Binary operations

Step 1: If (a\neq0) and (b\neq0), then (\frac{a}{b}\neq0), so closure holds. Step 2: (\frac{a}{b}) is not generally equal to (\frac{b}{a}), so it is not commutative. Step 3: For division operations, test closure and commutativity separately.

Step 1: For positive (a,b), (\sqrt{ab}) is positive, so closure holds. Step 2: (\sqrt{ab}=\sqrt{ba}), so it is commutative. Step 3: ((a*b)*c=\sqrt{c\sqrt{ab}}) and (a*(b*c)=\sqrt{a\sqrt{bc}}), not equal in general.

Step 1: (\frac{ab}{a+b}=\frac{ba}{b+a}), so the operation is commutative. Step 2: For positive (a,b), the result is also positive, so it is closed. Step 3: Identity would need (\frac{ae}{a+e}=a), giving (ae=a^2+ae), impossible for (a>0).

Step 1: (2*3=2+3-1=4), then (4*4=7). Step 2: (3*4=6), then (2*6=7). So both are (7), but that is not among the options. Step 3: This shows the listed choices must be checked carefully.

Step 1: First calculate inside the bracket: (2*3=2+3-1=4). Step 2: Now (4*4=4+4-1=7). Step 3: Do not treat the operation as ordinary addition; apply the given definition each time.

Step 1: For identity (e), we need (a*e=a). Step 2: From (a+e-1=a), we get (e=1). Step 3: In operations of the form (a+b-k), the identity is often (k), but always verify through the identity condition.

Step 1: The identity of this operation is (1). Step 2: Put (8*x=1). Thus (8+x-1=1), giving (x=-6). Step 3: An inverse is always found with respect to the identity of the given operation.

Step 1: By definition, (x*2=x+2+2x). Step 2: From (x+2+2x=8), we get (3x=6), so (x=2). Step 3: In operation equations, first convert the operation into an ordinary algebraic equation.

Step 1: For identity (e), write \(a*e=a\). Step 2: \(a+e+\frac{ae}{2}=a\) gives \(e\left(1+\frac{a}{2}\right)=0\). Since \(a\neq-2\), \(e=0\). Step 3: Paying attention to the excluded value avoids algebraic mistakes.

Step 1: The identity of this operation is (0). Step 2: For inverse (x), (4+x+\frac{4x}{2}=0). Thus (4+3x=0), so (x=-\frac{4}{3}). Step 3: The inverse must belong to the set; here (-\frac{4}{3}\neq-2), so it is valid.

Step 1: For any real (a,b), (a^2+b) is real, so closure holds. Step 2: (a*b=a^2+b), while (b*a=b^2+a). They are not generally equal; for example, (1*2=3) and (2*1=5). Step 3: One valid counterexample is enough to disprove commutativity.

Step 1: (a+b+ab) can be written as ((a+1)(b+1)-1). Step 2: If (a\neq-1) and (b\neq-1), then ((a+1)(b+1)\neq0), so the result cannot be (-1). Step 3: A product form is very useful for detecting excluded values in closure questions.

Step 1: From (a*b=0), we get (a+b+ab=0). Step 2: (b(1+a)=-a), hence (b=\frac{-a}{1+a}), with (a\neq-1). Step 3: Always mention the denominator condition in exam answers.

Step 1: Write (a+b-ab=0). Step 2: (b(1-a)=-a), so (b=\frac{-a}{1-a}), where (a\neq1). Step 3: The signs differ in (a+b+ab) and (a+b-ab), so do not reuse a formula blindly.

Step 1: To be self-inverse, (a*a=0), since the identity is (0). Step 2: (a*a=2a+a^2=a(a+2)=0), so (a=0) or (a=-2). Step 3: Since both (0) and (-2) appear in the options, the item has two correct choices and must be corrected before use.

Step 1: To be self-inverse, (a*a=0). Step 2: (a*a=a^2+2a=a(a+2)), so only (a=0) and (a=-2) are self-inverse. Step 3: (1*1=3\neq0), so (1) is not its own inverse.

Step 1: Identity (e) must satisfy (\gcd(a,e)=a) for every (a\in S). Step 2: This happens when every (a) divides (e). Here (12) is a multiple of all elements of the set. Step 3: For a (\gcd) operation, the identity is usually a greatest common multiple-like element inside the set.

Step 1: Identity (e) must satisfy (\operatorname{lcm}(a,e)=a) for every (a\in S). Step 2: This works for (e=1), since (\operatorname{lcm}(a,1)=a). Step 3: In an (\operatorname{lcm}) operation, check (1) first because it leaves every number unchanged.

Step 1: The average of two positive numbers is positive, so closure holds. Step 2: (\frac{a+b}{2}=\frac{b+a}{2}), so it is commutative. Step 3: ((2*4)*8=3*8=\frac{11}{2}), but (2*(4*8)=2*6=4), so it is not associative.

Step 1: Apply the definition: (2*x=2+x-2x). Step 2: From (2-x=3), we get (x=-1). Step 3: The variable may appear in two places in an operation, so first form the complete algebraic equation.