Math 115B Homework 2sol

# Math 115B Homework 2sol - MAT 115B HOMEWORK 2 SOLUTION GK...

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MAT 115B HOMEWORK 2 SOLUTION GK 2.1 (a) Suppose that 0 is friends with b in Z /p . We have by definition, 0 = - b or 0 = ± ib . In all the cases, b = 0. Hence 0 is friends with itself. Now suppose that a 6 = 0 and p > 3 is a prime. Observe that elements in { a, ia, i 2 a = - a, i 3 a = - ia } are friends with each other. Furthermore, we can check that they are all distinct. Eg: if a = - a then 2 a = 0, have a contradiction a = 0 since p is an odd prime. This proves that friends come in set of four. (b) Assume i Z /p . Part (a) shows that elements of Z /p can be grouped into sets of four elements expect for 0. It follows Z /p has p = 4 k + 1 elements. So p 3 (mod 4) is not possible to have a such i . GK 2.2 Let a = ( . . . a i . . . a 1 a 0 ) p and b = ( . . . b j . . . b 1 b 0 ) p in Z p such that ab = 0. Want to show that a = 0 or b = 0. Assume a 6 = 0 and b 6 = 0. Let a i , b j be the last non zero digit in a, b respectively. Then ab a i b i 6≡ 0 (mod p i + j +1 ). Contradiction with ab = 0. (b) If a is an idempotent other than 0 or 1, then 1 - a 6 = 0 and(1 - a ) 2 = 1 - 2 a + a 2 = (1 - a ).

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