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September 27th, 2010, 01:56 PM   #1
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Cardinality of P


I would really appreciate if you could help me!
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September 27th, 2010, 02:16 PM   #2
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Re: Cardinality of P

I can't make sense of your notation.
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September 27th, 2010, 02:19 PM   #3
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Re: Cardinality of P

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Originally Posted by butabi

I would really appreciate if you could help me!
Your notation is confusing, especially the part to the right of | in the expression for P={.....}.
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September 27th, 2010, 02:22 PM   #4
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Re: Cardinality of P

It's somewhere between |N| and |R|, probably closer to |R|...
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September 28th, 2010, 09:07 AM   #5
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Re: Cardinality of P

To tell the truth I don't really understand it, either.
But I think:

P is a set of polynomials
n+1 is a set, because every natural number can be considered as a set
k is element of n+1
a_k is an integer

I really hope that the problem is clearer for you than for me...
Any help would be appreciated.
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September 28th, 2010, 09:18 AM   #6
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Re: Cardinality of P

OK, so . Good, that wasn't obvious to me.

What is n? A fixed, unknown natural number?

What coding do you use for natural numbers (or whatever sort of object n is)? We need to know since you want to money with the elements of a number, and numbers aren't usually thought of as having elements. Perhaps this means k in {0, ..., n} or k in {1, ..., n} as it would with two popular encodings.
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September 28th, 2010, 10:08 AM   #7
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Re: Cardinality of P

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OK, so . Good, that wasn't obvious to me.

What is n? A fixed, unknown natural number?

What coding do you use for natural numbers (or whatever sort of object n is)? We need to know since you want to money with the elements of a number, and numbers aren't usually thought of as having elements. Perhaps this means k in {0, ..., n} or k in {1, ..., n} as it would with two popular encodings.
Sorry, you are right, I forgot to mention that n is a fixed, unknown natural number.
And you are right again, I suppose that it means that k is in {0,....,n} ( We definied natural numbers as sets, and the elements of n natural numbers are 0, 1, 2,....,n-1)

Sorry that my notation was confusing, but to tell the truth in the begining I can't understand this notation, either. But then I glanced through my notes and it cleard up a little bit. Unfortunatelly I can't solve the task yet anyway it has been already useful.
I hope you can help me. I would really appreciate it.
Thanks.
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September 28th, 2010, 10:18 AM   #8
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Re: Cardinality of P

So you're trying to count the degree-n polynomials with integer coefficients. That's , which is just .
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September 29th, 2010, 01:54 AM   #9
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Re: Cardinality of P

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Originally Posted by CRGreathouse
So you're trying to count the degree-n polynomials with integer coefficients. That's , which is just .
Thank you. You helped me a lot.
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