Binary operations

Step 1: For identity (e), we need (a*e=a). Step 2: From (a+e+ae=a), we get (e(1+a)=0). Since (a\neq -1), (e=0). Step 3: In identity questions, always apply the condition with a general element.

Step 1: The identity is (0). Step 2: For inverse (b), (a*b=0), so (a+b+ab=0). Thus (b(1+a)=-a), giving (b=\frac{-a}{1+a}). Step 3: To find inverse, set (a*b=e).

Step 1: Commutativity needs (a*b=b*a). Step 2: Here (a^{\log b}=b^{\log a}). Taking (\log) on both sides gives ((\log a)(\log b)=(\log b)(\log a)), always true. Step 3: For exponential operations, logarithms often simplify comparison.

Step 1: Compare ((a*b)*c) and (a*(b*c)). Step 2: Both sides become (a+b+c+kab+kac+kbc+k^2abc). Hence associativity holds for every real (k). Step 3: In associativity checks, expand fully before deciding.

Step 1: For identity (e), (a+e-2=a), so (e=2). Step 2: For inverse (x) of (7), (7+x-2=2), so (x=-3). Step 3: Always find the identity before finding an inverse.

Step 1: (\max(a,b)=\max(b,a)), so it is commutative. Step 2: (\max(\max(a,b),c)=\max(a,\max(b,c))), so it is associative. Step 3: In (\mathbb{N}), the least element is (1), hence (\max(a,1)=a).

Step 1: Identity (e) must satisfy (\min(a,e)=a) for every (a\in\mathbb{N}). Step 2: This means (e) must be greater than or equal to every natural number, but (\mathbb{N}) has no greatest element. Step 3: A (\min) operation has identity only when the set has a greatest element.

Step 1: We need (\min(a,e)=a) for every (a). Step 2: This is possible only when (e) is the greatest element of the set. Here the greatest element is (5). Step 3: For finite sets, connect (\min) and (\max) identities with greatest and least elements.

Step 1: (a*b=a^2+b^2=b^2+a^2=b*a), so it is commutative. Step 2: But ((1*2)*3=5*3=34), while (1*(2*3)=1*13=170). So it is not associative. Step 3: One counterexample is enough to disprove associativity.

Step 1: Put (a*e=a). Step 2: (\frac{ae}{2}=a). Since (a\neq0), (e=2). Step 3: In division-type operations, pay attention to nonzero restrictions.

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.