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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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Answer» 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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