Binary operations

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.

Step 1: (1+(a*b)=1+a+b+ab=(1+a)(1+b)). Step 2: If (a\neq -1) and (b\neq -1), this product is non-zero, so (a*b\neq -1). Step 3: Since (a*0=a), (0) is also the identity.

Step 1: First (b*c=b+c+bc). Step 2: (a*(b*c)=a+(b+c+bc)+a(b+c+bc)). Simplifying gives (a+b+c+ab+bc+ca+abc). Step 3: While expanding, write every term carefully.

Step 1: (1+(a*b)=1+a+b+ab=(1+a)(1+b)). Step 2: Multiplication is associative, so changing brackets in ((1+a)(1+b)(1+c)) does not change the value. Step 3: Recognizing a hidden structure saves lengthy calculation.

Step 1: In multiplication modulo (5), the identity is (1). Step 2: We need (2*x) to have remainder (1) on division by (5). Since (2\cdot 3=6), the remainder is (1). Step 3: A multiplicative inverse must make the product congruent to (1).

Step 1: The identity of this operation is (1). Step 2: (0*x=0) for every (x), so it can never become (1). Step 3: Zero has no multiplicative inverse, even in a remainder-based operation.

Step 1: For positive (a,b), (\frac{a}{b}) is positive, so closure holds. Step 2: ((a*b)*c=\frac{a}{bc}), while (a*(b*c)=\frac{ac}{b}), which are not equal in general. Step 3: Division-type operations are often not associative.

Step 1: (a*b=a+b+ab). Step 2: (b*a=b+a+ba). In real numbers, addition and multiplication are commutative, so both are equal. Step 3: Check how changing order affects every term of the operation.

Step 1: Write (a+b+ab=a). Step 2: (b+ab=0\Rightarrow b(1+a)=0). Since (a\neq -1), (b=0). Step 3: Such a question often hides the identity element condition.

Step 1: (x*1=x+1+x=2x+1). Step 2: (2x+1=5\Rightarrow 2x=4\Rightarrow x=2). Step 3: Once the operation is converted into an algebraic expression, the equation becomes easy.

Step 1: The identity element is (0) because (a*0=a). Step 2: (5*x=0\Rightarrow 5+x+5x=0\Rightarrow 6x=-5\Rightarrow x=-\frac{5}{6}). Step 3: Now (x*2=-\frac{5}{6}+2-\frac{10}{6}=-\frac{1}{2}). In exams, after finding the inverse, substitute it carefully into the next operation.