Binary operations

Step 1: The identity of this operation is (2). Step 2: For inverse (x), (5*x=2), so (\frac{5x}{2}=2). Hence (x=\frac{4}{5}). Step 3: Do not confuse operation inverse with ordinary reciprocal.

Step 1: Write (a*e=a). Step 2: From (a+e+1=a), we get (e=-1). Then (a*x=-1) gives (x=-a-2), a real number. Step 3: Finding the identity first makes inverse existence clear.

Step 1: We can write (a*b=1-(1-a)(1-b)). Step 2: If (a\neq1) and (b\neq1), then ((1-a)(1-b)\neq0), so (a*b\neq1). Step 3: For closure, rewriting into a product form can reveal excluded values.

Step 1: Put (a*e=a). Step 2: (a+e-ae=a) gives (e(1-a)=0). Since (a\neq1), (e=0). Step 3: Use the excluded-value condition while solving the identity equation.

Step 1: The identity is (0). Step 2: For inverse (x), (a+x-ax=0). Hence (x(1-a)=-a), so (x=\frac{-a}{1-a}). Step 3: Watch signs carefully because (1-a) and (a-1) differ by a negative sign.

Step 1: The operation is closed in (\mathbb{Z}) and associative. Step 2: The identity is (0). For inverse, (a*x=0) gives (x=\frac{-a}{1+a}). Step 3: For (a=1), (x=-\frac{1}{2}), not an integer, so not every element has an inverse.

Step 1: (a*b=a+b-ab). Step 2: (b*a=b+a-ba=a+b-ab), so (a*b=b*a). Step 3: For commutativity, compare the value after swapping the order.

Step 1: Expanding ((a*b)*c) gives (a+b-ab+c-c(a+b-ab)). Step 2: It simplifies to (a+b+c-ab-ac-bc+abc). The same expression comes from (a*(b*c)). Step 3: To prove associativity, reduce both sides to the same form.

Step 1: (0*0=0), (0*1=1), (1*0=1), and (1*1=1). Step 2: These are exactly the outputs of logical OR. Step 3: For finite sets, a small table helps identify the operation quickly.

Step 1: In multiplication, (1) is the identity because (a\cdot1=a). Step 2: The inverse of (1) is (1), but (0*x=1) is impossible. So (0) has no inverse. Step 3: In multiplication-like operations, check zero carefully.

Step 1: The left side is (a*(b\circ c)=a+b+c+a(b+c)). Step 2: The right side is ((a*b)\circ(a*c)=a+b+ab+a+c+ac=2a+b+c+ab+ac). Equality needs (a=2a), so (a=0). Step 3: In distributive questions, expand both sides separately.

Step 1: (a\circ(b*c)=a(b+c)=ab+ac). Step 2: ((a\circ b)*(a\circ c)=ab+ac), so (\circ) distributes over (*). Step 3: Conversely, (a*(bc)=a+bc) and ((a+b)(a+c)) are not generally equal.

Step 1: The difference of two real numbers is real, so closure holds. Step 2: (a-b) is not generally equal to (b-a), so it is not commutative. Step 3: ((a-b)-c) and (a-(b-c)) are generally different, so it is not associative.

Step 1: For identity (e), (a+e+3=a), so (e=-3). Step 2: Put (4*x=-3). Then (4+x+3=-3), so (x=-10). Step 3: For an operation (a+b+k), the identity is (-k).

Step 1: If (5) is the identity, then (a*5=a). Step 2: From (a+5+\lambda=a), we get (\lambda=-5). Step 3: When identity is given, substitute it directly into the identity condition.

Step 1: Commutativity requires (a+\mu b=b+\mu a) for all (a,b). Step 2: This can be written as ((1-\mu)a=(1-\mu)b). For all (a,b), this is possible only when (\mu=1). Step 3: In parameter questions, equality must hold for all elements.

Step 1: ((a*b)*c=(a+\mu b)+\mu c=a+\mu b+\mu c). Step 2: (a*(b*c)=a+\mu(b+\mu c)=a+\mu b+\mu^2c). Equality needs (\mu c=\mu^2c) for every (c), so (\mu=\mu^2). Step 3: Thus (\mu=0) or (\mu=1).

Step 1: (a*b=pa+qb) and (b*a=pb+qa). Step 2: For equality for every (a,b), the coefficients of (a) and (b) must match. Hence (p=q). Step 3: For linear operations, comparing coefficients is the fastest method.

Step 1: ((a*b)*c=p(pa+qb)+qc=p^2a+pqb+qc). Step 2: (a*(b*c)=pa+q(pb+qc)=pa+pqb+q^2c). Step 3: Equality for every (a,c) requires (p^2=p) and (q^2=q).

Step 1: For this type of operation, the identity is (0). Step 2: Put (2*(-1)=0): (2-1-2\alpha=0), so (\alpha=\frac{1}{2}). Step 3: Check: (3*(-\frac{3}{2})=3-\frac{3}{2}-\frac{9\alpha}{2}), which is (0) when (\alpha=\frac{1}{2}).