# Adv Alegbra HW Solutions 14 - commutative rings 1.30 Show...

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14 (v) Let f ( x ) = ax 2 + bx + c , where a , b , c are real numbers. If z is a root of f ( x ) , then z is also a root of f ( x ) . Solution. True. (vi) Let f ( x ) = ax 2 + bx + c , where a , b , c are complex numbers. If z is a root of f ( x ) , then z is also a root of f ( x ) . Solution. False. (vii) The primitive 4th roots of unity are i and i . Solution. True. 1.29 Prove that the binomial theorem holds for complex numbers: if u and v are complex numbers, then ( u + v) n = n X r = 0 ³ n r ´ u n r v r . Solution. The proof of the binomial theorem for real numbers used only properties of the three operations: addition, multiplication, and division. These operations on complex numbers have exactly the same properties. (Division enters in only because we chose to expand ( a + b ) n by using the formula for ( 1 + x ) n ; had we not chosen this expository path, then division would not have been used. Thus, the binomial theorem really holds for
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Unformatted text preview: commutative rings.) 1.30 Show that the binomial coef f cients are “ symmetric ” : ³ n r ´ = ³ n n − r ´ for all r with 0 ≤ r ≤ n . Solution. By Lemma 1.17, both ( n r ) and ( n n − r ) are equal to n ! r ! ( n − r ) ! . 1.31 Show, for every n , that the sum of the binomial coef f cients is 2 n : ³ n ´ + ³ n 1 ´ + ³ n 2 ´ + ··· + ³ n n ´ = 2 n . Solution. By Corollary 1.19, if f ( x ) = ( 1 + x ) n , then there is the expan-sion f ( x ) = ³ n ´ + ³ n 1 ´ x + ³ n 2 ´ x 2 + ··· + ³ n n ´ x n . Evaluating at x = 1 gives the answer, for f ( 1 ) = ( 1 + 1 ) n = 2 n ....
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