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Ansh Chawla posted an Question

May 17, 2021 • 20:28 pm 30 points

Example : let v be the set of all ordered pairs whether v(f) is a vector space or not or the following defined operat v vity in v] nmutativity in v) by associat

Example : Let V be the set of all ordered pairs whether V(F) is a vector space or not or the following defined operat V vity in V] nmutativity in V) by associativity in V associativ [by Examine wheth (a, b) + (C, d) = (0, b+d); al numbers and F be the vity in V] WTvers (a) (a, b)+(C, d) = (a +C, b + d); (b) (a, b) + (c, d) = (a + C, b+ d: eld of real pla, b) = (pa, pb) pla, b) = (0, pb) pla, b) = (Pa, p? b) i (a, b) + (C, d) = (a, b); pla, b) = (pa, pb) (d) (e) (a, b) + (C, d) = (a +c, b + d); pla, b) = ( Pla, | plb) V a, b, C, d, p e R t Pxist. If suppose any element (C, d) e V is taken as additive identity for vector addition, then Solution. (a) It can be easily seen that the additive identity for the defined vector addition does nol exist. If. (a, b) + (c, d) = (0, b + d) (a, b) Therefore (V, +) is not an abelian group. Consequently V(F) is not a vector space. (b) In this case, (V, +) is an abelian group but space axiom V, is not satisfied, because 1(a, b) = (0, 1b) = (O, b) (a, b) Therefore V(F) is not a vector space. (c) In this case, (V +) is an abelian group but space axiomV, is not satisfied, because (p+g)(a, b) = {(p + g)* a, (p + q) b}; P, geR (p a, p b) + (q^ a, qe b) (p+ q) (a, b) # p(a, b) +q(a, b) Iherefore V(F) is not a vector space. 9 in this case the defined vector addition is not commutative, because for (a, b); (c, d) e V (a, b) + (C, d) = (a, b) and (c, d) + (a, b) = (c, d) (a, b)+ (C, d) (c, d) +(a, b) erefore (V, +) is not an abelian group. Therefore nsequently, V(E) is not a vector space. V,)IS an abelian grouD but space axiom V, is not satisfied, becaUse p+9)(a, b) = (l p + q | a, Ip + g| D) e In this case. P, eR (pla, Ip | b) + (1 q| a, | q| b) Tp +g|* |p |+ lql. p, q e R (a, b) « P(a, b) + q(a, b) Us: Websitn Maduncle.com Call Toll Free :1800-120-1021 792

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