hw8math109

hw8math109 - x 9 (mod 11) is the unique solution to 3 x 5...

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MATH 109 Assignment 8 Due: 6/1/11 #22.3 Proof. To see that f is a surjection, suppose that n Z . By trichotomy, we have n > 0, n = 0, or n < 0. If n > 0, then n + 1 Z + and f ( n + 1 , 1) = n . If n = 0, then f (1 , 1) = 0 = n . If n < 0, then - n + 1 Z + and f (1 , - n + 1) = n . This proves that f is a surjection. The equivalence classes of the relation defined by ( x 1 , x 2 ) ( y 1 , y 2 ) f ( x 1 , x 2 ) = f ( y 1 , y 2 ) are of the form { ( x 1 , x 2 ) : x 1 - x 2 = k } for some k Z . #19.2 Proof. We are given that m | ( a 1 - a 2 ) and m | ( b 1 - b 2 ) so let x, y Z be the integers such that a 1 - a 2 = mx and b 1 - b 2 = my respectively. Since ( a 1 - b 1 ) - ( a 2 - b 2 ) = ( a 1 - a 2 ) + ( b 1 - b 2 )( - 1) = mx - my = m ( x - y ) , we see that m | ( a 1 - b 1 ) - ( a 2 - b 2 ) so a 1 - b 1 a 2 - b 2 (mod m ). #20.1 (i) Solution: To solve the congruence 3 x 5 (mod 11) we first solve the linear Diophantine equation 3 x + 11 y = 5. We have 11 = 3(3) + 2 3 = 2(1) + 1 2 = 1(2) + 0 . Rearranging these equations yields 1 = 3 - 2 = 3 - (11 - 3(3)) = 11( - 1) + 3(4) 5 = 11( - 5) + 3(20) . Hence x = 20 is a solution to 3 x 5 (mod 11). Since 20 9 (mod 11) and gcd(3,11)=1, we see that
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Unformatted text preview: x 9 (mod 11) is the unique solution to 3 x 5 (mod 11) by Theorem 20.1.7. #21.3 Below is the multiplication table mod 12. We see that 5 is invertible and is its own inverse. Similarly, 7 is invertible and is its own inverse. Finally 11 is invertible and is its own inverse. X 1 2 3 4 5 6 7 8 9 10 11 1 1 2 3 4 5 6 7 8 9 10 11 2 2 4 6 8 10 2 4 6 8 10 3 3 6 9 3 6 9 3 6 9 4 4 8 4 8 4 8 4 8 5 5 10 3 8 1 6 11 4 9 2 7 6 6 6 6 6 6 6 7 7 2 9 4 11 6 1 8 3 10 5 8 8 4 8 4 8 4 8 4 9 9 6 3 9 6 3 9 6 3 10 10 8 6 4 2 10 8 6 4 2 11 11 10 9 8 7 6 5 4 3 2 1 #23.2 Solution: We have 4148 = 2 2 17 61 and 7684 = 2 2 17 113. Hence gcd(4148,7684) = 2 2 * 17 = 68....
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hw8math109 - x 9 (mod 11) is the unique solution to 3 x 5...

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