Binary operations

Step 1: A binary operation needs closure for every ordered pair. Step 2: ((-1)*(-1)=-1-1-1=-3), and (-3 \notin A). Step 3: If closure fails even once, the operation is not binary on the set.

Step 1: (a+b+ab=b+a+ba), so the operation is commutative. Step 2: Since (1+(a*b)=(1+a)(1+b)), regrouping three elements behaves like ordinary multiplication. Step 3: For harder associativity checks, a smart transformation saves time.

Step 1: The identity of this operation is (0). Step 2: For inverse (b), (a+b+ab=0), so (b(1+a)=-a). Step 3: When (a=-1), (1+a=0), so no inverse exists.

Step 1: From (a*e=a), the identity is (0). Step 2: For inverse (b), (3+b-6b=0), so (3-5b=0). Step 3: Hence (b=\frac{3}{5}). Always equate the operation result to the identity.

Step 1: The identity is (0). Step 2: From (a+b-2ab=0), we get (b(1-2a)=-a). Step 3: For (a=\frac{1}{2}), (1-2a=0), so an inverse is not possible.

Step 1: The identity for this form is (0). Step 2: Since (2) and \(-\frac{1}{2}\) are inverses, \(2-\frac{1}{2}+k\cdot2\cdot\left(-\frac{1}{2}\right)=0\). Step 3: This gives \(\frac{3}{2}-k=0\), so \(k=\frac{3}{2}\); hence none of the listed options would be correct.

Step 1: For identity (e), we need (a*e=a). Step 2: (a+e-ae=a\Rightarrow e(1-a)=0), so (e=0) works for every (a\in A). Step 3: In exams, always verify both (a*e=a) and (e*a=a).

Step 1: The identity is (0). Step 2: For inverse (x), (a*x=0), so (a+x-ax=0\Rightarrow x(1-a)=-a\Rightarrow x=\frac{a}{a-1}). Step 3: Watch the sign carefully while matching options.

Step 1: The identity for this operation is (2). Step 2: (a*x=2\Rightarrow \frac{ax}{2}=2\Rightarrow x=\frac{4}{a}). Step 3: Find the identity first, then find the inverse.

Step 1: Closure means if (a,b) are integers, then (a*b) must also be an integer. Step 2: (a+b+ab) is formed using integer addition and multiplication, so it is an integer. Step 3: Closure checks the set of the result, not commutativity.

Step 1: Put (a*e=a). Step 2: (a+e+ae=a\Rightarrow e(1+a)=0), so (e=0) works for all (a). Step 3: The same identity must work for the whole set.

Step 1: The identity is (0). Step 2: (a*x=0\Rightarrow a+x+ax=0\Rightarrow x=\frac{-a}{a+1}). For integer (x), (a+1) must divide (a), hence it must also divide (-1). Thus (a+1=\pm1), so (a=0) or (a=-2). Step 3: For inverse questions on integers, divisibility is the key check.

Step 1: Identity (e) must satisfy (\min(a,e)=a) for every (a\in A). Step 2: This happens when (e) is the greatest element of the set, which is (4). Step 3: For a (\min) operation, the identity is the greatest element.

Step 1: We need (\max(a,e)=a) for every (a\in A). Step 2: This is possible when (e) is the smallest element, which is (1). Step 3: For a (\max) operation, the identity is the smallest element.

Step 1: For identity (e), (a+e+2=a\Rightarrow e=-2). Step 2: For inverse (x), (a+x+2=-2\Rightarrow x=-a-4). Step 3: In shifted addition operations, do not directly choose (-a).

Step 1: (a*b=b*a) because (a+b+2=b+a+2). Step 2: ((a*b)*c=a+b+c+4) and (a*(b*c)=a+b+c+4), so it is associative. Step 3: Check commutativity and associativity separately.

Step 1: (a+b+ab) is unchanged when (a) and (b) are interchanged, so it is commutative. Step 2: Since (a*b=(a+1)(b+1)-1), both groupings give ((a+1)(b+1)(c+1)-1). Step 3: Rewriting the operation often simplifies associativity checks.

Step 1: (a^b) is a natural number for natural (a,b), so closure holds. Step 2: But (2^3=8) while (3^2=9), so (a*b\neq b*a). Step 3: One counterexample is enough to disprove commutativity.

Step 1: Associativity requires ((a*b)*c=a*(b*c)). Step 2: ((2*3)*2=8^2=64), but (2*(3*2)=2^9=512). They are not equal. Step 3: With exponent operations, changing brackets often changes the value.

Step 1: (a-b) is generally not equal to (b-a), so it is not commutative. Step 2: ((a*b)*c=(a-b)-c), while (a*(b*c)=a-(b-c)=a-b+c), so it is not associative. Step 3: For subtraction-based operations, check both properties carefully.

Step 1: (a+b-ab) is unchanged by interchanging (a,b), so it is commutative. Step 2: Since (1-(a*b)=(1-a)(1-b)), both bracketings lead to the same product form, so it is associative. Step 3: Complement-type rewriting can make associativity quick.

Step 1: A binary operation must be defined for every ordered pair of elements of the set. Step 2: If (a=1,b=-1), both are in the set, but (a+b=0), so the value is undefined. Step 3: One undefined pair is enough to reject it as a binary operation.

Step 1: The identity is (0) because (a*0=a). Step 2: For inverse (x), (a+x+ax=0\Rightarrow x(a+1)=-a\Rightarrow x=\frac{-a}{a+1}). Step 3: Since (a\neq -1), the denominator is not zero.

Step 1: For closure, the result must belong to (A), so it must not be (-1). Step 2: (a*b+1=(a+1)(b+1)). Since (a,b\neq -1), both factors are non-zero, so (a*b\neq -1). Step 3: For restricted sets, check that the forbidden value cannot occur.

Step 1: The product of two non-zero real numbers is non-zero, so closure holds. Step 2: (1) is identity and every (a\neq0) has inverse (\frac{1}{a}). Step 3: Even familiar multiplication should be checked against the given set.