Binary operations

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.

Step 1: First calculate inside the bracket: (1*2=2\cdot1+2=4). Step 2: Now (4*3=2\cdot4+3=11). Step 3: Compare your calculation with the options. The correct calculation is (11), so the options would be invalid if (11) is absent.

Step 1: For associativity, compare ((a*b)*c) and (a*(b*c)). Step 2: ((a*b)*c=a+2b+2c), while (a*(b*c)=a+2b+4c). Step 3: Since these are not generally equal, the operation is not associative.

Step 1: (a*b=a+b-ab=b+a-ba=b*a), so it is commutative. Step 2: Since (1-(a*b)=(1-a)(1-b)), grouping three elements does not change the result. Step 3: Useful algebraic rewriting can make associativity easier to verify.

Step 1: First find the identity from (a*e=a): (a+e-4=a), so (e=4). Step 2: Put (7*b=4): (7+b-4=4), hence (b=1). Step 3: To find an inverse, equate the operation result to the identity.

Step 1: For identity (e), we need (a*e=a). Step 2: The least common multiple of any (a) with (1) is (a) itself. Step 3: For an LCM operation, check (1) first as the identity.