By Hefferon J.
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12 To verify it is one-to-one, assume that f1 (c1 x + c2 y + c3 z) = f1 (d1 x + d2 y + d3 z). Then c1 + c2 x + c3 x2 = d1 + d2 x + d3 x2 by the definition of f1 . Members of P2 are equal only when they have the same coefficients, so this implies that c1 = d1 and c2 = d2 and c3 = d3 . Therefore f1 (c1 x + c2 y + c3 z) = f1 (d1 x + d2 y + d3 z) implies that c1 x + c2 y + c3 z = d1 x + d2 y + d3 z, and so f1 is one-to-one. To verify that it is onto, consider an arbitrary member of the codomain a1 + a2 x + a3 x2 and observe that it is indeed the image of a member of the domain, namely, it is f1 (a1 x + a2 y + a3 z).
As with the prior question, most of the checks consist only of remarking that property is so familiar that it does not need to be proved. 3 There is no multiplicative inverse for 2, so the integers do not satisfy condition (5). 4 These checks can be done by listing all of the possibilities. For instance, to verify the commutativity of addition, that a + b = b + a, we can easily check it for all possible pairs a, b, because there are only four such pairs. Similarly, for associativity, there are only eight triples a, b, c, and so the check is not too long.
34 See the next subsection. 35 (a) Most of the conditions in the definition of a vector space are routine. We here sketch the verification of part (1) of that definition. For closure of U ×W , note that because U and W are closed, we have that u1 +u2 ∈ U and w1 + w2 ∈ W and so (u1 + u2 , w1 + w2 ) ∈ U × W . Commutativity of addition in U × W follows from commutativity of addition in U and W . (u1 , w1 ) + (u2 , w2 ) = (u1 + u2 , w1 + w2 ) = (u2 + u1 , w2 + w1 ) = (u2 , w2 ) + (u1 , w1 ) 44 Linear Algebra, by Hefferon The check for associativity of addition is similar.