Binary operations

Step 1: (a^2+b^2=b^2+a^2), so it is commutative. Step 2: Use a counterexample: ((2*1)*1=5*1=26), while (2*(1*1)=2*2=8). Therefore it is not associative. Step 3: To disprove associativity, choose a counterexample carefully.

Step 1: (|a-b|=|b-a|), so it is commutative. Step 2: ((1*3)*6=2*6=4), while (1*(3*6)=1*3=2). They are not equal. Step 3: Even with absolute value, changing brackets can change the result.

Step 1: To disprove commutativity, one counterexample is enough. Step 2: For (a=-2,b=3), (a*b=|-2|3=6), while (b*a=|3|(-2)=-6). Step 3: The operation need not fail for every pair; one unequal pair is enough.

Step 1: Under usual multiplication, the identity is (1). Step 2: ((-1)\cdot(-1)=1), so (-1) is its own inverse. Step 3: In a small set, inverse can be found by direct multiplication.

Step 1: Usual multiplication is commutative, so (ab=ba). Step 2: It is also associative, so ((ab)c=a(bc)). Step 3: In a finite set, check closure first; here products of (1) and (-1) remain in (A).

Step 1: Natural numbers are generally (1,2,3,\ldots). Step 2: (1*1=1+1-2=0), and (0) is not in this set. Hence closure fails. Step 3: For closure checks, always test the smallest elements.

Step 1: Put (a*e=a). Step 2: (a+e-2=a\Rightarrow e=2). Also (e*a=e+a-2=a). Step 3: The identity balances the constant subtracted in the operation.

Step 1: The identity of this operation is (2). Step 2: (5*x=2\Rightarrow 5+x-2=2\Rightarrow x=-1). Step 3: While finding inverse, the target is the identity of that operation, not always (0).

Step 1: For closure in (A), the result must not be (1). Step 2: (1-(a*b)=(1-a)(1-b)). Since (a,b\neq1), both factors are non-zero, so (a*b\neq1). Step 3: Showing that the forbidden element cannot occur proves closure neatly.

Step 1: The identity is (0). Step 2: For inverse (x), (a+x-ax=0\Rightarrow x=\frac{-a}{1-a}). When (a=1), the denominator becomes zero and no solution exists. Step 3: Always check where the inverse formula becomes undefined.

Step 1: To test commutativity, interchange (a) and (b). Step 2: (a+b+\lambda ab=b+a+\lambda ba), which is true for every (\lambda). Step 3: In parameter questions, first check whether the expression is symmetric.

Step 1: Put (a*e=a). Step 2: (a+e+\lambda ae=a\Rightarrow e(1+\lambda a)=0). (e=0) works for every (a). Step 3: The value that works for all (a) is the identity.

Step 1: The identity is (0). Step 2: (a*x=0\Rightarrow a+x+\lambda ax=0\Rightarrow x(1+\lambda a)=-a). If (1+\lambda a=0), (x) is not defined. Step 3: The most important check in inverse questions is whether the denominator becomes zero.

Step 1: The identity is (0). Step 2: For inverse (x), \(a+x+\lambda ax=0\Rightarrow x(1+\lambda a)=-a\). Therefore \(x=\frac{-a}{1+\lambda a}\). Step 3: The set excludes exactly the value that would make the denominator zero.

Step 1: For identity (e), we need (a*e=a) for every positive (a). Step 2: (\sqrt{ae}=a\Rightarrow ae=a^2\Rightarrow e=a), so (e) depends on (a). Step 3: An identity must be one fixed element, so no identity exists here.

Step 1: If (a,b) are real, then (a+2b) is also real, so closure holds. Step 2: Commutativity would require (a+2b=b+2a) always, but for (a=1,b=2), (1*2=5) and (2*1=4). Step 3: One valid counterexample is enough to disprove commutativity.

Step 1: First evaluate inside the bracket: (3*4=3+4-12=-5). Step 2: Now (2*(-5)=2+(-5)-2(-5)=7). Step 3: In binary operation questions, follow the bracket order carefully.

Step 1: For identity (e), we need (a*e=a). Step 2: From (a+e-ae=a), we get (e(1-a)=0), so for every (a\neq 1), (e=0). Step 3: In exams, start identity questions by writing (a*e=a).

Step 1: First find the identity; from (a*e=a), (e=0). Step 2: For inverse (b), use (a*b=0), so (a+b+ab=0). Step 3: (b(1+a)=-a), hence (b=\frac{-a}{1+a}); always confirm the identity before finding inverse.

Step 1: A binary operation needs closure. Step 2: If (a,b\in\mathbb{Z}), then (a+b+2ab) is also an integer. Step 3: In such questions, check whether the output returns to the same set.