781hw10sol

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online