Talk:Nimber

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Nice article! I didn't understand this:

(2) The nimber square of a Fermat 2-power x is equal to the standard value of 3x/2.

What does "standard value of 3x/2" mean? AxelBoldt 01:50 Jan 5, 2003 (UTC)

Thanks. The part that you asked about has been edited. It means the value of 3x/2 under the ordinary multiplication of natural numbers, so that the nimber square of 2 is 3, the nimber square of 4 is 6, the nimber square of 16 is 24, the nimber square of 256 is 384, etc. David McAnally 21:22 Jul 1, 2004 (AEST)

Other than the clumsy workaround 2^{2^n} or some sort of graphic (such as appears on the Fermat number page), is there any way to get Wikipedia to deal with nested exponents? I changed to the above workaround, as the previous format looked like 2²n in my browser. MJSS 20:29, 3 Sep 2004 (UTC)

If worse comes to worst you could resort to TeX and write: . Formerly, TeX looked good when "displayed" and often looked terrible when embedded in lines of text. Now it may look less than beautiful, but usually is nowhere near as bad as it used to be. Of course, if you're going to "display" it, thus

then using TeX works well. Michael Hardy 22:17, 3 Sep 2004 (UTC)

Oh -- one more thing. The paragraph I wrote above was already indented, so I need to indent the "displayed" TeX twice. Michael Hardy 22:18, 3 Sep 2004 (UTC)

There is a similar problem with omega^omrga^omega at the point where it was pointed out that the set of nimbers less than omega^omega^omega is an algebraically closed field. Like 2^2^n, it does not look right, although it once did.

David McAnally 25 Septemeber, 2004

What does the following look like to you?

Specifically where in the article is the problem you're referring to?

Michael Hardy 00:14, 26 Sep 2004 (UTC)

[edit] i don't understand addition

α + β = mex{α ′ + β : α ′ < α, α + β ′ : β ′ < β},

This definition doesn't make any sense to me. There are too many colons and commas inside the set braces. Revolver 19:19, 27 July 2005 (UTC)

I have modified the definition to

α + β = mex({α ′ + β : α ′ < α} ∪ {α + β ′ : β ′ < β}).

This should make it more understandable. Figaro 3:01, 13 August, 2005 (EST)

Okay, I thought that may have been one intended meaning. I'm still not sure I understand addition, though. If α and β are finite ordinals, with α not zero, then it seems that

{α ′ + β : α ′ < α} = {β, ..., α + β − 1}

If α and β are both nonzero, then

({α ′ + β : α ′ < α} ∪ {α + β ′ : β ′ < β}) = {min{α, β}, ..., α + β − 1}

so that

mex({α ′ + β : α ′ < α} ∪ {α + β ′ : β ′ < β}) = 0

If α = β = 0, it's also = 0, since the argument of the mex function is the empty set.

If one of them is zero, the other not, say α not zero, but β = 0, then it should be = α

This seems to agree with your exclusive-or definition, except for the case where each is nonzero. Where am I making a mistake? Revolver 20:51, 13 August 2005 (UTC)

The definition of addition is recursive, so that, in the definition

α + β = mex({α ′ + β : α ′ < α} ∪ {α + β ′ : β ′ < β}),

the addition on the right hand side is not ordinal addition, but nimber addition. Your suggestions above use ordinal addition. It is possible to prove by induction that α + 0 = α for all nimbers α. Similarly, 0 + α = α for all nimbers α. It follows that

1 + 1 = mex{{0 + 1, 1 + 0}) = mex({1, 1}) = 0,

1 + 2 = mex(0 + 2, 1 + 0, 1 + 1}) = mex({2, 1, 0}) = 3,

2 + 1`= mex({0 + 1, 1 + 1, 2 + 0}) = mex({1, 0, 2}) = 3,

1 + 3 = mex({0 + 3, 1 + 0, 1 + 1, 1 + 2}) = mex({3, 1, 0, 3}) = 2,

3 + 1 = mex({0 + 1, 1 + 1, 2 + 1, 3 + 0}) = mex({1, 0, 3, 3}) = 2,

2 + 2 = mex({0 + 2, 1 + 2, 2 + 0, 2 + 1}) = mex({2, 3, 2, 3}) = 0,

and so on. The exclusive-or evaluation of the sum of two finite nimbers is provable by induction on the rwo nimbers. Figaro 19:23 17 August, 2005 (AEST)

Yup. My fault for missing the word "recursively"! Revolver 03:52, 18 August 2005 (UTC)