Is 3 Divisible by n if 3 is Divisible by n^2?

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Discussion Overview

The discussion revolves around the question of whether 3 being divisible by \( n^2 \) implies that 3 is also divisible by \( n \), where \( n \) is an integer. This inquiry is tied to proving the irrationality of the square root of three and involves mathematical reasoning and proof techniques.

Discussion Character

  • Mathematical reasoning
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the proof and seeks assistance in understanding the implication that \( 3|n^2 \) leads to \( 3|n \).
  • Another participant notes that since 3 is a prime number, if 3 is not a prime factor of \( n \), it cannot be a factor of \( n^2 \).
  • A different participant acknowledges the contrapositive proof suggested but requests further elaboration on the proof itself.
  • A participant reiterates their confusion and presents a modular arithmetic argument, suggesting that if \( n \equiv a \pmod{3} \) and \( a \neq 0 \), then \( n^2 \equiv a^2 \pmod{3} \) leads to a contradiction, implying \( 3 \nmid n^2 \).
  • Another participant proposes that the Fundamental Theorem of Arithmetic, which discusses the uniqueness of prime factorizations, might be relevant to the discussion.

Areas of Agreement / Disagreement

The discussion contains multiple competing views and remains unresolved, with participants expressing differing levels of understanding and approaches to the proof.

Contextual Notes

Participants have not reached a consensus on the proof's validity, and there are unresolved assumptions regarding the implications of divisibility and the application of modular arithmetic.

cubicmonkey
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I'm completely lost on this one. I need this to be able to solve that the square root of three is irrational. So it's a proof within a proof, but I like this way best. Please help me out.
This is what I need to know how to prove.

3|n^2 implies 3|n where n is some integer.
 
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3 is a prime number. If 3 were not a prime factor of n, it could not be a factor of n2.
 
I think I understand what you're saying and that looks like a contrapositive proof, but you don't actually prove it. Could you elaborate?
 
cubicmonkey said:
I'm completely lost on this one. I need this to be able to solve that the square root of three is irrational. So it's a proof within a proof, but I like this way best. Please help me out.
This is what I need to know how to prove.

3|n^2 implies 3|n where n is some integer.

Suppose :

n \equiv a \pmod 3 ~\text{and}~ a \neq 0
then :
n^2 \equiv a^2 \pmod 3 ~\text{and}~ a^2 \neq 0
hence :
3 \nmid n^2
contradiction .

Q.E.D.
 
I think the Fundamental Theorem of Arithmetic (uniqueness up to order of prime factorizations) might help?
 

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