Binary operations

Step 1: For a binary operation, every result for (a,b\in A) must lie in (A). Step 2: Here (a\ast b=a), and (a) always belongs to (A). Step 3: In exams, first check closure of the result.

Step 1: For a binary operation on (N), (a-b) must always be in (N). Step 2: (2-5=-3), which is not in (N). Step 3: One counterexample is enough to reject closure.

Step 1: The sum of integers is always an integer. Step 2: (a+b+1) is also an integer for every (a,b\in Z). Step 3: On (Z), expressions made using addition usually remain closed.

Step 1: An identity element (e) must satisfy (a\ast e=a) and (e\ast a=a). Step 2: In multiplication, (a\cdot1=a) and (1\cdot a=a). Step 3: For multiplication, quickly identify (1) as the identity.

Step 1: The identity element for addition is (0). Step 2: The inverse of (5) is a number (x) such that (5+x=0), so (x=-5). Step 3: For addition, the inverse is the negative of the number.

Step 1: The identity is (1) because multiplication by (1) does not change a number. Step 2: (1\cdot1=1), so (1) is its own inverse. Step 3: In a small set, checking options directly is helpful.

Step 1: For identity (e), (a\ast e=a) must hold. Step 2: From (a+e-2=a), we get (e=2). Also, (e+a-2=a) gives (e=2). Step 3: Always check identity from both sides.

Step 1: Put (a=0) and (b=5) in the given rule. Step 2: (0\ast5=0+5+0\cdot5=5). Step 3: Even in simple calculations, apply the operation rule carefully.

Step 1: The rule is (a\ast b=a+2b). Step 2: (3\ast4=3+2\cdot4=11). Step 3: Do not interchange the positions of (a) and (b).

Step 1: For commutativity, (a\ast b=b\ast a) must hold. Step 2: (a\ast b=2a+b), while (b\ast a=2b+a), which are not generally equal. Step 3: A simple example like (a=1,b=2) checks this quickly.

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.