Square Root Of 2 Irrationality Proof

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Discussion Overview

The discussion revolves around the proof of the irrationality of the square root of 2, specifically focusing on the reasoning and assumptions made in the proof regarding the prime factorization of squared integers. Participants are examining the logical steps and potential contradictions within the proof.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions the assumption that t² contains an odd number of 2's, suggesting that this is not clearly justified in the proof.
  • Another participant argues that the odd number of factors of 2 in t² is a consequence of earlier statements, not an assumption.
  • A participant attempts to clarify that t² must have an odd number of 2's because it equals s²/2, while also questioning why this does not lead to an immediate contradiction.
  • There is a discussion about the implications of s² being even and a perfect square, leading to the conclusion that s must also be even, which would imply that s² contains an even number of 2's.

Areas of Agreement / Disagreement

Participants express differing interpretations of the proof's logic, with some agreeing that contradictions arise while others challenge the clarity of the reasoning. The discussion remains unresolved regarding the justification of certain assumptions and the implications of the statements made in the proof.

Contextual Notes

Participants highlight potential ambiguities in the proof's phrasing and the logical connections between the statements made about the factors of 2 in s² and t². There is an acknowledgment of the complexity in understanding the implications of even and odd factors in the context of perfect squares.

moe darklight
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Hi, I'm having trouble understanding some statements in this proof from my textbook:

"Thus, 2 = s^2/t^2 and 2t^2 = s^2. Since s^2 and t^2 are squares, s^2 contains an even number of 2's as prime factors (This is our Q statement), and t^2 contains an even number of 2's. But then t^2 contains an odd number of 2's as factors. Since 2t^2 = s^2, s^2 has an odd number of 2's. (This is the statement ~Q.) This is a contradiction, because s2 cannot have both an even and an odd number of 2's asfactors. We conclude that sqrt(2) is irrational."

Why does he assume that t^2 contains an odd number of 2's all of a sudden? ... and even if it did, s^2 would still be even, because it is equal to 2t^2, not t^2. :confused:
 
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moe darklight said:
Why does he assume that t^2 contains an odd number of 2's all of a sudden?
There is something funny in the phrasing; did you copy it exactly? Anyways, he didn't assume, because having an odd number of factors of 2 is a consequence of the previous observations.
 
yes I copied it exactly. ok, let me see if I get it:

t^2 must have an odd number of 2's because it equals s^2/2, and s^2 has an even number of twos (because any squared number has an even number of twos). but why not stop at that? isn't that a contradiction already? ... and the second statement still makes no sense to me, because if you multiply a number with an odd number of twos as factors (2t) by two, then you have an even number of twos, so why does he say s^2 is odd after 2t^2 = s^2?.

I assume, as usual, I'm overlooking something very obvious :biggrin:
 
moe darklight said:
yes I copied it exactly. ok, let me see if I get it:

t^2 must have an odd number of 2's because it equals s^2/2, and s^2 has an even number of twos (because any squared number has an even number of twos). but why not stop at that? isn't that a contradiction already?
Yes it is.
 
Well s^2 is an even number right? (since it equals 2t^2 and t is an integer therefore s^2 = 2k, for k \in Z ) but it's also a perfect square which means s is an even number I believe. Well if s is an even number then s^2 must contain an even number of 2s.
 

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