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Let A = Q x Q , Q being the set of rationals . Let ‘*’ be a binary operation
on A , defined by (a, b) * (c , d) = ( ac , ad + b) . Show that
(i) ‘*’ is not commutative (ii) ‘*’ is associative
(iii The Identity element w.r.t ‘*’ is ( 1 , 0)
 one year ago
 one year ago
Let A = Q x Q , Q being the set of rationals . Let ‘*’ be a binary operation on A , defined by (a, b) * (c , d) = ( ac , ad + b) . Show that (i) ‘*’ is not commutative (ii) ‘*’ is associative (iii The Identity element w.r.t ‘*’ is ( 1 , 0)
 one year ago
 one year ago

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sirm3dBest ResponseYou've already chosen the best response.1
(i) (a,b) * (c,d) = (ac, ad + b) (c,d) * (a,b) = (ca,cb + d) (ac, ad + b) \(\neq\) (ca, cb + d) therefore it is not commutative
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
thanks bro and how i solve (ii) part?
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
compute (a,b) * ((c,d) * (e,f)) compute ((a,b) * (c,d)) * (e,f) compare the two ordered pairs if they are equal
 one year ago

sirm3dBest ResponseYou've already chosen the best response.1
(a,b) * ((c,d) * (e,f)) = (a,b) * (ce, cf + d) = (ace, a(cf + d) + b) ((a,b) * (c,d)) * (e,f) = (ac, ad + b) * (e,f) = (ace, acf + (ad + b)) the two ordered pairs are the same, therefore * is associative
 one year ago

rishabh.missionBest ResponseYou've already chosen the best response.0
thnanks again @sirm
 one year ago
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