uncool
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…no, NTuft, that is not why e^(i pi) = 1. Your blind substitution is not correct. You are claiming that e^(x*i*pi) = x cos(pi) + i x sin(pi) = x, if I’ve remade your equations correctly. This simply isn’t true. It has nothing to do with the actual justification of the original equation (which has to do with McLaurin series), and is selfcontradictory with a bit of thought.

…I have to admit, I’m…uneasy with this reasoning. It suggests that there is policy interest in who gets to reproduce based on speculation on future criminality, which has some connections with eugenics. Those connections strengthen when you add in the conservative framing of abortion being connected to the right of the fetus to life  under that framing, this argument suggests that some people don’t have a right to live because of speculative criminal activity. I’m not suggesting that this is what you meant to suggest. Just that it may have unintended implications, especially among people with other framings.

Worth noting it was used on both “sides”: by slavery supporters to say that their “rights” to own slaves shouldn’t end when they visited free states (or when slaves successfully ran away to free states) and by abolitionists to say that African Americans should have the same right to be free in all states

Additionally, the Senate could refuse to confirm any extra Supreme Court justices if it changed its mind after allowing more. It would be very unlikely, but the Senate has both direct power over the number of justices (through legislation) and indirect control (through confirmation). What are you defining as “rights” here? Does e.g. the right to buy alcohol count as a “right”? The right to buy guns? The existence of a patchwork of rights sounds like the same thing as different states having different laws. What you are looking for is probably covered by the “Full Faith and Credit” clause of the Constitution.

"Disproving" Cantor's hypothesis  it's trivial, anyway
uncool replied to NTuft's topic in Speculations
Again: the set of algebraic numbers is countable. There is a bijection between the set of integers and the set of algebraic numbers. And by making an infinite set of exponents, you are again making a map from a countable set (the set of pairs of bases and exponents). The image of a countable set under any map is countable. This isn’t that hard to prove  it’s not an axiom, it’s a theorem. 
"Disproving" Cantor's hypothesis  it's trivial, anyway
uncool replied to NTuft's topic in Speculations
Part of the point of “forcing” is that it is not constructing new sets within the same model  which is your approach  but instead constructing a new model of the real numbers. 
"Disproving" Cantor's hypothesis  it's trivial, anyway
uncool replied to NTuft's topic in Speculations
The set of algebraic numbers is countable; therefore, if the set of powers you use is also countable, then the set you get as a result will again be countable. In fact, the output set will be of the same cardinality as the exponent set, for whatever exponent set you choose (possibly subject to some rare exceptions). 
The difference between competitive sports and politics is that in competitive sports there’s a clear, objective measure of “better”. Tiger Woods has won because he did better on the golf course. There is no single objective “better” when it comes to the judiciary. There is no “best judge” that can be objectively determined, merely a pool of judges with better qualifications, among whom the President picks. So by picking a black woman from that pool, the President isn’t deliberately choosing a worse candidate  merely a particular one from among a pool of similarly qualified candidates.

"Disproving" Cantor's hypothesis  it's trivial, anyway
uncool replied to NTuft's topic in Speculations
again: if a set can be indexed by pairs of natural numbers, then it is countable. i’ is countable, so it can be indexed by the set of countable numbers, so the set of all powers of the form xy where x and y are in i’ can be indexed by pairs of natural numbers, so that set is also countable. 
"Disproving" Cantor's hypothesis  it's trivial, anyway
uncool replied to NTuft's topic in Speculations
Still countable. Can be indexed by a triple of integers  numerator, denominator, and exponent  and the set of triples of integers is countable (for largely the same reason that the set of pairs of integers is countable). Again, there is nothing “inherently” uncountable about transcendental numbers. I see no specific reason to assume it would, unless you assume the negation of the continuum hypothesis to begin with. Which is your right, but you would then only have a circular argument. 
"Disproving" Cantor's hypothesis  it's trivial, anyway
uncool replied to NTuft's topic in Speculations
So it is a set that can be indexed by ordered pairs of integers, namely the base and the prime whose square root is the exponent. It will again be countable, as the set of ordered pairs of natural numbers has a bijection with the set of natural numbers. You keep trying to use the fact that these numbers are transcendental as if that gives you a natural connection to uncountability. It doesn’t. The fact that the set of all transcendentals is large doesn’t mean that any particular subset will be large. 
"Disproving" Cantor's hypothesis  it's trivial, anyway
uncool replied to NTuft's topic in Speculations
What do you mean by {ℕi'}? If you mean the set of all i’indexed sequences of natural numbers, that set has cardinality equal to that of the continuum. It clearly has the same cardinality as ℕℕ, and it’s not hard to establish that that has the same cardinality as the reals. 
"Disproving" Cantor's hypothesis  it's trivial, anyway
uncool replied to NTuft's topic in Speculations
The set of all transcendentals has cardinality equal to that of the reals. The set of all square roots of primes has cardinality equal to that of the natural numbers, by rather obvious bijection. Neither of those have “intermediate” cardinality. If you are claiming to have constructed a set of intermediate cardinality, what is it? And your point is? Cardinality is not about underlying members. Just because the elements of a set are transcendentals and there are lots of transcendentals doesn’t mean that set is large. The cardinality of the set containing just pi is still 1. 
"Disproving" Cantor's hypothesis  it's trivial, anyway
uncool replied to NTuft's topic in Speculations
It’s trivial to construct a bijection between the set of prime numbers and the set of square roots of prime numbers. The set of prime numbers is obviously countable, so the set of square roots of prime numbers is countable, too. It sounds like you are trying to talk about the set of digits of the square roots of prime numbers. This is still countable, though slightly less obviously, but it is not the same as the set of square roots of prime numbers. It should be, if you can make clear statements. My point is that you have not done so. I have been: I can’t determine what statements you are proposing to be true.