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victor
Another shot: (c$)^(c$) , where c is the cardinality of the continuum.


07 Mar 2008, 04:49 

victor
Quote: ...it is only equivalent to your previous "Beth two"... Beth two = 2^c c$ = (c!)^(c!)^(c!)^... Hence, (c$)^(c$) should be much, much bigger than Beth two. 

07 Mar 2008, 05:07 

revolution
victor wrote:
Let's break it down: 1. c$ = (c!)^(c!)^(c!)^... 2. c! = c 3. c^c = Beth1^Beth1 = Beth2 4. c^c^c = Beth1^Beth1^Beth1 = (Beth1^Beth1)^Beth1 = Beth2^Beth1 = Beth2 So no matter how many more ^c terms you add you still get C$ = Beth2 Therefore (c$)^(c$) = (Beth2)^(Beth2) = Beth3 See here 

07 Mar 2008, 05:36 

Tomasz Grysztar
revolution wrote: 2. c! = c Is this defined at all? 

07 Mar 2008, 06:05 

edfed
lim ∑x^x
x>oo or simply: x Last edited by edfed on 07 Mar 2008, 19:40; edited 2 times in total 

07 Mar 2008, 06:17 

revolution
Tomasz Grysztar wrote:
2. c! = c*(c1)*(c2)*...*1, 3. c! = 1*c*(c1)*(c2)*...*2, ;1*c=c 4. c! = c*(c1)*(c2)*...*2, 5. c! = 2*c*(c1)*(c2)*...*3, ;2*c=c 6. c! = c*(c1)*(c2)*...*3, etc. 7. c! = c*(c1), 8. c! = c 

07 Mar 2008, 06:18 

revolution
edfed wrote: lim sigma(x^x) Take your pick: ∑Σ 

07 Mar 2008, 06:24 

Tomasz Grysztar
revolution wrote:
Are you trying to do a transfinite induction here? Then you have to assume axiom of choice and make some total order on c. And still, you also need to define some kind of transfinite multiplication in order to equations above make any sense at all. 

07 Mar 2008, 06:42 

revolution
Tomasz Grysztar wrote: Are you trying to do a transfinite induction here? Then you have to assume axiom of choice and make some total order on c. And still, you also need to define some kind of transfinite multiplication in order to equations above make any sense at all. 

07 Mar 2008, 06:49 

Tomasz Grysztar
revolution wrote: Okay, I don't know. I used the idea of things like this as my basis. Basically it suggests that u*v=max(u,v). But I may have applied it wrongly. Well, this multiplication takes just two arguments  and is associative, so you can extend it to any finite number of arguments. But that's about all. Making infinite products is not such an easy task. 

07 Mar 2008, 06:54 

revolution
I can find nothing to support a different answer than c!=c. How about this, c^c = 2^c = Beth2, we are given this, and c! < c^c, and by GCH there are no cardinals strictly between c and 2^c, therefore, with c! less than c^c it can only take the value c.


07 Mar 2008, 08:44 

Tomasz Grysztar
revolution wrote: I can find nothing to support a different answer than c!=c. How about this, c^c = 2^c = Beth2, we are given this, and c! < c^c, and by GCH there are no cardinals strictly between c and 2^c, therefore, with c! less than c^c it can only take the value c. The hasn't be to be answer at all. The value of c! doesn't even exist until we define the transfinite meaning of "!". And we may have different definitions leading to different results. How do you know that c!<c^c when you haven't even defined the c! yet? 

07 Mar 2008, 08:58 

revolution
Tomasz Grysztar wrote: The value of c! doesn't even exist until we define the transfinite meaning of "!". Does the following hold? c^c = 2^c c^(c1) < c^c and therefore, c^(c1) < 2^c Perhaps the "c1" thing is also not defined/allowed? Anybody know? 

07 Mar 2008, 09:05 

Tomasz Grysztar
revolution wrote: Perhaps the "c1" thing is also not defined/allowed? Anybody know? Even for ordinals this wouldn't work  there are ordinals that don't have a preceding element, they are called the limit ordinals. Transfinite induction doesn't work with just a +1 steps, as they cannot get you to the points after the omega. 

07 Mar 2008, 09:11 

revolution
So does that mean I have to disqualify the entry "(c$)^(c$)" because it is not well defined?


07 Mar 2008, 09:13 

Tomasz Grysztar
Yes, unless we are given some acceptable definition of $ operator for cardinals.
Still, as I already wrote above, I don't think that allowing both ordinals and cardinals is a good idea. 

07 Mar 2008, 09:19 

revolution
Tomasz Grysztar wrote: Still, as I already wrote above, I don't think that allowing both ordinals and cardinals is a good idea. Code: nonAleph: 5. 9$$$$$$$$ bitRAKE 4. 9($^9$$$) Tomasz Grysztar 3. G($^G$$$) Tomasz Grysztar (belatedly accepted) 2. BB(BB(9)) Tomasz Grysztar 1. (BB^9)(9) Tomasz Grysztar (belatedly accepted) Purely mathematical: 5. BethZ$^Z bitRAKE 4. Aleph9$$ Tomasz Grysztar 3. AlephF$$ MHajduk 2. AlephZ$$ MHajduk 1. BethZ$$$ bitRAKE 

07 Mar 2008, 09:25 

Tomasz Grysztar
To take further the discussion of problems I mentioned: let's now take the ring of 2adic integers, Z_2. Since there's no natural order there, we have to make up some order, otherwise we wouldn't be able to tell whether some number is bigger compared to the other one.
I propose this very simple and intuitive order: For 2adic integers a=[a_0 a_1 a_2 ...] and b=[b_0 b_1 b_2 ...] we've got a<=b iff there exists i such that a_j=b_j for j<i and a_i<=b_i. So 0=[000...]<1=[1000...], 2=[01000...]<3=[11000...], but 1=[1000...]>2=[01000...] Now what biggest number can we get here? Well, it's 1=[11111...]. It's the maximal element in this order, and thus there exist no element greater that it. 

07 Mar 2008, 10:33 

revolution
Okay, you're describing a bitreflected version of standard integer binary. But even a nonbitreflected version still has 1 as the "greatest" element.
Can you point out if I have made a mistake in this puzzle? Specifically have I made a comparison error in the two lists? Maybe comparing things that can't be compared? Or allowing something that makes no sense? Or should I define an extra parameter (a rule) that further submissions must meet? 

07 Mar 2008, 10:43 

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