Binary operations

Step 1: Add (1) to the operation. Step 2: (1+(a*b)=1+a+b+ab=(1+a)(1+b)). Step 3: This form is very useful in inverse and associativity questions.

Step 1: Write (1-a*b=1-(a+b-ab)). Step 2: This becomes (1-a-b+ab=(1-a)(1-b)). Step 3: Such forms connect the operation with ordinary multiplication and simplify proofs.

Step 1: The identity for this operation is (0). Step 2: (1*b=1+b-b=1), which can never become (0). Step 3: If the inverse equation cannot reach the identity, that element has no inverse.

Step 1: Put (a*e=a). Step 2: From (a+e-2ae=a), we get (e(1-2a)=0). Step 3: For all (a), this is true when (e=0), so the identity is (0).

Step 1: The identity is (0). Step 2: For inverse (b), (a+b-2ab=0), so (b(1-2a)=-a). Step 3: If (a=\frac{1}{2}), then (1-2a=0), so the inverse equation cannot be solved.

Step 1: The identity is (0). Step 2: Put (2*b=0): (2+b-4b=0), so (2-3b=0). Step 3: Thus (b=\frac{2}{3}); solve the inverse equation carefully.

Step 1: Write (1+(a*b)=(1+a)(1+b)). Step 2: Ordinary multiplication is associative, so ((1+a)(1+b)(1+c)) is obtained in both groupings. Step 3: Transformation reduces long associativity calculations.

Step 1: If (e\in A) were identity, then (1*e=1) must hold. Step 2: But (1+e\ge2), so (\min(1+e,4)) can never be (1). Step 3: Failure for one element proves that no identity exists.

Step 1: The identity is (0), since the remainder of (a+0) modulo (4) is (a). Step 2: We need (3*b=0), so (3+b) must be divisible by (4). Step 3: With (b=1), the sum is (4), giving remainder (0).

Step 1: The identity for this operation is (1). Step 2: (3*3) gives the remainder of (9) on division by (4), which is (1). Step 3: In modular multiplication, elements relatively prime to the modulus have inverses.

Step 1: (a^2+b) is a real number, so the operation is closed on (\mathbb{R}). Step 2: (a*b=a^2+b), while (b*a=b^2+a), generally not equal. Step 3: Closure and commutativity are different properties; test them separately.

Step 1: Identity must work from both sides. Step 2: (0*a=a) is true, but (a*0=a^2), which is not (a) for every (a). Step 3: One-sided identity is not enough; both sides must return the same element.

Step 1: The identity of this operation is (0). Step 2: For inverse (b), (a+b+kab=0), hence (b(1+ka)=-a). Step 3: When (a=-\frac{1}{k}), (1+ka=0), so no inverse exists.

Step 1: The identity is (0). Step 2: From (2*b=0), (2+b+6b=0), so (2+7b=0). Step 3: (b=-\frac{2}{7}); combine the terms containing (b) carefully.

Step 1: Solve (a+b+3ab=0) for inverse. Step 2: (b(1+3a)=-a), so division is possible only when (1+3a\neq0). Step 3: For (a=-\frac{1}{3}), the denominator fails, so this element has no inverse.

Step 1: Write (a+b-ab=0). Step 2: (b(1-a)=-a), so (b=\frac{-a}{1-a}), when (a\neq1). Step 3: Always note the denominator condition because it decides existence of inverse.

Step 1: Compare (a*b) and (b*a). Step 2: (a*b=a+b-ab) and (b*a=b+a-ba), which are equal. Step 3: Since addition and multiplication of real numbers are commutative, this operation is also commutative.

Step 1: Write (1-(a*b)=(1-a)(1-b)). Step 2: For three elements, the result follows (1-{(a*b)*c}=(1-a)(1-b)(1-c)). Step 3: Ordinary multiplication is associative, so both groupings give the same result.

Step 1: Since (a,b\in A), (a\neq1) and (b\neq1). Step 2: (1-a*b=(1-a)(1-b)), and both factors are non-zero, so the result cannot be (1). Step 3: For closure, it is enough to show the result remains in the same set.

Step 1: The identity is (0). Step 2: From (3*b=0), (3+b-3b=0), so (3-2b=0). Step 3: (b=\frac{3}{2}), and it belongs to the set because it is not (1).