Binary operations

Step 1: For closure, (a*b) must belong to (S) for all (a,b \in S). Step 2: The remainder on division by (4) is always one of (0,1,2,3). Step 3: In exams, compare the possible remainders with the given set.

Step 1: For identity (e), we need (a*e=a). Step 2: (a*e=a+e+2=a) gives (e=-2). Step 3: In such questions, write (a*e) first and compare it with (a).

Step 1: First find the identity. From (a*e=a+e-3=a), we get (e=3). Step 2: For the inverse of (5), set (5*x=3), so (5+x-3=3) and (x=1). Step 3: Always find the identity before finding an inverse.

Step 1: Put (a*e=a). Step 2: (a+e+ae=a) gives (e(1+a)=0), so (e=0) works for all (a). Step 3: The identity must work for every element, not just for one value.

Step 1: The identity for this operation is (0). Step 2: For inverse (x), we set (2*x) equal to the identity (0). Step 3: In inverse questions, the product under the operation must equal the identity.

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

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

Step 1: For commutativity, (a*b=b*a) must hold. Step 2: Here (a*b=a-b) and (b*a=b-a), which are not equal in general. Step 3: A counterexample such as (3*1 \ne 1*3) is enough in exams.

Step 1: Write (a*b=a+b+ab). Step 2: (b*a=b+a+ba), and addition and multiplication do not change by order. Step 3: If the expression is symmetric in (a) and (b), commutativity is often easy to identify.

Step 1: Find ((a*b)*c), which is ((a+b+1)+c+1=a+b+c+2). Step 2: Find (a*(b*c)=a+(b+c+1)+1=a+b+c+2). Step 3: Since both are equal, the operation is associative.

Step 1: Notice that (a*b=(a+1)(b+1)-1). Step 2: Thus ((a*b)*c+1=(a+1)(b+1)(c+1)), and (a*(b*c)+1) gives the same value. Step 3: Rewriting the operation can make associativity checks shorter.

Step 1: Put (a*e=a). Step 2: (\frac{ae}{2}=a), and since (a>0), (e=2). Step 3: In multiplication-type operations, pay attention to the constant divisor.

Step 1: The identity of this operation is (2). Step 2: Put (6*x=2). Then (\frac{6x}{2}=2), so (3x=2) and (x=\frac{2}{3}). Step 3: For inverse, the operation result must be the identity.

Step 1: The identity element is (0). Step 2: Put (-1*x=0). This gives (-1+x-x=0), or (-1=0), which is impossible. Step 3: Not every element must have an inverse.

Step 1: Substitute (a=1) and (b=1) in the given formula. Step 2: (1*1=1+1-1=1). Step 3: For small sets, direct substitution quickly gives the answer.

Step 1: Check (a*e=a). Step 2: (a*0=a+0-a\cdot0=a) for every (a \in A). Step 3: In a two-element set, checking both possible values is a safe method.

Step 1: Associativity needs ((a*b)*c=a*(b*c)). Step 2: Take (a=1,b=2,c=3). Then ((1*2)*3=5*3=34), while (1*(2*3)=1*13=170). Step 3: One counterexample is enough to disprove associativity.

Step 1: Write (a*b=a+2b) and (b*a=b+2a). Step 2: They are not equal in general. For example, (1*2=5), but (2*1=4). Step 3: For commutativity, changing the order must not change the result.

Step 1: Put (a*e=a). Step 2: (a+e-ae=a) gives (e(1-a)=0), so (e=0) works for all (a). Step 3: Do not check only one value of (a); the identity must work for all values.

Step 1: The identity element is (0). Step 2: Put (2*x=0). Then (2+x-2x=0), so (2-x=0) and (x=2). Step 3: Sometimes an element can be its own inverse.