Binary operations

Step 1: For identity (e), we need (a*e=a) for every (a \in A). Step 2: Here (a*e=\min(a,e)), so (e) must be the greatest element of the set. Step 3: In a minimum operation on a finite ordered set, the greatest element works as the identity.

Step 1: For identity (e), (\max(a,e)=a) must hold for all (a \in A). Step 2: This is possible only when (e) is the least element of (A). Step 3: In a maximum operation, the least element is the identity.

Step 1: Let the identity be (e). Then (a*e=a). Step 2: From (a+e+ae=a), we get (e(1+a)=0), so (e=0) works for all (a). Step 3: To find identity, use a general element and form the equation.

Step 1: First find the identity. From (a*e=a), (a+e-2=a), so (e=2). Step 2: Put (3*b=2). Then (3+b-2=2), so (b=1). Step 3: Always find the identity before finding an inverse.

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

Step 1: (a*b=a-b) and (b*a=b-a), which are not equal in general. Step 2: ((a*b)*c=a-b-c) but (a*(b*c)=a-b+c), so it is not associative. Step 3: For subtraction-type operations, order matters a lot.

Step 1: For identity (e), we need (a*e=a). Step 2: (\frac{ae}{2}=a) gives (e=2). Step 3: While simplifying, note that (a) is positive and hence non-zero.

Step 1: For commutativity, check whether (a*b=b*a). Step 2: (a*b=a^2+b^2) and (b*a=b^2+a^2), which are equal. Step 3: The order of terms in addition does not change the value.

Step 1: For commutativity, (a+2b=b+2a) must hold. Step 2: Simplifying gives (a=b). This is not true for all pairs. Step 3: Equality for special pairs does not make the whole operation commutative.

Step 1: The operation is based on the remainder modulo (3). Step 2: The remainder of (a+0) remains (a), so (a*0=a). Step 3: For addition modulo a number, (0) is usually the identity.

Step 1: The identity for this operation is (0). Step 2: We need (3*b=0). Since (3+1=4), the remainder modulo (4) is (0). Step 3: In modulo addition, the inverse makes the sum a multiple of the modulus.

Step 1: Put (a*e=a), giving (a+e-ae=a). Step 2: This gives (e(1-a)=0). For all (a), (e=0) works. Step 3: A single element must work for every element to be the identity.

Step 1: The identity of this operation is (0). Step 2: For inverse (b), write (a+b-ab=0). Step 3: (b(1-a)=-a), so (b=\frac{a}{a-1}). In exams, always equate the operation to the identity.

Step 1: A binary operation needs closure. Step 2: The sum of two natural numbers is again a natural number. Step 3: Having inverses is not necessary for being a binary operation.

Step 1: In a binary operation, the result must remain in the same set for every pair. Step 2: (1*2=1-2=-1), which is not a natural number. Step 3: One counterexample is enough to break closure.

Step 1: (a*b=ab+1) and (b*a=ba+1). Step 2: Since (ab=ba), both values are equal. Step 3: The commutativity of multiplication often helps in checking new operations.

Step 1: For associativity, compare ((a*b)*c) and (a*(b*c)). Step 2: ((a*b)*c=abc+c+1), while (a*(b*c)=abc+a+1). Step 3: These are not equal in general, so the operation is not associative.

Step 1: (a*b=\frac{a+b}{2}=\frac{b+a}{2}=b*a), so it is commutative. Step 2: ((a*b)*c=\frac{a+b+2c}{4}), while (a*(b*c)=\frac{2a+b+c}{4}). They are not equal in general. Step 3: Average-type operations often fail associativity.

Step 1: The identity for this operation is (0). Step 2: Put (2*b=0), so (2+b+2b=0). Step 3: (2+3b=0), hence (b=-\frac{2}{3}).

Step 1: For inverse (b), we need (a+b+ab=0). Step 2: This gives (b(1+a)=-a). If (a=-1), then (1+a=0), so no inverse exists. Step 3: Always check where the denominator becomes zero.

Step 1: (a*b=a) and (b*a=b), so they are generally not equal. The operation is not commutative. Step 2: ((a*b)*c=a*c=a) and (a*(b*c)=a*b=a), so it is associative. Step 3: Apply the rule directly to both sides.

Step 1: (a*b=b) and (b*a=a), so it is not commutative in general. Step 2: ((a*b)*c=b*c=c) and (a*(b*c)=a*c=c), so it is associative. Step 3: For projection-type operations, carefully observe which element is returned.

Step 1: For identity (e), (a+e-3=a) must hold. Step 2: This gives (e=3). Step 3: In operations like (a+b-k), the number (k) often becomes the identity.

Step 1: The identity of this operation is (3). Step 2: Put (5*b=3), so (5+b-3=3). Step 3: Hence (b=1). The inverse is the element that gives the identity after the operation.

Step 1: Put (a=1) and (b=2) in the given rule. Step 2: (1*2=1+2+1\cdot2=5). Step 3: Even in simple evaluation, do not treat the operation symbol as ordinary multiplication.