How Do We Prove \(\forall n\in\mathbb{N}\;\varphi(n)\mid n\) Is False?

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Discussion Overview

The discussion revolves around the statement \(\forall n\in\mathbb{N}\;\varphi(n)\mid n\) and its validity. Participants explore how to prove that this statement is false, examining specific cases and the implications of contradictions in the context of mathematical reasoning.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants clarify the statement as \(\forall n\in\mathbb{N}\;\varphi(n)\mid n\) and note that it is false for certain values, specifically citing \(\varphi(3)\!\not\,\,\mid3\).
  • There is a suggestion that the statement only holds for specific integers such as 1, 2, 4, 6, 8, 12, and 16, referencing a sequence for further context.
  • One participant questions the method of deriving a contradiction to demonstrate the statement's falsehood, indicating that a contradiction was provided in an earlier post.
  • Another participant expresses confusion regarding the interpretation of the original question, highlighting the distinction between proving the statement itself and proving its falsehood.
  • There is a consensus that a contradiction can be used to show that \(\forall n\in\mathbb{N}\;\varphi(n)\mid n\) fails, although the specifics of how to prove this remain under discussion.

Areas of Agreement / Disagreement

Participants generally agree that the statement \(\forall n\in\mathbb{N}\;\varphi(n)\mid n\) is false, but there is no consensus on the method of proving this falsehood or the interpretation of the original question.

Contextual Notes

Some participants express uncertainty regarding the arithmetic involved in proving the contradiction, specifically questioning the value of \(\varphi(3)\) and its implications for the broader statement.

hadi amiri 4
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[tex]\foralln\inN\varphi(n)/mid/n[/tex]
 
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i made a mistake in writing
 


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.
 


how we prove
[tex] \forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex]
 
Last edited by a moderator:


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


how we prove the statement in post 3
 


[tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex]

hadi amiri 4 said:
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, ... = http://www.research.att.com/~njas/sequences/A007694 .
 
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Why can’t we derive a contradiction in order to show that it’s false?
 


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

I gave a contradiction, 3, in my first post.
 
  • #10


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 said:
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 [itex]\phi(3)[/itex]?
 
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  • #11


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.

Ah. I took that to mean 'How do we prove the statement "[tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex]" from post #3', rather than 'How do we prove the statement "[tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex] [...] is false" from post #3'. To me, "[tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex]" was the only mathematical statement in post #3; "(which is false[...])" is a nonrestrictive clause. '"[tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex]" 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 [tex]\forall n\in\mathbb{N}\;\varphi(n)\mid n[/tex] fails.
 

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