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July 10th, 2018, 01:22 PM   #1
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Prime number question

1) Show there are infinitely many odd prime numbers $\displaystyle p=2n-1$
2) Show that number two (2) is the only even prime number $\displaystyle p=2$

Last edited by skipjack; July 17th, 2018 at 10:10 AM.
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July 10th, 2018, 03:55 PM   #2
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There are an infinite number of primes (well-known proof). All even numbers are divisible by 2, so 2 is the only even prime.

Why are you asking?

Last edited by skipjack; July 17th, 2018 at 10:11 AM.
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July 17th, 2018, 06:04 AM   #3
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Quote:
Originally Posted by idontknow View Post
1) Show there are infinitely many odd prime numbers $\displaystyle p=2n-1$
2) Show that number two (2) is the only even prime number $\displaystyle p=2$
1)
Every prime number can be expressed as x + (x+1) or y + (y-1) (example: 6+7=13 or 8+9=17), and x, (x+1) and y, (y-1) can not have common factors because they add up to create prime number which by definition can't have dividers except for itself and 1. Since every prime number can be expressed in this way, we have y + (y-1) = 2y -1 and the conclusion is that every odd prime number can be expressed as 2n - 1. Similarly, every odd prime number can be expressed as 2n + 1 since x + (x+1) = 2x + 1.

2)
If there is an even prime number different from 2, then that number can be expressed as 2*x, which makes it divisible by 2, which implies that the number is composite, and that is contradiction.

Last edited by skipjack; July 17th, 2018 at 09:32 PM.
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July 17th, 2018, 08:51 AM   #4
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Quote:
Originally Posted by 1ucid View Post
1)
Every prime number can be expressed as x + (x+1) or y + (y-1) (example: 6+7=13 or 8+9=17), and x, (x+1) and y, (y-1) can not have common factors because they add up to create prime number which by definition can't have dividers except for itself and 1. Since every prime number can be expressed in this way we have y + (y-1) = 2y -1 and the conclusion is that every odd prime number can be expressed as 2n - 1. Similarly every odd prime number can be expressed as 2n + 1 since x + (x+1) = 2x + 1.
I think this proof is faulty because the initial assertion is neither proved nor necessary.

$q \text { is an odd number } \iff \exists \ r \in \mathbb Z \text { such that } 2r - 1 = q.$

$\therefore p \text { is odd and prime } \implies p \text { is odd } \implies$

$\exists \ n \in \mathbb Z \text { such that } 2n - 1 = p.$

In other words, for any odd prime p, 2n - 1 = p follows from the definition of odd and is not peculiar to odd primes.

To prove that there are an infinite number of odd primes then follows from the proof of the infinitude of primes and the proof that there is only one even prime.

Last edited by JeffM1; July 17th, 2018 at 08:56 AM.
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July 17th, 2018, 10:47 AM   #5
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My proof is also true, but yours is simpler.
Is there any way to see all the code for these symbols you use?

Last edited by skipjack; July 17th, 2018 at 10:53 AM.
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July 17th, 2018, 10:56 AM   #6
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Either use the "Quote" button for the post or, if you can, right click on the relevant code and explore the menu system generated.
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July 17th, 2018, 12:27 PM   #7
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Originally Posted by 1ucid View Post
My proof is also true, but yours is simpler.
It baffles me that you felt the need to prove odd primes have the form $2n - 1$, but felt it was okay to assume (without proof) that they have the form $x + (x + 1)$ or $y + (y - 1)$...
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July 17th, 2018, 12:41 PM   #8
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i just wanted to say that $2n -1$ is the same as $n + n - 1 = n + (n - 1)$
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July 17th, 2018, 09:43 PM   #9
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Question (1) didn't ask for a proof of that. It asked for a proof that there are infinitely many odd primes, and you didn't even attempt to provide that, link to such a proof or indicate how what you posted could contribute to such a proof.
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July 18th, 2018, 03:15 AM   #10
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Every odd number can be represented as a sum of n + (n - 1), and every prime number is odd number so i don't know what is the issue.
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