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1.2-1.4

# 1.2-1.4 - 115A HW SELECTED SOLUTIONS Sec 1.2 1 True/False(b...

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115A HW SELECTED SOLUTIONS Sec 1.2. 1. True/False (b). F (Corollary 1) (c). F ( x =0) (d). F ( a =0) (h). F (deg( f + g ) n ) (k). T (Example 3) 7. Let S = { 0 , 1 } and F = R . In F ( S, R ), show that f = g and f + g = h , where f ( t ) = 2 t + 1, g ( t ) = 1 + 4 t - 2 t 2 , and h ( t ) = 5 t + 1. Recall that two functions (“vectors”) f and g in the vector space F ( S, R ) are equal precisely when f ( s ) = g ( s ) for all s S . Hence we need only check that f (0) = g (0), f (1) = g (1), f (0) + g (0) = h (0), and f (1) + g (1) = h (1). 18. Let V = { ( a 1 , a 2 ): a 1 , a 2 R } . For ( a 1 , a 2 ) , ( b 1 , b 2 ) V and c R , define (0.0.1) ( a 1 , a 2 ) + ( b 1 , b 2 ) = ( a 1 + 2 b 1 , a 2 + 3 b 2 ) , c ( a 1 , a 2 ) = ( ca 1 , ca 2 ) . We must check the vector space properties VS 1-VS 8. 1 Since scalar multiplication is defined coordinate-wise, it follows at once that VS 5 and VS 6 hold. However, VS 1 and VS 2 fail. Indeed, VS 1 fails since ( a 1 , a 2 ) + ( b 1 , b 2 ) = ( a 1 + 2 b 1 , a 2 + 2 b 2 ) ( b 1 , b 2 ) + ( a 1 , a 2 ) = ( b 1 + 2 a 1 , b 2 + 2 a 2 ) , and these two quantities are not equal in general. Property VS 2 does not hold since (( a 1 , a 2 ) + ( b 1 , b 2 )) + ( c 1 , c 2 ) = ( a 1 + 2 b 1 , a 2 + 3 b 2 ) + ( c 1 , c 2 ) = ( a 1 + 2 b 1 + 2 c 1 , a 2 + 3 b 2 + 3 c 2 ) but ( a 1 , a 2 ) + ( b 1 , b 2 ) + ( c 1 , c 2 )) = ( a 1 , a 2 ) + ( b 1 + 2 c 1 , b 2 + 3 c 2 ) = ( a 1 + 2 b 1 + 4 c 1 , a 2 + 3 b 2 + 9 c 2 ) . The vector (0 , 0) V is the zero vector, so VS 3 holds. For VS 4, observe that ( a 1 , a 2 ) + ( - a 1 2 , - a 2 3 ) = ( a 1 + 2( - a 1 2 ) , a 2 + 3( - a 2 3 )) = (0 , 0) .

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