Binary operations

Step 1: For identity (e), we need (a*e=a). Step 2: (a+e+ae=a) gives (e(1+a)=0), so (e=0) because (a\neq -1). Step 3: In exams, also check the other side; here (e*a=a) holds.

Step 1: The identity for this operation is (0). Step 2: For inverse (x), (a*x=0), so (a+x+ax=0). Step 3: Hence (x(1+a)=-a), giving (x=\frac{-a}{1+a}); always identify the identity first.

Step 1: (a*b=a+b-3=b+a-3=b*a), so it is commutative. Step 2: ((a*b)*c=a+b+c-6) and (a*(b*c)=a+b+c-6), so it is associative. Step 3: For linear operations, expand both sides and compare.

Step 1: First find the identity (e): (a+e-3=a), hence (e=3). Step 2: For inverse (x), (a+x-3=3), so (x=6-a). Step 3: Do not assume the usual additive identity (0) for a new operation.

Step 1: Since (ab=ba), (a*b=b*a), so the operation is commutative. Step 2: ((a*b)*c=\sqrt{c\sqrt{ab}}), while (a*(b*c)=\sqrt{a\sqrt{bc}}), not equal in general. Step 3: Values like (a=1,b=4,c=16) quickly disprove associativity.

Step 1: This is addition modulo (4), whose identity is (0). Step 2: (2*x) means the remainder of (2+x) on division by (4), and it must be (0). Step 3: Since (2+2=4), the remainder is (0), so (2) is its own inverse.

Step 1: For identity (e), (\max(a,e)=a) for every (a). Step 2: This happens only when (e) is the smallest element of the set. Step 3: Here the smallest element is (1), so it is the identity.

Step 1: For identity (e), (\min(a,e)=a) for every (a). Step 2: This is possible when (e) is the largest element of the set. Step 3: Here (4) is the largest, so for a minimum operation look at the largest element.

Step 1: Put (a*e=a), so (\frac{ae}{2}=a). Step 2: Since (a\neq 0), divide by (a) to get (e=2). Step 3: Under a new operation, the identity is not always the usual multiplicative (1).

Step 1: The identity for this operation is (2). Step 2: For inverse (x), (\frac{ax}{2}=2). Step 3: Thus (ax=4), so (x=\frac{4}{a}); find identity before inverse.

Step 1: For identity (e), (a+e+1=a), so (e=-1). Step 2: For inverse (x), (a+x+1=-1). Step 3: Hence (x=-a-2); handle the constant term carefully.

Step 1: From (a*e=a), we get (a+e+ae+2=a). Step 2: This gives (e(1+a)+2=0), which cannot hold for all (a) with one fixed (e). Step 3: An identity must be a fixed element, not dependent on (a).

Step 1: (2*3=2^3=8) and (3*2=3^2=9), so it is not commutative. Step 2: ((2*3)*2=8^2=64), while (2*(3*2)=2^9=512), so it is not associative. Step 3: A good counterexample can disprove both properties quickly.

Step 1: Put (a*e=a), giving (a+e-ae=a). Step 2: Thus (e(1-a)=0), and (e=0) works for every (a). Step 3: Also (e*a=e+a-ea=a), so (0) is the identity.

Step 1: (1-(a*b)=1-a-b+ab=(1-a)(1-b)). Step 2: Since (a\neq 1) and (b\neq 1), this is non-zero, so (a*b\neq 1); hence closure holds. Step 3: Identity is (0), and for (a\neq 1), the inverse is (x=\frac{a}{a-1}).

Step 1: Sums and products of integers are still integers. Step 2: (a+b+ab) is a sum of three integers, so it belongs to (\mathbb{Z}). Step 3: Closure requires the result to remain in the same set, not necessarily positive.

Step 1: For a binary operation, the result must again belong to (A). Step 2: (2) and (4) are even, but (\frac{2+4}{2}=3), which is not even. Step 3: Once closure fails, it is not a binary operation on that set.

Step 1: Closure requires every result to lie in (A). Step 2: (1+7=8), whose remainder on division by (8) is (0). Step 3: Since (0\notin A), one counterexample is enough.

Step 1: Commutativity checks whether (a*b=b*a). Step 2: (a*b=a^2+b^2) and (b*a=b^2+a^2), equal by commutativity of usual addition. Step 3: Do not confuse commutativity with associativity or inverse.

Step 1: Associativity requires ((a*b)*c=a*(b*c)). Step 2: ((1*2)*3=5*3=25+9=34), while (1*(2*3)=1*13=1+169=170). Step 3: Being commutative does not guarantee associativity.

Step 1: Expanding ((a*b)*c) gives (a+b+c+kab+kc(a+b+kab)). Step 2: Expanding (a*(b*c)) gives (a+b+c+kbc+ka(b+c+kbc)). Step 3: Both equal (a+b+c+kab+kac+kbc+k^2abc), so it is associative for every (k).

Step 1: Put (a*e=a), giving (a+e+kae=a). Step 2: Thus (e(1+ka)=0). In the given set (1+ka\neq 0), so (e=0). Step 3: The excluded element is important for closure and inverse behavior.

Step 1: The identity is (0). Step 2: From (a*x=0), we get (a+x+2ax=0). Step 3: Hence (x(1+2a)=-a), so (x=\frac{-a}{1+2a}); the excluded value avoids zero denominator.

Step 1: If (a,b) are real, (a+2b) is real, so closure holds. Step 2: (1*2=5) and (2*1=4), so it is not commutative. Step 3: ((1*2)*3=11), but (1*(2*3)=17), so it is not associative.

Step 1: A right identity would require (a*e=2a+e=a), giving (e=-a), dependent on (a). Step 2: A left identity would require (e*a=2e+a=a), giving (e=0). Step 3: Since there is no single two-sided identity, no identity exists.