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Unformatted text preview: ~v , (3) if ~u + ~v = ~u + ~w then ~v = ~w , (4) ~v = ~ 0 for all ~v V , (5) r ~ 0 = ~ 0 for all scalars r R , (6) ( r ) ~v = r ( ~v ) = ( r~v ) for all r R and for all ~v V . 3.1 Vector Spaces 3 Proof of (1) and (3). Suppose that there are two additive identities, ~ 0 and ~ . Then consider: ~ 0 = ~ 0 + ~ (since ~ is an additive identity) = ~ (since ~ 0 is an additive identity). Therefore, ~ 0 = ~ and the additive identity is unique. Suppose ~u + ~v = ~u + ~w . Then we add ~u to both sides of the equation and we get: ~u + ~v + ( ~u ) = ~u + ~w + ( ~u ) ~v + ( ~u ~u ) = ~w + ( ~u ~u ) ~v + ~ 0 = ~w + ~ ~v = ~w The conclusion holds. QED Example. Page 189 number 14 and page 190 number 24....
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 Fall '04
 IgorDolgachev
 Addition, Multiplication, Vectors, Scalar, Vector Space

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