Re: If the operation @ is defined for all integers a and b by a@b = a + b
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24 Nov 2021, 10:41
We have that: \(a@b=a+b-ab\)
I. \(a@b = b@a\) --> \(a@b=a+b-ab\) and \(b@a=b+a-ab\) --> \(a+b-ab=b+a-ab\), results match;
II. \(a@0 = a\) --> \(a@0=a+0-a*0=0\) --> \(0=0\), results match;
III. \((a@b)@c = a@(b@c)\) --> \((a@b)@c=a@b+c-(a@b)*c=(a+b-ab)+c-(a+b-ab)c=a+b+c-ab-ac-bc+abc\) and \(a@(b@c)=a+b@c-a*(b@c)=a+(b+c-bc)-(b+c-bc)a=a+b+c-bc-ab-ac+abc\), results match.
Answer: E.