Binary operations

Step 1: (a,b) are non-zero rational numbers, so (b\neq0). Step 2: (\frac{a}{b}) is rational and non-zero, so it lies in (A). Step 3: Division becomes a problem only when the divisor is zero, which is excluded here.

Step 1: (\frac{2}{3}\neq\frac{3}{2}), so it is not commutative. Step 2: ((8*4)*2=2*2=1), while (8*(4*2)=8*2=4), so it is not associative. Step 3: For division operations, small numerical examples are very useful.

Step 1: Put (a*e=a). Step 2: (a+e-1=a\Rightarrow e=1). Also (e*a=e+a-1=a). Step 3: In shifted addition, balance the constant term to find identity.

Step 1: The identity is (1). Step 2: (a*x=1\Rightarrow a+x-1=1\Rightarrow x=2-a). Step 3: Inverse is always found with respect to the identity element.

Step 1: This is addition modulo (3). Step 2: The remainder of (a+0) on division by (3) is (a) itself. So (0) is identity. Step 3: In remainder-based addition, test (0) first.

Step 1: The identity is (0). Step 2: (2*x=0) means the remainder of (2+x) should be (0). Since (2+1=3), the remainder is (0). Step 3: In small finite sets, inverses can be checked directly.

Step 1: Check whether (ab) can be divisible by (5). Step 2: Since (5) is prime and none of (1,2,3,4) is divisible by (5), the product is not divisible by (5). So remainder (0) never occurs. Step 3: If the remainder always returns to the set, closure holds.

Step 1: This is multiplication modulo (5). Step 2: The remainder of (a\cdot1) on division by (5) is (a). Hence (1) is identity. Step 3: In modular multiplication, test (1) as identity first.

Step 1: The identity is (1). Step 2: We need the remainder of (3*x) to be (1). Since (3\cdot2=6), the remainder on division by (5) is (1). Step 3: A quick modular multiplication table helps in such questions.

Step 1: Try to rewrite the expression as a product. Step 2: ((a+1)(b+1)=ab+a+b+1), exactly equal to (a*b). Step 3: Algebraic rewriting makes property checking easier.

Step 1: Put (a*e=a). Step 2: (ae+a+e+1=a\Rightarrow e(a+1)+1=0\Rightarrow e=\frac{-1}{a+1}), which depends on (a). Step 3: An identity must be one fixed element for all (a), so it does not exist here.

Step 1: The result is real, so closure holds. Also (2a+2b=2b+2a), so it is commutative. Step 2: ((a*b)*c=4a+4b+2c), while (a*(b*c)=2a+4b+4c), not generally equal. Step 3: For associativity, compare the complete algebraic forms.

Step 1: ((a*b)*c=k(ka+kb)+kc=k^2a+k^2b+kc). Step 2: (a*(b*c)=ka+k(kb+kc)=ka+k^2b+k^2c). Equality for all (a,c) requires (k^2=k). Step 3: (k(k-1)=0), so (k=0) or (k=1).

Step 1: For identity (e), (a+e+k=a\Rightarrow e=-k). Step 2: For inverse (x), (a+x+k=-k\Rightarrow x=-a-2k). Step 3: First identify how the constant (k) shifts the identity.

Step 1: The product of two non-zero real numbers is non-zero, so (\frac{ab}{2}) is also non-zero. Step 2: (\frac{ae}{2}=a\Rightarrow e=2), which belongs to the set. Step 3: For non-zero sets, ensure the result cannot become zero.

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.