Binary operations

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.

Step 1: For identity (e), we need (a*e=a). Step 2: From (a+e+ae=a), we get (e(1+a)=0). Since (a\neq -1), (e=0). Step 3: In identity questions, always apply the condition with a general element.

Step 1: The identity is (0). Step 2: For inverse (b), (a*b=0), so (a+b+ab=0). Thus (b(1+a)=-a), giving (b=\frac{-a}{1+a}). Step 3: To find inverse, set (a*b=e).

Step 1: Commutativity needs (a*b=b*a). Step 2: Here (a^{\log b}=b^{\log a}). Taking (\log) on both sides gives ((\log a)(\log b)=(\log b)(\log a)), always true. Step 3: For exponential operations, logarithms often simplify comparison.

Step 1: Compare ((a*b)*c) and (a*(b*c)). Step 2: Both sides become (a+b+c+kab+kac+kbc+k^2abc). Hence associativity holds for every real (k). Step 3: In associativity checks, expand fully before deciding.

Step 1: For identity (e), (a+e-2=a), so (e=2). Step 2: For inverse (x) of (7), (7+x-2=2), so (x=-3). Step 3: Always find the identity before finding an inverse.

Step 1: (\max(a,b)=\max(b,a)), so it is commutative. Step 2: (\max(\max(a,b),c)=\max(a,\max(b,c))), so it is associative. Step 3: In (\mathbb{N}), the least element is (1), hence (\max(a,1)=a).

Step 1: Identity (e) must satisfy (\min(a,e)=a) for every (a\in\mathbb{N}). Step 2: This means (e) must be greater than or equal to every natural number, but (\mathbb{N}) has no greatest element. Step 3: A (\min) operation has identity only when the set has a greatest element.

Step 1: We need (\min(a,e)=a) for every (a). Step 2: This is possible only when (e) is the greatest element of the set. Here the greatest element is (5). Step 3: For finite sets, connect (\min) and (\max) identities with greatest and least elements.

Step 1: (a*b=a^2+b^2=b^2+a^2=b*a), so it is commutative. Step 2: But ((1*2)*3=5*3=34), while (1*(2*3)=1*13=170). So it is not associative. Step 3: One counterexample is enough to disprove associativity.

Step 1: Put (a*e=a). Step 2: (\frac{ae}{2}=a). Since (a\neq0), (e=2). Step 3: In division-type operations, pay attention to nonzero restrictions.