781hw10sol

# 781hw10sol - 18.781 Fall 2007 Problem Set 10 Solutions to...

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18.781, Fall 2007 Problem Set 10 Solutions to Selected Problems Problem 1 We have the following identity : sin 2 (2 θ ) = 4 sin 2 θcos 2 θ = 4 sin 2 θ (1 - sin 2 θ ) Applying θ = π 12 and let α = sin ( π 12 ) . Using sin π 6 = 1 2 , we have 1 4 = 4 α 2 (1 - α 2 ) . That is, 16 α 4 - 16 α 2 + 1 = 0 . Therefore, α = sin ( π 12 ) is algebraic. 2 Problem 2 (a) First, let’s prove following claim. There is no element α Z [ - 5] such that N ( α ) = 2 or 3. To prove this, it is enough to show that there is no integral solution ( x, y ) satisfying x 2 +5 y 2 = 2 or 3. This is clear. For α = 2 , 3 , 1 + - 5 , 1 - - 5, we have N (2) = 4 , N (3) = 9 , N (1 + - 5) = N (1 - - 5) = 6 . Thus for each α , if there is any further factorization α = βγ with N ( β ) , N ( γ ) > 1(i.e. each of β, γ is not a unit), since N ( α ) = N ( β ) N ( γ ), N ( β ) or N ( γ ) should be 2 or 3, which is absurd by the claim. Hence, 6 is not factorized uniquely, and this implies that

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## This note was uploaded on 04/28/2009 for the course MATH 18.781 taught by Professor Brubaker during the Spring '09 term at MIT.

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781hw10sol - 18.781 Fall 2007 Problem Set 10 Solutions to...

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