HW5sol

HW5sol - MATH 3360 Applicable Algebra HW 5 Solutions(1...

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MATH 3360 – Applicable Algebra. HW 5 Solutions (1) Problem 2 on page 130. Solution: By definition - ( - a ) is the inverse of ( - a ). Since a + ( - a ) = 0 and inverses are unique, we must have - ( - a ) = a . (2) Problem 7 on page 130. Solution: We know that 1 + 0 = 1 b · (1 + 0) = b · 1 b · 1 + b · 0 = b · 1 b · 1 + b · 0 - b · 1 = b · 1 - b · 1 b · 0 = 0. (3) Problem 11 on page 130. Solution: Since a is a nonzero element of a field, it has multiplicative inverse, say a - 1 . So we have ar = as a - 1 · ar = a - 1 · as 1 · r = 1 · s r = s . (4) Problem 23 on page 138. (graded) Solution: Since a is a zero divisor, b , 0 such that a · b = 0. Also, a · 0 = 0 by problem 7 of this assignment. Thus, there is more than one solution to ax = 0 when a is a zero divisor. (5) Problem 31 on page 139. (graded) Solution: The inverse of [9] is [3], the inverse of [11] is [19], the inverse of [17] is [23]. Note that 22 does not have an inverse, since (22 , 26) = 2. (6) Problem 35 on page 139. Solution: (i) 0 , 1 , 2 , i , 1 + i , 2 + i , 2 i , 1 + 2 i , 2 + 2 i . (ii) Pair the elements with their inverses.
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