Math 115B Homework 6solution

# Math 115B Homework 6solution - MAT 115B HOMEWORK 6 SOLUTION...

• jjammal221
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MAT 115B HOMEWORK 6 SOLUTION 14.1.4: (a) No, since | α | - | β | . (b) Yes, since (2 + i )(6 - 3 i ) = 15. 14.1.10: Solve the equation a + bi = u ( a - bi ) for u ∈ { 1 , i, - 1 , - i } and we get α is a multiple of 1 , 1 + i, i or 1 - i . 14.1.16: α β = 79 - 8 i 13 . Take γ = 6 - i , then ρ = α - βγ = - 1 + i . Verify that N ( - 1 + i ) = 2 < 13 = N (2 + 3 i ). 14.1.38: Observe that N ( G k ) = f 2 k + f 2 k +1 . You can prove by induction that f 2 k +1 = f 2 k +1 + f 2 k and f 2 k = f 2 k +1 + f 2 k - 1 for all non-negative integers k . You can also use the general term of Fibonacci number f k = 1 5 1+ 5 2 k - 1 5 1 - 5 2 k to show this question. 14.2.7: Consider for example α = 2+ i and β = 2 - i . Observe that 2+ i = (1+ i )(2 - i ) - 1, so α and β are relatively prime Gaussian integers. But N ( α ) = N ( β ) = 5, so N ( α ) and N ( β ) are not relatively prime. 14.2.8: γ | α N ( γ ) | N ( α ). γ | β N ( γ ) | N ( β ). Therefore N ( γ ) | gcd ( N ( α ) , N ( β )). 14.2.12: (a) 2 - 11 i = ( - 1 - i )(7 + 8 i ) + 1 - 4 i , 7 + 8 i = (2 - i )(1 + 4 i ) + 1 + i , 1 + 4 i = (3 + 2 i )(1 + i ) - i , 1 + i = ( - 1 + i )( - i ). So gcd (2 - 11 i, 7 + 8 i ) = - i , which is a unit. Therefore, they are relatively prime. (b) μ = - 1 + 9 i and ν = - 8 + 5 i . GK 6.1: see attached picture. GK 6.2 and 6.3: see the pattern of the napkin in the picture on the course website.

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• Fall '08
• Shapiro,I
• Math, Number Theory, Negative and non-negative numbers, Prime number, Euclidean algorithm, finitely, 5 2 k

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