Phi

[hadi amiri 4]

\foralln\inN\varphi(n)/mid/n

[hadi amiri 4]

i made a mistake in writing

[CRGreathouse]

I imagine you meant
\forall n\in\mathbb{N}\;\varphi(n)\mid n (which is false; \varphi(3)\!\not\,\,\mid3)
but I'm not sure what the question is.

[hadi amiri 4]

how we prove

\forall n\in\mathbb{N}\;\varphi(n)\mid n

[hadi amiri 4]

How we prove that?
\forall n\in\mathbb{N}\;\varphi(n)\mid n

[hadi amiri 4]

how we prove the statement in post 3

[CRGreathouse]

\forall n\in\mathbb{N}\;\varphi(n)\mid n

how we prove the statement in post 3

You can't, it's false. It only holds for 1, 2, 4, 6, 8, 12, 16, ... = A007694 (http://www.research.att.com/~njas/sequences/A007694).

[roam]

Why can’t we derive a contradiction in order to show that it’s false?

[CRGreathouse]

Why can’t we derive a contradiction in order to show that it’s false?

I gave a contradiction, 3, in my first post.

[HallsofIvy]

I imagine you meant
\forall n\in\mathbb{N}\;\varphi(n)\mid n (which is false; \varphi(3)\!\not\,\,\mid3)
but I'm not sure what the question is.

how we prove the statement in post 3

\forall n\in\mathbb{N}\;\varphi(n)\mid n



You can't, it's false. It only holds for 1, 2, 4, 6, 8, 12, 16, ... = A007694 (http://www.research.att.com/~njas/sequences/A007694).

Why can’t we derive a contradiction in order to show that it’s false?

I gave a contradiction, 3, in my first post.
CRGreathouse, he asked how you prove the contradiction you gave in post 3 and you answered "You can't, it's false. "! You were, of course, referring to his original post, not the post you quoted.

roam, you prove the contradiction by doing the arithmetic. What is \phi(3)?

[CRGreathouse]

CRGreathouse, he asked how you prove the contradiction you gave in post 3 and you answered "You can't, it's false. "! You were, of course, referring to his original post, not the post you quoted.

Ah. I took that to mean 'How do we prove the statement "\forall n\in\mathbb{N}\;\varphi(n)\mid n" from post #3', rather than 'How do we prove the statement "\forall n\in\mathbb{N}\;\varphi(n)\mid n [...] is false" from post #3'. To me, "\forall n\in\mathbb{N}\;\varphi(n)\mid n" was the only mathematical statement in post #3; "(which is false[...])" is a nonrestrictive clause. '"\forall n\in\mathbb{N}\;\varphi(n)\mid n" is false' would have been a mathematical statement, but one I only implied. That's why I was so confused by post #6.

Of course a contradiction is an easy way to show that \forall n\in\mathbb{N}\;\varphi(n)\mid n fails.