Binary operations

Step 1: (a+b-ab) is unchanged by interchanging (a,b), so it is commutative. Step 2: Since (1-(a*b)=(1-a)(1-b)), both bracketings lead to the same product form, so it is associative. Step 3: Complement-type rewriting can make associativity quick.

Step 1: A binary operation must be defined for every ordered pair of elements of the set. Step 2: If (a=1,b=-1), both are in the set, but (a+b=0), so the value is undefined. Step 3: One undefined pair is enough to reject it as a binary operation.

Step 1: The identity is (0) because (a*0=a). Step 2: For inverse (x), (a+x+ax=0\Rightarrow x(a+1)=-a\Rightarrow x=\frac{-a}{a+1}). Step 3: Since (a\neq -1), the denominator is not zero.

Step 1: For closure, the result must belong to (A), so it must not be (-1). Step 2: (a*b+1=(a+1)(b+1)). Since (a,b\neq -1), both factors are non-zero, so (a*b\neq -1). Step 3: For restricted sets, check that the forbidden value cannot occur.

Step 1: The product of two non-zero real numbers is non-zero, so closure holds. Step 2: (1) is identity and every (a\neq0) has inverse (\frac{1}{a}). Step 3: Even familiar multiplication should be checked against the given set.

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.