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Show that the opeation * on Q -{1} defined gby a*b=a+b-ab for all a, b in Q- {1} Satifies (i) the closure property (ii) the associative law (iii) the commutative alw (iv) what is theidentity element ? (v)for each a in Q-{1} find the inverse of a |
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Answer» Solution :(i) closure porperty Let a `in` Q-{1} and b in Q-{1} We know that Q is closed for addition substraction and MULTIPLICATION `therefore`a+b-ab in Q but a*b =1 `rarr` a+b-ab=1 `rarr` a(1-b)=(1-b)`rarra`=1 which is a contradication since 1 `in` Q-{1} therefore a*b `ne` 1 Thus a `in` Q-{1} ,b `in` Q-{1} `rarr` a*b in Q-{1} `therefore` * is a binary operation on Q-{1} (II) Associative law Let a,b,c =(a+b-ab)*c =(a+b-ab)+c-(a+b-ab)c =(a+b+c)-(ab+BC+ac)+bc And a*(b*c)=a*(b+c-bc) =a(b+c-bc)-a(b+c-bc) =(a+b+c)-(ab+bc+ac)+abc `therefore` (a*b)*c=a*(b*c) Hence * is associative (iii) commutative Law Let a,b `in`Q-{1} Then a*b=a+b-ab ltrbgt =b+a-ab =b*a (iv)Existence of identity element Let e be the identity element Then for all a in Q-{1} we have a* e =a `rarr` a + e - ae =a `rarr` e(1-a)=x `rarre=0 in` Q-{1} Nowa*0=a+0-ax0=a And 0*a=0+a-0xa=a ltrbgt Thus o is the identity element in Q-{1} Existence of inverse Let a in Q-{1} and `le t a^(-1)=b`then a*b=0 `rarra`+b-ab=0 `rarr` a=ab-b=(a-1)b `RARRB=(a)/(a-1)in Q-{1}` `therefore a^(-1)=(a)/(a-1)in Q-{1}` Thus each a in Q-{1} has its iverse in Q-{1} |
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