Binary operations

Step 1: Write (a\ast b=a+b+3) and (b\ast a=b+a+3). Step 2: Since (a+b=b+a) for real numbers, both are equal. Step 3: Operations based on addition often become commutative.

Step 1: For commutativity, (a-b=b-a) must hold for every pair. Step 2: (5-2=3), but (2-5=-3). They are not equal. Step 3: Treating subtraction as commutative is a common mistake.

Step 1: (a\ast b=ab+1) and (b\ast a=ba+1). Step 2: Since (ab=ba), both values are equal. Step 3: In multiplication-based expressions, check whether order changes the value.

Step 1: First, (2\ast3=2+2\cdot3=8). Step 2: Then, (3\ast2=3+2\cdot2=7). Step 3: In such operations, changing the order can change the answer.

Step 1: Put (a\ast e=a). Step 2: From (a+e-1=a), we get (e=1). Also, (e\ast a=e+a-1=a) gives (e=1). Step 3: To find identity, assume (e) and form an equation.

Step 1: For (a\ast e=a), write (a+e+ae=a). Step 2: This gives (e(1+a)=0), which works for every (a) when (e=0). Also, (0\ast a=a). Step 3: An identity must work for every element, not just one value.

Step 1: First find the identity: (a+e+2=a), so (e=-2). Step 2: The inverse (x) of (3) satisfies (3+x+2=-2), so (x=-7). Step 3: Always find the identity correctly before finding an inverse.

Step 1: The identity (e) satisfies (a+e+2=a), so (e=-2). Step 2: Let the inverse of (3) be (x), so (3+x+2=-2). Thus (x=-7). Step 3: An inverse is the element that takes the given element to the identity.

Step 1: (\min(a,b)) gives the smaller of the two numbers. Step 2: Between (3) and (4), the smaller number is (3), so (3\ast4=3). Step 3: For maximum and minimum operations, read the rule directly.

Step 1: For identity (e), (\max(a,e)=a) must hold for every (a\in A). Step 2: This happens when (e) is the smallest element. The smallest element of (A) is (1). Step 3: For a (\max) operation, the identity is usually the smallest element of the set.

Step 1: For identity (e), (\min(a,e)=a) must hold for every (a\in A). Step 2: This happens when (e) is the largest element. The largest element in (A) is (4). Step 3: For a (\min) operation, the identity is usually the largest element of the set.

Step 1: For associativity, ((a\ast b)\ast c=a\ast(b\ast c)) must hold. Step 2: In addition, ((a+b)+c=a+(b+c)). Step 3: Associativity of addition is a basic property to recognize quickly.

Step 1: Use an example to check associativity. Step 2: ((5\ast2)\ast1=(5-2)-1=2), but (5\ast(2\ast1)=5-(2-1)=4). They are not equal. Step 3: Subtraction is generally not associative.

Step 1: In multiplication, (ab=ba), so it is commutative. Step 2: Also, ((ab)c=a(bc)), so it is associative. Step 3: Recognizing standard multiplication saves time.

Step 1: Put (a=1) and (b=2) in the rule. Step 2: (1\ast2=1+2+1\cdot2=5). Step 3: Include both the addition and multiplication parts of the rule.

Step 1: (a\ast0=a+0-a\cdot0=a). Step 2: (0\ast a=0+a-0\cdot a=a). So (0) does not change the number from either side. Step 3: Check identity from both directions.

Step 1: First calculate (2+2=4). Step 2: The remainder when (4) is divided by (3) is (1). Thus, (2\ast2=1). Step 3: In remainder operations, the final answer should lie in the set.

Step 1: With identity (e), the remainder of (a+e) should be (a) again. Step 2: If (e=0), the remainder of (a+0) remains (a). Step 3: In remainder-based addition, (0) is usually the identity.

Step 1: The identity for this operation is (0). Step 2: The inverse (x) of (1) must make the remainder of (1+x) equal to (0). Since (1+2=3), the remainder is (0). Step 3: In remainder operations, make the sum a multiple of the modulus.

Step 1: Calculate (3+3=6). Step 2: The remainder when (6) is divided by (4) is (2). Hence, (3\ast3=2). Step 3: In such questions, always take the remainder after adding.