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Unformatted text preview: 1 ,y 2 ,y 3 ) S . Thus, x 1 + x 2 + x 3 = 0 and y 1 + y 2 + y 3 = 0. We want to show that x + y = ( x 1 + y 1 ,x 2 + y 2 ,x 3 + y 3 ) S . Well this is true only if ( x 1 + y 1 ) + ( x 2 + y 2 ) + ( x 3 + y 3 ) = 0, which clearly true because ( x 1 + y 1 )+( x 2 + y 2 )+( x 3 + y 3 ) = ( x 1 + x 2 + x 3 )+( y 1 + y 2 + y 3 ) = 0+0 = 0. Thus S is closed under addition. X closed under scalar multiplication: Let x = ( x 1 ,x 2 ,x 3 ) S and let be scalar. Thus, x 1 + x 2 + x 3 = 0 and R . We want to show that x = ( x 1 ,x 2 ,x 3 ) S . This is true only if x 1 + x 2 + x 3 = 0, which is true because x 1 + x 2 + x 3 = ( x 1 + x 2 + x 3 ) = 0 = 0 . X Thus, S is a subspace of R 3 . 1...
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 Spring '10
 PANTANO
 Real Numbers, Addition, Multiplication, Scalar

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