Square Root Of 2 Irrationality Proof

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The discussion revolves around the proof of the irrationality of the square root of 2, particularly focusing on the contradiction arising from the prime factorization of squares. Participants clarify that the assumption about t^2 having an odd number of 2's is derived from the relationship between s^2 and t^2, specifically that s^2 equals 2t^2. This leads to the conclusion that if s^2 has an even number of 2's, t^2 must have an odd number, creating a contradiction. The conversation highlights confusion regarding the implications of these factors and the nature of perfect squares. Ultimately, the proof illustrates that the square root of 2 cannot be rational due to this contradiction.
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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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