alsuren ([info]alsuren) wrote,
@ 2007-03-30 11:51:00
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lots of nothing
Not posted in a while. I've been messing about with telepathy, Windows vista and kde4 (details can be found on my wiki at http://www.srcf.ucam.org/~dl325/moin/moin.cgi/RecentChanges ), and not dancing and not doing any work. I feel that dancing and doing work may be linked. I always feel productive when I've been dancing and gotten a good night's sleep.

I've also been going to see my nan in hospital. She had a brain tumour removed a while ago, and her mood swings somehow remind me of my mum's laptop, when it halts and catches fire every so often.

I was also thinking of something that Gavin said at past midnight sometime during term. I think it was something about not having any philosophical objection to certain hypotheses relating to the set of real numbers being impossible to prove or disprove, because the set of real numbers contains numbers that are impossible to define (if you remember the proof for the fact that the set of real numbers is uncountably infinite).

This got me to thinking whether you could produce a set of numbers that could do most of the things you normally want to do with the set of real (or complex) numbers, but is countably infinite. Let's call my set Cc (the countable set of complex numbers)

So what do I want my set of numbers to do?

1) be countable.
2) contain the rational numbers
3) satisfy f(x,y) in Cc for all x,y in Cc (for functions involving powers, like f(x,y)=x^y )
4) (for bonus marks) contain pi, and other "interesting values" like e

Is it possible to create such a set, or is there some proof that says "this is impossible"?

My first idea was to create a set of the "rational powers of rational numbers" (letting you write numbers of the form x=(i/j)^(k/l), which would be countably infinite). This doesn't even satisfy 3) though, because 2^(2^(1/2)) is an "irrational power of a rational number". My instinct says it might not even be guaranteed to satisfy 3) for functions like f(x,y)=x*y. This is Not Good.

Any mathmos fancy coming to my rescue?



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(Anonymous)
2007-04-01 12:34 pm UTC (link)
You're trying to say what your set is first and then check that it is closed under f. However this probably won't work as you have found. I think you can just declare your set to be the set a) containing the rationals, e, pi and anything else you like b) f(x,y) for any x, y in the set. I.e. you just declare that your set is closed under f (except: you can't divide by zero) and as long as you can prove the set is still countable then you're fine.

hothr.livejournal.com

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[info]alsuren
2007-04-01 01:43 pm UTC (link)
That's the whole thing: I don't know whether it's possible to produce a countable set that is "closed under f"

I don't completely understand the terminology, so I'm just going to abuse what I think you mean when you say "closed under f"

Naturals are "closed under" the following operators:
*
^
+

Integers are "closed under":
*
+
-

Rationals are "closed under":
*
/
+
-

Reals are "closed under":
*
/
+
-
(and possibly some other operators that I've missed)
but they are uncountable.

Complex numbers are "closed under":
*
/
+
-
^
[ (-1)^(1/2) requires complex numbers ]
but they are also uncountable.

What I want is a set that is closed under *,/,+,-,and^, but is countable. To be honest, I don't care whether e and pi are in there, but I suspect they will have to be.

The more I think about it, the more I think it's impossible. The only reason I'm "trying to say what your set is first and then check that it is closed under f" is because that's the only way I know to prove that it's countable. How would you go about proving that a set is countable if you don't know what the set is?

Such a set would be so much nicer to use than the set of reals, because (by virtue of them being countable) it should be possible to write down a definition for every number in the set. It should also be possible to define some identities to simplify calculations.

Your point about zero and infinity is also interesting. I think that in defining a set of numbers, I would probably follow IEEE's example and include +/-infinity and +/-zero. That's just personal preference though.

(Reply to this) (Parent)(Thread)


[info]hothr
2007-04-02 01:02 pm UTC (link)
No, it's not immediately obvious that the set would be countable. The main thing you need to prove this sort of thing is that a union of a countable number of countable sets is countable. A set declared to be closed under certain binary operations is the union of the following:

a) The rationals, e, pi, etc.
b) f(x, y) where x, y are in a)
c) f(x, y) where x, y are in a) or b)
d) f(x, y) where x, y are in a) or b) or c)
.
.
.

It is needed that each of b), c), ... are themselves countable. This isn't too difficult.

(You'd also need Not a Number for things like oo - oo).

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