My Math Forum Cardinality of P

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 September 27th, 2010, 01:56 PM #1 Member   Joined: Sep 2010 Posts: 60 Thanks: 0 Cardinality of P $\begin{equation}P=\{\sum\limits_{k=0}^{n} a_kx^k | \forall k \epsilon n + 1 : a_k \epsilon Z \}\end{equation} |P|=?$ I would really appreciate if you could help me!
 September 27th, 2010, 02:16 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Cardinality of P I can't make sense of your notation.
September 27th, 2010, 02:19 PM   #3
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Re: Cardinality of P

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 Originally Posted by butabi $\begin{equation}P=\{\sum\limits_{k=0}^{n} a_kx^k | \forall k \epsilon n + 1 : a_k \epsilon Z \}\end{equation} |P|=?$ 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={.....}.

 September 27th, 2010, 02:22 PM #4 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Cardinality of P It's somewhere between |N| and |R|, probably closer to |R|...
 September 28th, 2010, 09:07 AM #5 Member   Joined: Sep 2010 Posts: 60 Thanks: 0 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.
 September 28th, 2010, 09:18 AM #6 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Cardinality of P OK, so $P\subseteq\mathbb{Z}[x]$. 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.
September 28th, 2010, 10:08 AM   #7
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Re: Cardinality of P

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 Originally Posted by CRGreathouse OK, so $P\subseteq\mathbb{Z}[x]$. 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.

 September 28th, 2010, 10:18 AM #8 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Cardinality of P So you're trying to count the degree-n polynomials with integer coefficients. That's $|\mathbb{Z}|^{n+1}$, which is just $\aleph_0$.
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 $|\mathbb{Z}|^{n+1}$, which is just $\aleph_0$.
Thank you. You helped me a lot.

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