Binary operations

Step 1: For commutativity, check whether (a*b=b*a). Step 2: Since (a^2+b^2=b^2+a^2), the operation is commutative. Step 3: Associativity is a different test; swapping (a) and (b) checks commutativity only.

Step 1: (a*b=a-b) and (b*a=b-a) are generally not equal. Step 2: ((a*b)*c=(a-b)-c), while (a*(b*c)=a-(b-c)), which are generally different. Step 3: In subtraction, both order and grouping can change the result.

Step 1: First evaluate the inner operation: (b*c=2b+3c). Step 2: Then (a*(b*c)=2a+3(2b+3c)=2a+6b+9c). Step 3: In nested operation questions, always work from inside to outside.

Step 1: ((a*b)*c=k(ka+b)+c=k^2a+kb+c). Step 2: (a*(b*c)=ka+(kb+c)=ka+kb+c). Step 3: Equality needs (k^2a=ka) for all (a), so (k^2=k); among the options, (k=1) works.

Step 1: For identity (e), use (a*e=a). Step 2: From (a+e+k=a), we get (e=-k). Step 3: In addition-type operations, balance the extra constant term carefully.

Step 1: First find the identity (e): (a+e+5=a), so (e=-5). Step 2: For inverse (b), use (a+b+5=-5). Step 3: Hence (b=-a-10); inverse is always found with respect to the identity element.

Step 1: Substitute (a=2) and (b=3) in the given operation. Step 2: (2*3=2+3-2\cdot3=5-6=-1). Step 3: Do not treat a new operation as ordinary addition or multiplication; apply its definition.

Step 1: The identity for this operation is (0). Step 2: Put (2*b=0): (2+b-2b=0), so (2-b=0). Step 3: Thus (b=2); an inverse must combine with the element to give the identity.

Step 1: For the greater-element operation, we need (a*e=a). Step 2: (e) must not become greater than any (a) and change the result. Step 3: The smallest element (0) works, so it is the identity.

Step 1: For the smaller-element operation, (a*e=a) is required. Step 2: (e) must not be smaller than any (a), otherwise it would change the result. Step 3: The greatest element (2) works, so the identity is (2).

Step 1: Identity (e) would require (\gcd(a,e)=a) for every (a\in\mathbb{N}). Step 2: Such (e) would have to be a multiple of every natural number, which is not possible as one fixed natural number. Step 3: In (\gcd), (1) often makes the result (1), so do not mistake it for identity.

Step 1: Identity (e) must satisfy (\operatorname{lcm}(a,e)=a). Step 2: (\operatorname{lcm}(a,1)=a) for every natural number (a). Step 3: In (\operatorname{lcm}), (1) does not change the element, so it is the identity.

Step 1: (a*b=a+b-1=b+a-1=b*a), so it is commutative. Step 2: ((a*b)*c=a+b+c-2) and (a*(b*c)=a+b+c-2), so it is associative. Step 3: In shifted addition, if the constant adjusts equally in both groupings, associativity holds.

Step 1: For identity (e), (a+e-1=a), so (e=1). Step 2: Put (5*b=1): (5+b-1=1), giving (b=-3). Step 3: While finding inverse, equate the result to the identity found, not automatically to zero.

Step 1: Write (a*e=a). Step 2: (\frac{ae}{2}=a), and since (a\neq0), (e=2). Step 3: In quotient-based operations, ensure the quantity you divide by is non-zero.

Step 1: The identity of this operation is (2). Step 2: For inverse (b), use (\frac{ab}{2}=2). Step 3: This gives (ab=4), so (b=\frac{4}{a}); finding identity first is essential.

Step 1: Closure means the result must remain in the same set. Step 2: (\frac{1}{2}*\frac{1}{2}=\frac{1}{2}+\frac{1}{2}+\frac{1}{4}=\frac{5}{4}), which is not in ((-1,1)). Step 3: One valid counterexample is enough to disprove closure.

Step 1: Add (1) to (a*b=a+b+ab). Step 2: (1+a+b+ab=(1+a)(1+b)), which helps prove associativity quickly. Step 3: Turning an unusual operation into a familiar product form saves time in exams.

Step 1: If (a,b\neq0), then (\frac{a}{b}\neq0), so closure holds. Step 2: ((a*b)*c=\frac{a}{bc}), while (a*(b*c)=\frac{ac}{b}), generally not equal. Step 3: Division may be closed on non-zero reals, but associativity and commutativity must be tested separately.

Step 1: The identity of this operation is (0). Step 2: From (-1*b=0), we get (-1+b-b=-1), which cannot be (0). Step 3: Do not assume every element has an inverse; solve the inverse equation.