From: Richard Carr
Subject: Commutativity in ordinals
Date: Mon, 20 Dec 1999 03:50:21 -0500
Newsgroups: sci.math
Keywords: characterizing commuting pairs of ordinals (addition, multiplication)
It is easy to show that for ordinals a,b one has a+b=b+a iff for some
ordinal c and some natural numbers d,e we have a=cd, b=ce.
I think it is also true that ab=ba iff (either ab is finite or there is
some ordinal c and some natural numbers d,e such that a=c^d, b=c^e)- I
think I did this once, but my proof may have been false.
Is the above statement about ab=ba true? Also is it true that a^b=b^a iff
(either a=b or {a,b}={2,4})? (The latter is a baseless conjecture, rooted
only in that it is true for natural numbers.)
Curiously,
Richard Carr.
==============================================================================
From: Fred Galvin
Subject: Re: Commutativity in ordinals
Date: Mon, 20 Dec 1999 11:20:06 -0600
Newsgroups: sci.math
I was too tired to look this stuff last night. Here are the answers.
> On Mon, 20 Dec 1999, Fred Galvin wrote:
>
> :On Mon, 20 Dec 1999, Richard Carr wrote:
> :
> :> It is easy to show that for ordinals a,b one has a+b=b+a iff for some
> :> ordinal c and some natural numbers d,e we have a=cd, b=ce. I think it
> :> is also true that ab=ba iff (either ab is finite or there is some
> :> ordinal c and some natural numbers d,e such that a=c^d, b=c^e)- I
> :> think I did this once, but my proof may have been false.
Counterexample: a = omega^2+omega, b = omega^3+omega^2
> :> Is the above statement about ab=ba true? Also is it true that a^b=b^a
> :> iff (either a=b or {a,b}={2,4})? (The latter is a baseless conjecture,
> :> rooted only in that it is true for natural numbers.)
Counterexample: a = omega, b = (epsilon_0)*omega
> :If I needed to know stuff like that, the first place I'd look would be
> :Sierpinski's book _Cardinal and Ordinal Numbers_.
Problems of this kind are discussed in great detail, in sections titled
"On ordinal numbers commutative with respect to addition", "On ordinal
numbers commutative with respect to multiplication", and "On the equation
alpha^beta = beta^alpha", on pp. 346-367 of Waclaw Sierpinski, _Cardinal
and Ordinal Numbers_, Second Edition Revised, PWN, Warsaw, 1965. (This is
the great book on naive set theory. Lots of material about the axiom of
choice, but no other axioms are mentioned.)
==============================================================================
From: Richard Carr
Subject: Re: Commutativity in ordinals
Date: Mon, 20 Dec 1999 19:12:53 -0500
Newsgroups: sci.math
On Mon, 20 Dec 1999, Fred Galvin wrote:
[deletia -- see above. --djr]
:Counterexample: a = omega^2+omega, b = omega^3+omega^2
Yes, I just looked it up in the reference you gave. It is true if a,b are
successor ordinals but not for limit ordinals. (In general, one has that
for some natural numbers m,n we have a^m=b^n (if ab is not finite) but
there may not be a c such that a and b are finite powers of c- in the
above counter-example a^3=b^2. I haven't read the proofs yet and do not
recall whether a^m=b^n implies ab=ba as well- I think this was the case
but it is at least 10 minutes since I was in the library and I don't
trust my memory that far.)
:
:> :> Is the above statement about ab=ba true? Also is it true that a^b=b^a
:> :> iff (either a=b or {a,b}={2,4})? (The latter is a baseless conjecture,
:> :> rooted only in that it is true for natural numbers.)
:
:Counterexample: a = omega, b = (epsilon_0)*omega
:
There is an awful lot of counter-examples here, embarrassingly enough.
(For any limit ordinal a(>0, if you count 0 as a limit ordinal), let t
be an epsilon number >a and let b=ta. Then a^b=b^a. These are all the
examples- one direction is harder than the other.)
Thanks for the information.
[rest of above article quoted, too --djr]