1.

Let A=NxxN Define * on A by (a,b)*(c,d)=(a+c,b+d) Show that (i) A is closed for * (ii) * is commutative (iii) * is associative (iv) identify element does not exist in A

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Solution :(i) Let (a,b) in A and (c,d) inA Then a,b,c d in N
`(a,b)*(c,d)=(a+c,b+d)in A`[before a+c in N, b +d `in` N]
`therefore` A is CLOSED for*
(a,b)*(c,d)=(a+c,b+d)
`=(c+a,d+b) ["before" a+c=c+a and b+d=d+b]`
=(c,d)*(a,b)
(III)(a,b)*(c,d)*(e,F)=(a+c,b+d)*(e,f)
=[(a+c)+e,(b+d)+f]
=([a+(c+e),b+(d+f)]
=(a,b)*[(c,de,d+f)
`=(a,b)*[(c,d)*(e,f)]`
(IV) (a,b)*(0,0)=(a+0,b+0)=(a,b)
But ,(0,0)`ne` A since one N
So identity element does not BELONG to A


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