# LS10-1 - S UMMARY OF 10/1 L ECTURE We began lecture by...

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Unformatted text preview: S UMMARY OF 10/1 L ECTURE We began lecture by correcting an earlier proof. Theorem 1. Q × is not cyclic. The original proof we gave of this fact was in terms of our initial (incorrect) definition of the word “cyclic”. Because the definition was wrong, so was our proof. Here are a couple of ways to prove it correctly. Proof. Handwaving approach : by contradiction. Suppose Q × were cyclic with generator a q . Then all elements of Q × would have the form a n q n or q n a n with n ≥ . But there are clearly other fractions out there: for example, choose any integer k which is coprime to both a and q . Then k cannot be of the prescribed form. Rigorous approach : We will prove that (1) n √ 2 6∈ Q × for every n ≥ 2 . On the other hand, if Q × were cyclic, generated by α , say, then α n would have to equal 2 for some n ∈ Z (since 2 ∈ Q × ). Without loss of generality, we can assume n ≥ (otherwise, take 1 α to be the generator instead). From (1) we see that we must have n = 0 or 1 . But this would imply that α = 1 or 2 , neither of which generates Q × . Therefore, to prove the theorem it suffices to prove (1). Suppose (2) a b n = 2 for some non-zero rational number a b , with n ≥ 2 . Without loss of generality, we can assume both a and b are positive integers (i.e. if there exists a rational number which is theare positive integers (i....
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LS10-1 - S UMMARY OF 10/1 L ECTURE We began lecture by...

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