Binary operations

Step 1: First evaluate the inner operation: (4\circ 5=4+5-2=7). Step 2: Now (3*7=3+7+2=12). Step 3: In mixed-operation questions, solve the bracketed operation first.

Step 1: The remainder is always (0) or (1), so the operation is closed. Step 2: The remainder of (a+b) does not change with order, and associativity follows from addition modulo (2). Step 3: For small sets, an operation table quickly verifies properties.

Step 1: For identity (e), (\operatorname{lcm}(a,e)=a) for every positive integer (a). Step 2: (\operatorname{lcm}(a,1)=a), so (e=1). Step 3: In the least common multiple operation, (1) acts as the identity.

Step 1: If (e) is the identity, then (\gcd(a,e)=a) for every positive integer (a). Step 2: Such an (e) would need to be a multiple of every positive integer, which no finite positive integer can be. Step 3: Pay attention to the set; if (0) were included, the conclusion could change.

Step 1: (f(a*b)=a+b+ab+1). Step 2: This can be written as ((a+1)(b+1)), which is (f(a)f(b)). Step 3: Sometimes a binary operation hides an ordinary multiplication structure.

Step 1: First (1*2=1+2+2=5). Step 2: Now (5*3=5+3+15=23). Step 3: If associativity is not already established, evaluate according to the given brackets.

Step 1: For inverse (x), we need (\gcd(4,x)=6). Step 2: For any (x\in A), (\gcd(4,x)) can be at most (4), so it cannot be (6). Step 3: An inverse exists only if the operation can reach the identity element.

Step 1: For inverse (x), (\operatorname{lcm}(6,x)=1) is required. Step 2: The least common multiple with (6) can never be (1). Step 3: Having an identity does not mean every element has an inverse.

Step 1: (a*0=a+0+\lambda a0=a). Step 2: (0*a=0+a+\lambda 0a=a). Both hold for every (\lambda). Step 3: A term multiplied by zero often makes the parameter irrelevant for 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), no inverse is obtained. Step 3: Identify the value that makes the denominator zero.

Step 1: Since (a*0=a), the identity is (0). Step 2: (a*x=0\Rightarrow a+x-ax=0\Rightarrow x(1-a)=-a). If (a=1), no value of (x) can satisfy it. Step 3: While finding inverse, check where factors like (1-a) become zero.

Step 1: (2+2=4). Step 2: The remainder when (4) is divided by (3) is (1), so (2*2=1). Step 3: In a remainder-based operation, the final answer must lie in the given set.

Step 1: The identity is (0) because (a*0=a). Step 2: (a*x=0\Rightarrow a+x-4ax=0\Rightarrow x(1-4a)=-a). At (a=\frac{1}{4}), (1-4a=0), so no solution exists. Step 3: Check the denominator before substituting special values.

Step 1: By definition, (x*2=x+2+2x). Step 2: (x+2+2x=7\Rightarrow 3x=5\Rightarrow x=\frac{5}{3}). Step 3: In equation-based problems, first convert the operation into an algebraic equation.

Step 1: (1*2=2(1)+3(2)=8). Step 2: (2*1=2(2)+3(1)=7), so the values are not equal. Step 3: A linear operation with different coefficients is generally not commutative.

Step 1: For a right identity, (a*e=a\Rightarrow 2a+3e=a), giving (3e=-a), which depends on (a). Step 2: No fixed (e) works for all (a). Step 3: An identity must be a fixed element, not a value depending on (a).

Step 1: Since (5) is the identity, (a*5=a). Step 2: (a+5+m=a\Rightarrow m=-5). Step 3: Substitute the given identity directly into the operation to find the parameter.

Step 1: The identity is (0). Step 2: We need (2*(-1)=0). Thus (2-1+m(2)(-1)=0\Rightarrow 1-2m=0\Rightarrow m=\frac{1}{2}). Step 3: In inverse-related parameter problems, set the operation equal to the identity.

Step 1: Write (a+b+ab=0). Step 2: (b(1+a)=-a), so (b=-\frac{a}{a+1}), where (a\neq -1). Step 3: Collect the (b)-terms and factor them for a quick solution.

Step 1: (1-(a*b)=1-a-b+ab=(1-a)(1-b)). Step 2: If (a\neq 1) and (b\neq 1), then ((1-a)(1-b)\neq 0), so (a*b\neq 1). Step 3: To check closure on an excluded-value set, show that the excluded value cannot appear as an output.