Uncountability of the Rationals?
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From: David R Tribble on 14 Jun 2010 15:08 Tony Orlow wrote: >> Actually, in Bigulosity, there are much larger countably infinite >> sets. The square roots of the naturals have a higher "density" than >> the naturals. > David R Tribble wrote: >> How is that? The two sets should be exactly the same size >> (your size(), not just cardinality). Every natural has a square root, >> and every natural square root has a corresponding natural. >> (I assume you're only dealing with the positive roots.) > Tony Orlow wrote: > Holy cannoli! There exists a bijection, f:N->S and g:S->N, so indeed > they have the same cardinality. They do not share the same bigulosity, > since f and g are not the identity function. The number of square > roots over omega is omega^2. Sorry, but it's not clear what you mean by "the number of square roots over omega is omega^2". Do you mean: 1. the number of naturals in omega that are square roots is omega^2? 2. the number of naturals in omega that have square roots is omega^2? 3. the set of the squares roots of all the naturals in omega has omega^2 members? 4. or something else? When you said | The square roots of the naturals have a higher "density" than | the naturals. I assumed you meant that the "square roots of the naturals" is the set S = { x | x*x = n, for all n in N } Is this not what you mean by the "square roots of the naturals"?
From: FredJeffries on 14 Jun 2010 15:09 On Jun 14, 4:49 am, Tony Orlow <t...(a)lightlink.com> wrote: > > Actually, you're right. N should always start with 1 for IFR. I made > an error of 1 by including 0. If we include 0, then we may say E=N/2+1 > and O=N/2, because of that extra even element. Apologies. > Let me see if I've got this right: E=N/2+1 and O=N/2 so N = E + O = (N/2+1) + (N/2) = N+1
From: David R Tribble on 14 Jun 2010 15:10 Tony Orlow wrote: >> In terms of natural density, they probably have a measure of oo? > David R Tribble wrote: >> The set of positive natural square roots is not a subset >> of the naturals, so it does not have a natural density. > Tony Orlow wrote: > That is probably correct. The squares have a natural density of 0. If > natural density is actually restricted to naturals, then there is no > natural density, but qactually natural density allows for elements > outside of N, such as the multiples of 1/2 with a density of 2. So, > your objection is based ona false presumption about a theory that's > not my invention. My guess is that this set would have an infinite > natural density if any. What false presumption? Natural density is defined only for subsets of the naturals. You can look it up on Wikipedia if you don't believe me. http://en.wikipedia.org/wiki/Natural_density
From: David R Tribble on 14 Jun 2010 15:23 Brian Chandler wrote: >> I certainly don't think it would be correct to say that the reciprocal >> of zero is the empty set. > Tony Orlow wrote: > In standard mathematics it is "undefined", both positively AND > negatively infinite. It doesn't exist. The set of solutions at 0 is > empty, ot perhaps a pair of hyperreals. A hyperreal can be returned by a real-valued function?
From: Virgil on 14 Jun 2010 15:49
In article <49f89dd7-4c7f-434d-af8c-391194e15b31(a)a30g2000yqn.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Nope. They simply say that IF N has a size, it cannot be a member of > N. Note that for the von Neumann naturals, the 'size' of a natural is never a member of that natural. So why should it be any different for the set of all of them? |