Binary operations

Step 1: For identity (e), (a+e-3=a), so (e=3). Step 2: Put (8*b=3): (8+b-3=3), so (b=-2). Step 3: In shifted addition, the inverse is not simply (-a).

Step 1: First (2*5=2+5-3=4). Step 2: Then (4*7=4+7-3=8). Step 3: When brackets are given, follow that order even if the operation is associative.

Step 1: First (b*c=b+c-3). Step 2: (a*(b*c)=a+(b+c-3)-3=a+b+c-6). Step 3: Substitute the complete value of the inner operation into the next step.

Step 1: For identity (e), (a+e+\lambda ae=a). Step 2: This gives (e(1+\lambda a)=0). Step 3: The fixed element satisfying this for all (a) is (e=0).

Step 1: The identity is (0). Step 2: For inverse (b), (a+b+\lambda ab=0), so (b(1+\lambda a)=-a). Step 3: Such (b) exists only when (1+\lambda a\neq0).

Step 1: With (\lambda=2), (a*b=a+b+2ab). Step 2: From (4*b=0), (4+b+8b=0), so (4+9b=0). Step 3: (b=-\frac{4}{9}); substitute the parameter first to simplify the equation.

Step 1: (\operatorname{lcm}(a,1)=a), so (1) is the identity. Step 2: An inverse (b) of (a) must satisfy (\operatorname{lcm}(a,b)=1). Step 3: This is possible only for (a=1) and (b=1), so only (1) has an inverse.

Step 1: For identity (e), we need (a*e=a). Step 2: (a+e+ae=a) gives (e(1+a)=0), so (e=0) because (a\neq -1). Step 3: In exams, also check the other side; here (e*a=a) holds.

Step 1: The identity for this operation is (0). Step 2: For inverse (x), (a*x=0), so (a+x+ax=0). Step 3: Hence (x(1+a)=-a), giving (x=\frac{-a}{1+a}); always identify the identity first.

Step 1: (a*b=a+b-3=b+a-3=b*a), so it is commutative. Step 2: ((a*b)*c=a+b+c-6) and (a*(b*c)=a+b+c-6), so it is associative. Step 3: For linear operations, expand both sides and compare.

Step 1: First find the identity (e): (a+e-3=a), hence (e=3). Step 2: For inverse (x), (a+x-3=3), so (x=6-a). Step 3: Do not assume the usual additive identity (0) for a new operation.

Step 1: Since (ab=ba), (a*b=b*a), so the operation is commutative. Step 2: ((a*b)*c=\sqrt{c\sqrt{ab}}), while (a*(b*c)=\sqrt{a\sqrt{bc}}), not equal in general. Step 3: Values like (a=1,b=4,c=16) quickly disprove associativity.

Step 1: This is addition modulo (4), whose identity is (0). Step 2: (2*x) means the remainder of (2+x) on division by (4), and it must be (0). Step 3: Since (2+2=4), the remainder is (0), so (2) is its own inverse.

Step 1: For identity (e), (\max(a,e)=a) for every (a). Step 2: This happens only when (e) is the smallest element of the set. Step 3: Here the smallest element is (1), so it is the identity.

Step 1: For identity (e), (\min(a,e)=a) for every (a). Step 2: This is possible when (e) is the largest element of the set. Step 3: Here (4) is the largest, so for a minimum operation look at the largest element.

Step 1: Put (a*e=a), so (\frac{ae}{2}=a). Step 2: Since (a\neq 0), divide by (a) to get (e=2). Step 3: Under a new operation, the identity is not always the usual multiplicative (1).

Step 1: The identity for this operation is (2). Step 2: For inverse (x), (\frac{ax}{2}=2). Step 3: Thus (ax=4), so (x=\frac{4}{a}); find identity before inverse.

Step 1: For identity (e), (a+e+1=a), so (e=-1). Step 2: For inverse (x), (a+x+1=-1). Step 3: Hence (x=-a-2); handle the constant term carefully.

Step 1: From (a*e=a), we get (a+e+ae+2=a). Step 2: This gives (e(1+a)+2=0), which cannot hold for all (a) with one fixed (e). Step 3: An identity must be a fixed element, not dependent on (a).

Step 1: (2*3=2^3=8) and (3*2=3^2=9), so it is not commutative. Step 2: ((2*3)*2=8^2=64), while (2*(3*2)=2^9=512), so it is not associative. Step 3: A good counterexample can disprove both properties quickly.