hadi amiri 4
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[tex]\foralln\inN\varphi(n)/mid/n[/tex]
hadi amiri 4 said:how we prove the statement in post 3
roam said:Why can’t we derive a contradiction in order to show that it’s false?
CRGreathouse said:I imagine you meant
[tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex] (which is false; [itex]\varphi(3)\!\not\,\,\mid3[/itex])
but I'm not sure what the question is.
hadi amiri 4 said:how we prove the statement in post 3
CRGreathouse said:[tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex]
You can't, it's false. It only holds for 1, 2, 4, 6, 8, 12, 16, ... = http://www.research.att.com/~njas/sequences/A007694 .
roam said:Why can’t we derive a contradiction in order to show that it’s false?
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.CRGreathouse said:I gave a contradiction, 3, in my first post.
HallsofIvy said: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.