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a. N1 = {(i1, i2, . . . , in) : ik ∈ Ik for all k ∈ {1, 2, . . . , n}} and b. N2 = {(x1, x2, . . . , xn) : ∑n k=1 xk = 0}. Proof. To prove (a), it suffices to show, by Proposition 1, that N1 is nonempty and x + ry ∈ N1 for all r ∈ R and all x, y ∈ N1. For the first condition, (0, 0, . . . , 0) ∈ N1 since Ik is a subgroup of R containing the additive identity 0 for all k ∈ {1, 2, . . . , n}. Tha...
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ژورنال
عنوان ژورنال: Electronic Markets
سال: 2010
ISSN: 1019-6781,1422-8890
DOI: 10.1007/s12525-010-0040-0