The world needs a better free encyclopedia. Help create it.
All unapproved articles are subject to a disclaimer; please read.
Feed the Servers! Contribute to the Citizendium.

User:Jitse Niesen/sandbox

From Citizendium, the Citizens' Compendium

Jump to: navigation, search

This web page shows different possibilities to display mathematics on Citizendium.

Current setup (120 dpi)

Euclid proved that for any finite set of prime numbers, there is always another prime number which is not in that set. Choose any finite set of prime numbers \{ p_1, p_2, p_3, \ldots, p_n \}. Then the number

N = p_1 p_2 \cdots p_n +1

presents a problem. On the one hand, the number N-1 = p_1 p_2 \cdots p_n is a multiple of p1, and multiples of the same prime number are never next to each other, so N itself can't be a multiple of p1. The same argument actually shows that N itself cannot be a multiple of any of the prime numbers p_1, p_2, p_3, \dots, p_n. On the other hand, every number N > 1 is divisible by some prime.

Some more complicated formulas to probe readability. If N = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n} is one factorization in primes (i.e., the pi are distinct primes and \alpha_i\in\mathbb{N}), and N = \tilde p_1^{\tilde\alpha_1} \tilde p_2^{\tilde\alpha_2} \ldots \tilde p_{\tilde n}^{\tilde\alpha_{\tilde n}} is another, then n=\tilde n and there exists a permutation \sigma\in S_n such that p_i = \tilde p_{\sigma(i)} and \alpha_i = \tilde\alpha_{\sigma(i)} for i=1,\ldots,n.

Changing resolution to 110 dpi

Euclid proved that for any finite set of prime numbers, there is always another prime number which is not in that set. Choose any finite set of prime numbers \{ p_1, p_2, p_3, \ldots, p_n \}. Then the number

N = p_1 p_2 \cdots p_n +1

presents a problem. On the one hand, the number N-1 = p_1 p_2 \cdots p_n is a multiple of p1, and multiples of the same prime number are never next to each other, so N itself can't be a multiple of p1. The same argument actually shows that N itself cannot be a multiple of any of the prime numbers p_1, p_2, p_3, \dots, p_n. On the other hand, every number N > 1 is divisible by some prime.

Some more complicated formulas to probe readability. If N = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n} is one factorization in primes (i.e., the pi are distinct primes and \alpha_i\in\mathbb{N}), and N = \tilde p_1^{\tilde\alpha_1} \tilde p_2^{\tilde\alpha_2} \ldots \tilde p_{\tilde n}^{\tilde\alpha_{\tilde n}} is another, then n=\tilde n and there exists a permutation \sigma\in S_n such that p_i = \tilde p_{\sigma(i)} and \alpha_i = \tilde\alpha_{\sigma(i)} for i=1,\ldots,n.

Changing resolution to 100 dpi

Euclid proved that for any finite set of prime numbers, there is always another prime number which is not in that set. Choose any finite set of prime numbers \{ p_1, p_2, p_3, \ldots, p_n \}. Then the number

N = p_1 p_2 \cdots p_n +1

presents a problem. On the one hand, the number N-1 = p_1 p_2 \cdots p_n is a multiple of p1, and multiples of the same prime number are never next to each other, so N itself can't be a multiple of p1. The same argument actually shows that N itself cannot be a multiple of any of the prime numbers p_1, p_2, p_3, \dots, p_n. On the other hand, every number N > 1 is divisible by some prime.

Some more complicated formulas to probe readability. If N = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n} is one factorization in primes (i.e., the pi are distinct primes and \alpha_i\in\mathbb{N}), and N = \tilde p_1^{\tilde\alpha_1} \tilde p_2^{\tilde\alpha_2} \ldots \tilde p_{\tilde n}^{\tilde\alpha_{\tilde n}} is another, then n=\tilde n and there exists a permutation \sigma\in S_n such that p_i = \tilde p_{\sigma(i)} and \alpha_i = \tilde\alpha_{\sigma(i)} for i=1,\ldots,n.

Changing resolution to 90 dpi

Euclid proved that for any finite set of prime numbers, there is always another prime number which is not in that set. Choose any finite set of prime numbers \{ p_1, p_2, p_3, \ldots, p_n \}. Then the number

N = p_1 p_2 \cdots p_n +1

presents a problem. On the one hand, the number N-1 = p_1 p_2 \cdots p_n is a multiple of p1, and multiples of the same prime number are never next to each other, so N itself can't be a multiple of p1. The same argument actually shows that N itself cannot be a multiple of any of the prime numbers p_1, p_2, p_3, \dots, p_n. On the other hand, every number N > 1 is divisible by some prime.

Some more complicated formulas to probe readability. If N = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n} is one factorization in primes (i.e., the pi are distinct primes and \alpha_i\in\mathbb{N}), and N = \tilde p_1^{\tilde\alpha_1} \tilde p_2^{\tilde\alpha_2} \ldots \tilde p_{\tilde n}^{\tilde\alpha_{\tilde n}} is another, then n=\tilde n and there exists a permutation \sigma\in S_n such that p_i = \tilde p_{\sigma(i)} and \alpha_i = \tilde\alpha_{\sigma(i)} for i=1,\ldots,n.

Changing resolution to 80 dpi

Euclid proved that for any finite set of prime numbers, there is always another prime number which is not in that set. Choose any finite set of prime numbers \{ p_1, p_2, p_3, \ldots, p_n \}. Then the number

N = p_1 p_2 \cdots p_n +1

presents a problem. On the one hand, the number N-1 = p_1 p_2 \cdots p_n is a multiple of p1, and multiples of the same prime number are never next to each other, so N itself can't be a multiple of p1. The same argument actually shows that N itself cannot be a multiple of any of the prime numbers p_1, p_2, p_3, \dots, p_n. On the other hand, every number N > 1 is divisible by some prime.

Some more complicated formulas to probe readability. If N = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n} is one factorization in primes (i.e., the pi are distinct primes and \alpha_i\in\mathbb{N}), and N = \tilde p_1^{\tilde\alpha_1} \tilde p_2^{\tilde\alpha_2} \ldots \tilde p_{\tilde n}^{\tilde\alpha_{\tilde n}} is another, then n=\tilde n and there exists a permutation \sigma\in S_n such that p_i = \tilde p_{\sigma(i)} and \alpha_i = \tilde\alpha_{\sigma(i)} for i=1,\ldots,n.

Current setup, with \scriptstyle

Euclid proved that for any finite set of prime numbers, there is always another prime number which is not in that set. Choose any finite set of prime numbers \scriptstyle \{ p_1, p_2, p_3, \ldots, p_n \}. Then the number

N = p_1 p_2 \cdots p_n +1

presents a problem. On the one hand, the number \scriptstyle N-1 = p_1 p_2 \cdots p_n is a multiple of p1, and multiples of the same prime number are never next to each other, so N itself can't be a multiple of p1. The same argument actually shows that N itself cannot be a multiple of any of the prime numbers \scriptstyle p_1, p_2, p_3, \dots, p_n. On the other hand, every number N > 1 is divisible by some prime.

Some more complicated formulas to probe readability. If \scriptstyle N = p_1^{\alpha_1} p_2^{\alpha_2} \ldots p_n^{\alpha_n} is one factorization in primes (i.e., the \scriptstyle p_i are distinct primes and \scriptstyle \alpha_i\in\mathbb{N}), and \scriptstyle N = \tilde p_1^{\tilde\alpha_1} \tilde p_2^{\tilde\alpha_2} \ldots \tilde p_{\tilde n}^{\tilde\alpha_{\tilde n}} is another, then \scriptstyle n=\tilde n and there exists a permutation \scriptstyle \sigma\in S_n such that \scriptstyle p_i = \tilde p_{\sigma(i)} and \scriptstyle \alpha_i = \tilde\alpha_{\sigma(i)} for \scriptstyle i=1,\ldots,n.

Wikicode 

Euclid proved that for any finite set of prime numbers, there is always another prime number which is not in that
set. Choose any finite set of prime numbers <math>\{ p_1, p_2, p_3, \ldots, p_n \}</math>. Then the number

:<math>N = p_1 p_2 \cdots p_n +1</math>

presents a problem. On the one hand, the number <math>N-1 = p_1 p_2 \cdots p_n</math> is a multiple of
<math>p_1</math>, and multiples of the same prime number are never next to each other, so <math>N</math> itself
can't be a multiple of <math>p_1</math>. The same argument actually shows that <math>N</math> itself cannot be a
multiple of any of the prime numbers  <math>p_1, p_2, p_3, \dots, p_n</math>. On the other hand, every number
<math>N>1</math> is divisible by ''some'' prime.

Some more complicated formulas to probe readability. If <math>N = p_1^{\alpha_1} p_2^{\alpha_2} \ldots
p_n^{\alpha_n}</math> is one factorization in primes (i.e., the <math>p_i</math> are distinct primes and
<math>\alpha_i\in\mathbb{N}</math>), and <math>N = \tilde p_1^{\tilde\alpha_1} \tilde p_2^{\tilde\alpha_2} \ldots
\tilde p_{\tilde n}^{\tilde\alpha_{\tilde n}}</math> is another, then <math>n=\tilde n</math> and there exists a
permutation <math>\sigma\in S_n</math> such that <math>p_i = \tilde p_{\sigma(i)}</math> and <math>\alpha_i =
\tilde\alpha_{\sigma(i)}</math> for <math>i=1,\ldots,n</math>.
Retrieved from "http://en.citizendium.org/wiki/User:Jitse_Niesen/sandbox"