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HW01_solution

# HW01_solution - 1 a0 a0 = a(0 0 by(VS7 = a0 by(VS3 = a0 0...

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1. a 0 + a 0 = a (0 + 0 ) by (VS7) = a 0 by (VS3) = a 0 + 0 by (VS3) by theorem 1.1 a 0 = 0 2. (1) It is not a vector space, since the conditions (VS2) (VS3) (VS4) (VS8) do not hold. (VS2) ( a 1 , a 2 ) , ( b 1 , b 2 ) , ( c 1 , c 2 ) V (( a 1 , a 2 )+( b 1 , b 2 ))+( c 1 , c 2 ) = ( a 1 + b 1 , - a 2 - b 2 )+( c 1 , c 2 ) = ( a 1 + b 1 + c 1 , a 2 + b 2 - c 2 ) , ( a 1 , a 2 )+(( b 1 , b 2 )+( c 1 , c 2 )) = ( a 1 , a 2 )+( b 1 + c 1 , - b 2 - c 2 ) = ( a 1 + b 1 + c 1 , - a 2 + b 2 + c 2 ) (( a 1 , a 2 ) + ( b 1 , b 2 )) + ( c 1 , c 2 ) 6 = ( a 1 , a 2 ) + (( b 1 , b 2 ) + ( c 1 , c 2 )) (VS3) 0 does not exist. supposed exist, let 0 = ( x 1 , x 2 ) V ( a 1 , a 2 ) + ( x 1 , x 2 ) = ( a 1 + x 1 , - a 2 - x 2 ) = ( a 1 , a 2 ) ( x 1 , x 2 ) = (0 , - 2 a 2 ) similarly choose ( b 1 , b 2 ) V and b 2 6 = a 2 we get ( x 1 , x 2 ) = (0 , - 2 b 2 ) 6 = (0 , - 2 a 2 ) = ( x 1 , x 2 ) contradiction 0 does not exist. (VS4) because 0 does not exist. (VS8) ( a 1 , a 2 ) V and c 1 , c 2 R ( c 1 + c 2 )( a 1 , a 2 ) = (( c 1 + c 2 ) a 1 , a 2 ) = ( c 1 a 1 + c 2 a 1 , a 2 ) c 1 ( a 1 , a 2 ) + c 2 ( a 1 , a 2 ) = ( c 1 a 1 , a 2 ) + ( c 2 a 1 , a 2 ) = ( c 1 a 1 + c 2 a 1 , - 2 a 2 ) ( c 1 + c 2 )( a 1 , a 2 ) 6 = c 1 ( a 1 , a 2 ) + c 2 ( a 1 , a 2 ) (2) It is not a vector space, since the conditions (VS5) (VS6) (VS7) (VS8) do not hold. (VS5) ( a 1 , a 2 ) V 1( a 1 , a 2 ) = ( a 1 + 1 , a 2 + 1) 6 = ( a 1 , a 2 ) (VS6) ( a 1 , a 2 ) V and c 1 , c 2 ∈ < ( c 1 c 2 )( a 1 , a 2 ) = ( c 1 c 2 a 1 + 1 , c 1 c 2 a 2 + 1) c 1 ( c 2 ( a 1 , a 2 )) = c 1 ( c 2 a 1 + 1 , c 2 a 2 + 1) = ( c 1 c 2 a 1 + c 1 + 1 , c 1 c 2 a 2 + c 1 + 1) ( c 1 c 2 )( a 1 , a 2 ) 6 = c 1 ( c 2 ( a 1 , a 2 )) (VS7) ( a 1 , a 2 ) , ( b 1 , b 2 ) V and c ∈ < c (( a 1 , a 2 )+( b 1 , b 2 )) = c ( a 1 + b 1 +1 , a 2 + b 2 +1) = ( ca 1 + cb 1 + c +1 , ca 2 + cb 2 + c +1) c ( a 1 , a 2 )+ c ( b 1 , b 2 ) = ( ca 1 +1 , ca 2 +1)+( cb 1 +1 , cb 2 +1) = ( ca 1 + cb 1 +3 , ca 2 + cb 2 +3) c (( a 1 , a 2 ) + ( b 1 , b 2 )) 6 = c ( a 1 , a 2 ) + c ( b 1 , b 2 ) (VS8) ( a 1 , a 2 ) V and c 1 , c 2 ∈ < 1

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( c 1 + c 2 )( a 1 , a 2 ) = ( c 1 a 1 + c 2 a 1 + 1 , c 1 a 2 + c 2 a 2 + 1) c 1 ( a 1 , a 2 ) + c 2 ( a 1 , a 2 ) = ( c 1 a 1 + 1 , c 1 a 2
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HW01_solution - 1 a0 a0 = a(0 0 by(VS7 = a0 by(VS3 = a0 0...

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