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Anonymous Sun 26 Sep 22:27:57 2021 No.13686543 View Reply Original Report

Quoted By: >>13686546 >>13686565 >>13686577

Why isn't 1 a prime number? Is it just because there are formulas that produce only prime numbers and said formulas don't produce 1?

Anonymous

Anonymous Sun 26 Sep 2021 22:29:26 No.13686546 Report

Quoted By: >>13686555 >>13686559 >>13686605

>>13686543
Same reason 0 isn't even

Anonymous

Anonymous Sun 26 Sep 2021 22:32:15 No.13686555 Report

Quoted By: >>13686581

>>13686546
But 0 is even. And infinity is odd.

Anonymous

Anonymous Sun 26 Sep 2021 22:32:38 No.13686559 Report

Quoted By: >>13686617

>>13686546
is it though ?
even by def is 2n
while prime by def is a number that only divides itself. By that logic 1 should be prime. How am I wrong ?

Anonymous

Anonymous Sun 26 Sep 2021 22:34:02 No.13686564 Report

Quoted By:

When you define that 1 is not prime, you can prove
the theorem that every natural number has a unique
factorization as a product of primes.
You might have defined that 1 is prime. But then you
would not have this theorem.

Anonymous

Anonymous Sun 26 Sep 2021 22:34:22 No.13686565 Report

Quoted By:

>>13686543
Some older mathematicians considered it a prime number. It's mostly just a matter of convenience so we can speak of a unique prime decomposition.

Anonymous

Anonymous Sun 26 Sep 2021 22:39:57 No.13686577 Report

Quoted By: >>13686588

>>13686543
Because if 1 was a prime number, then integers wouldn't have a unique prime factorization, because for any prime factorization $k=p_1^{a_1}...p_n^{a_n}$ of an integer k, we have $k=p_1^{a_1}. . .p_n^{a_N}=p_1^{a_1}. . .p_n^{a_N}\times1$ , so that the

Anonymous

Anonymous Sun 26 Sep 2021 22:40:10 No.13686581 Report

Quoted By:

>>13686555
The infinity retards that keep infesting /sci need to get shot, so that we can
return to business as usual.

Anonymous

Anonymous Sun 26 Sep 2021 22:43:48 No.13686588 Report

Quoted By:

>>13686577
Oops.

Because if 1 was a prime number, then integers wouldn't have a unique prime factorization, because for any prime factorization $k=p_1^{a_1}...p_n^{a_n}$ of an integer k, we have $k=p_1^{a_1}...p_n^{a_n}=p_1^{a_1}...p_n^{a_n}\times 1$ , so that the prime factorization is not unique.

Anonymous

Anonymous Sun 26 Sep 2021 22:49:02 No.13686605 Report

Quoted By: >>13686628 >>13686633

>>13686546
Isn't even what?

Anonymous

Anonymous Sun 26 Sep 2021 22:55:02 No.13686617 Report

Quoted By:

>>13686559
n=0

Anonymous

Anonymous Sun 26 Sep 2021 22:57:15 No.13686628 Report

Quoted By:

>>13686605
He means:
when you divide 1 by 1 you get remainder 0
when you divide 0 by 2 you get remainder 0.
Therefore 1 is not prime is the same as 0 is not even.

Anonymous

Anonymous Sun 26 Sep 2021 22:58:41 No.13686633 Report

Quoted By:

>>13686605
The rest of my expression

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