Binary operations

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.

Step 1: For identity (e), (\gcd(a,e)=a) must hold for every (a \in A). Step 2: This happens for (e=12), because all elements of (A) divide (12). Step 3: For a GCD operation on divisors of a number, the largest element is often the identity.

Step 1: Write (a*b=a^2+b) and (b*a=b^2+a). Step 2: For example, (1*2=3), but (2*1=5). They are not equal. Step 3: One valid counterexample is enough to disprove commutativity.

Step 1: For closure, the result must stay in the same set. Step 2: If (a) and (b) are integers, then (a+b+ab) is also an integer. Step 3: Addition and multiplication are closed in integers, so such a combination also gives an integer.

Step 1: For a binary operation, the result for every pair must be a natural number. Step 2: (1*2=\frac{3}{2}), which is not a natural number. Step 3: A single counterexample can show that closure fails.

Step 1: Put (a=1) and (b=0) in the rule. Step 2: (1*0=1+0-1\cdot0=1). Step 3: Even for small sets, directly applying the rule is the safest method.

Step 1: Write (a*b=3a+3b-2ab) and (b*a=3b+3a-2ba). Step 2: Addition and multiplication do not change value when order is changed, so both are equal. Step 3: Symmetric-looking operations are often commutative.

Step 1: Put (a*e=a), giving (a+e+2ae=a). Step 2: This becomes (e(1+2a)=0). For all (a), (e=0) works. Step 3: The identity is the element that does not change any element under the operation.

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

Step 1: First add (2+4=6). Step 2: When (6) is divided by (5), the remainder is (1). Step 3: In remainder-based operations, the final answer must belong to the given set.

Step 1: In a multiplication-based remainder operation, we need (a*e=a). Step 2: The remainder of (a\cdot1) when divided by (5) is (a), since (a) already lies from (0) to (4). Step 3: For multiplication modulo a number, (1) is the identity.

Step 1: For a binary operation, the result must remain in (A). Step 2: (1*1=|1-1|=0), but (0 \notin A). Step 3: Closure must hold for every pair, not just some pairs.

Step 1: (a*b=|a-b|) and (b*a=|b-a|). Step 2: Because of absolute value, (|a-b|=|b-a|). Step 3: Absolute difference is commutative, but associativity must be checked separately.

Step 1: Put (a*e=a), giving (a+e-1=a). Step 2: This gives (e=1). Step 3: For a linear operation, (a*e=a) is the simplest way to find the identity.

Step 1: The identity in this operation is (1). Step 2: Put (4*b=1), so (4+b-1=1). Step 3: Hence (b=-2). An inverse is the element that gives the identity under the operation.

Step 1: (a+b+5=b+a+5), so the operation is commutative. Step 2: ((a*b)*c=a+b+c+10) and (a*(b*c)=a+b+c+10), so it is associative. Step 3: Even with a constant added, expand both sides completely.

Step 1: Put (a*e=a), giving (a+e+5=a). Step 2: This gives (e=-5). Step 3: In an operation like (a+b+k), the identity is usually (-k).

Step 1: Write (a*b=\sqrt{ab}) and (b*a=\sqrt{ba}). Step 2: Since (ab=ba), both square roots are equal. Step 3: Use the commutative property of multiplication to test the new operation.

Step 1: Test with an example. ((4*1)*1=\sqrt{\sqrt{4\cdot1}\cdot1}=\sqrt{2}). Step 2: (4*(1*1)=\sqrt{4\cdot1}=2). These are not equal. Step 3: One clear counterexample is enough to disprove associativity.

Step 1: Put (a*e=a), giving (a+e-2ae=a). Step 2: This gives (e(1-2a)=0). For all (a), (e=0) works. Step 3: The identity must be one fixed element that works for every real number.

Step 1: In this multiplication-based remainder operation, the identity is (1). Step 2: We need (2*b=1). Since (2\cdot2=4), and the remainder of (4) on division by (3) is (1), the inverse is (2). Step 3: While finding an inverse, always equate the operation result to the identity element.