Question

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Determine whether or not S = {(x, 2x)|x is a real number} is a vector space or not.

Answer 1

PAGE NO. DATE: Sol? Given Set s S = {(2,4) XER} we define 't' and M. (21,2%) + (%,,22,) = (x,+22, 262, +%s)) d(48,1,22.)) - (Q2, 224) = 2(x, qx) first we will check for (S,+) is abelian Group X, ER (I) closure An (x, , 2x) (22; 2X₂) 2.ER sit = (2,2x) + (x, 2x) (2+2 , 2 (x,+X,)) +2ER lie. 8,48 ES » Sit) closure Associative BE 24.9% dan (2,2%) B = (2₂, 223) XER ZER ER

PAGE NO. DATE: 1 (378x)+62 = (0,20,3 + (, ; 2%,)) + (x2, 243) (2, +2, 2, 2. (2, + 22 +23)) s, + (62783) = (x + 2 tag, 2(x + 2q +2₂) (s, t82) +62 = S + (6 + 6) i.e. (St) associative. Existance of identity wort site e=ce, 26) S = ( x2x2 GER & t e = 8 & (a,,2x) + (e, , 20) = (a +22) e=0 1 e =(0,oo (II) Existance of Inverse xer si (x, 2x2 - (2) sat i

PAGE NO. - (x,,2x) + (i gry) = (0,o) " ㅋ (in, 28) i (-2,-2x) Iverse of s is de wigit 6+) (v abelian S = (x,, 24) z (x2, 2x2) GER XLER si+82 = (2+2, 210, +22) Ag+ 8, = (2,+ x2 > 203, +23)) + 8 = Sa+s, (5+) is abelian Group (VI) X, ER Cheek LER B = (x 2 x) S₂ = (x2,2x) XER L(8, + h) = 2.2 + LB 2.1 (2,122,) +122,9%) x(x+2₂, 2(x,**)) x (x;; 2) + 2 (72,2%) = X 8, + X de

PAGE NO. DATE: check y & BER (2+8).$, = xb, + BB, = (x,120) GER (&+B). (2,22) = (2+3) 2,3 268+8).24) = (22,782, 9 922+284) & (2, 2x) + B (2, , 2X) - as, + BS (TIL) L, BER S = (a,, 22,) XER char (2.8):6 & (BB) (2.8) (x,,2x1) = (QBX, 28.8 x;) EX (BX,, 2 BX2 = a (8.812 1 EF where It =R 1 is Unity of UR, +, 1)

PAGE NO. DATE: Check I. 8, = 8) s=(2, 2x) X, EIR I (2, 2x) (2,2x) S = {(2, 2x) XEIR } stisfies all above I to (IX) Postulates. Hence (St.) is vector space over R. So, SCR) Vector space. Thanks
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