Binary operations

Step 1: For identity (e), write (a+e-ae=a). Step 2: This gives (e(1-a)=0), which is satisfied for all (a) by (e=0). Step 3: Subtracting (a) from both sides is a quick simplification.

Step 1: The identity for this operation is (0). Step 2: From (4*x=0), (4+x-4x=0), so (4-3x=0). Step 3: Hence (x=\frac{4}{3}); be careful with negative signs.

Step 1: From (a*e=a), we get (a+e+ae=a). Step 2: Thus (e(1+a)=0), and (a\neq -1). Step 3: Hence (e=0); always read the set condition carefully.

Step 1: The identity is (0). Step 2: From (2*x=0), (2+x+2x=0), so (3x=-2). Step 3: Hence (x=-\frac{2}{3}), and it also belongs to (A).

Step 1: The identity is (0). Step 2: From (a*x=0), (a+x+ax=0), so (x=\frac{-a}{1+a}). Step 3: At (a=-1), the denominator is zero, so inverse does not exist.

Step 1: For identity (e), (\frac{ae}{5}=a). Step 2: Since (a\neq0), divide by (a) to get (e=5). Step 3: In fraction-based operations, clear the denominator first.

Step 1: The identity is (5). Step 2: From (10*x=5), (\frac{10x}{5}=5), so (2x=5). Step 3: Hence (x=\frac{5}{2}); find inverse by equating to the identity.

Step 1: For identity (e), (\max(a,e)=a) for all (a\in A). Step 2: This happens only when (e) is the smallest element. Step 3: The smallest element of this set is (2).

Step 1: For identity (e), (\min(a,e)=a) for all (a\in A). Step 2: This is possible only when (e) is the greatest element. Step 3: Hence the identity here is (8).

Step 1: For identity (e), (\gcd(a,e)=a) for every (a\in A). Step 2: This means every (a) must divide (e). Step 3: (6) is a multiple of all given elements, so (e=6).

Step 1: For identity (e), (\operatorname{lcm}(a,e)=a). Step 2: For this, (e) must divide every (a). Step 3: (1) divides all elements, so it is the identity.

Step 1: Difference of two integers is an integer, so closure holds. Step 2: (5-2\neq2-5), so it is not commutative. Step 3: Do not accept identity only from (a*0=a); also check (0*a=a).

Step 1: For a binary operation, the result must remain in (\mathbb{Z}). Step 2: (1*2=\frac{3}{2}), which is not an integer. Step 3: One counterexample is enough to disprove closure.

Step 1: Since (a^2+b^2=b^2+a^2), it is commutative. Step 2: ((1*1)*2=8), while (1*(1*2)=26), so it is not associative. Step 3: Small values are very useful for testing associativity.

Step 1: For commutativity, (ka+b=kb+a) for all (a,b). Step 2: View it as ((k-1)a=(k-1)b). Step 3: This holds for all values only when (k=1).

Step 1: Expanding ((a*b)*c) gives (a+b+c+kab+kbc+kca+k^2abc). Step 2: Expanding (a*(b*c)) gives the same expression. Step 3: Hence it is associative for every real (k).

Step 1: First find identity: (a+e+c=a), so (e=-c). Step 2: From (p*x=e), (p+x+c=-c). Step 3: Hence (x=-p-2c), the inverse of (p).

Step 1: For identity (e), (ae=a). Step 2: Since (a>0), (e=1). Step 3: In usual multiplication, (1) is identity, but it must also belong to the given set.

Step 1: Dividing a positive number by a positive number gives a positive number, so closure holds. Step 2: (\frac{a}{b}) is generally not equal to (\frac{b}{a}). Step 3: In division operations, check left and right sides separately.

Step 1: For identity (e), (a+e+1=a), so (e=-1). Step 2: From (5*x=-1), (5+x+1=-1). Step 3: Hence (x=-7); always connect inverse with identity.