogonal to "h" because you have 1/0.gill1109 wrote: ↑Mon Dec 27, 2021 12:35 amNearly zero? Exactly zero, I think. For this reason, it does not matter how you define Bell's (4) when "a" and "h" are orthogonal. You could define it to always to equal +1 in that case. Or introduce a further hidden variable which is an independent fair coin toss [which actually never gets used!]. As Bell wrote "as the probability of this is zero we will not make special prescriptions for it". He doesn't bother to write down a special prescription because *any special prescription would do". He doesn't say that we should *discard outcomes from his mathematical model*.Joy Christian wrote: ↑Sun Dec 26, 2021 12:33 pmIf a happens to be orthogonal to h (the measure of that happening, by the way, is nearly zero), then sign(h.a) is assumed to be equal to the sign of the first nonzero component from the set {a_x, a_y, a_z}. Bell does not specify this in his 1964 paper, but he does so in a later paper, and so do GHSZ in their paper.
There is simply no point in giving further details.
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