Binary operations

Step 1: First find the identity. Since (a*0=a), the identity is (0). Step 2: Put (2*x=0). Then (2+x+2x=0\Rightarrow 3x=-2\Rightarrow x=-\frac{2}{3}). Step 3: In exams, always identify the identity before finding an inverse.

Step 1: From (a*e=a), we get (a+e-1=a), so (e=1). Step 2: Put (7*x=1). Then (7+x-1=1\Rightarrow x=-5). Step 3: For operations like (a+b-c), compare directly to find the identity.

Step 1: For associativity, ((a*b)*c=a*(b*c)) must hold. Step 2: Here ((a*b)*c=\sqrt{c\sqrt{ab}}) and (a*(b*c)=\sqrt{a\sqrt{bc}}), which are not equal in general. Step 3: A small counterexample is a quick way to disprove associativity.

Step 1: Adding (0) leaves every element unchanged, so the identity is (0). Step 2: (3*x=0) means (3+x) must have remainder (0) on division by (4). With (x=1), the sum is (4). Step 3: In remainder operations, the inverse makes the sum a multiple of the modulus.

Step 1: For commutativity, (a*b=b*a) must hold. Step 2: (2*3=2^3=8), but (3*2=3^2=9). Step 3: One valid counterexample is enough to disprove commutativity.

Step 1: Substitute (a=2) and (b=3). Step 2: (2*3=2+3+6k=11\Rightarrow 6k=6\Rightarrow k=1). Step 3: For unknown-constant questions, directly use the given operation value.

Step 1: For identity (e), (a*e=a) must hold. Step 2: (a+e-2ae=a\Rightarrow e(1-2a)=0). This is true for every (a) when (e=0). Step 3: An identity must work for all elements, not just one value.

Step 1: Put (a*e=a). Step 2: (\frac{ae}{2}=a). Since (a\neq 0), we get (e=2). Step 3: Before dividing by a variable, check that the set excludes zero.

Step 1: The identity for this operation is (2). Step 2: Put (5*x=2). Then (\frac{5x}{2}=2\Rightarrow x=\frac{4}{5}). Step 3: Inverses are always found with respect to the identity element.

Step 1: The identity is (0). Step 2: From (a*x=0), (a+x+ax=0\Rightarrow x(1+a)=-a). If (a=-1), no value of (x) can satisfy it. Step 3: If the denominator in an inverse formula becomes zero, check that element separately.

Step 1: For identity (e), (\max(a,e)=a) for every (a\in A). Step 2: This happens when (e) is the least element of the set. Here the least element is (1). Step 3: For a (\max) operation, the identity is the smallest element of the set.

Step 1: If (e) is the identity, (\min(a,e)=a) for every (a\in A). Step 2: This is possible only when (e) is the greatest element. In (A), the greatest element is (5). Step 3: For a (\min) operation, the identity is the greatest element.

Step 1: ((a*b)*c=a*c=a) and (a*(b*c)=a*b=a), so it is associative. Step 2: (1*2=1), but (2*1=2), so it is not commutative. Step 3: Check commutativity and associativity separately.

Step 1: Since (a*0=a), the identity is (0). Step 2: (3*x=0\Rightarrow 3+x-3x=0\Rightarrow 3-2x=0\Rightarrow x=\frac{3}{2}). Step 3: Be careful with signs because the (ab) term can change the inverse.

Step 1: (a*e=a\Rightarrow a+e-3=a), so (e=3). Step 2: (a*x=3\Rightarrow a+x-3=3\Rightarrow x=6-a). Step 3: In this type of operation, the subtracted constant becomes the identity.

Step 1: For closure, if (a,b\in \mathbb{Z}), then (a*b\in \mathbb{Z}). Step 2: (a+b+2ab) is formed using sums and products of integers, so it is always an integer. Step 3: To check closure, verify that the result stays in the same set.

Step 1: For a binary operation on natural numbers, the result must also be natural. Step 2: (1*2=1-2=-1), which is not a natural number. Step 3: One counterexample is enough to disprove closure.

Step 1: (a*b=a^2+b^2=b^2+a^2=b*a), so it is commutative. Step 2: ((1*1)*2=2*2=8), while (1*(1*2)=1*5=26), so it is not associative. Step 3: Test commutativity and associativity with separate checks.

Step 1: (a*e=a\Rightarrow a+e+1=a), so (e=-1). Step 2: (a*x=-1\Rightarrow a+x+1=-1\Rightarrow x=-a-2). Step 3: The identity may be negative, so solve the equation instead of guessing.

Step 1: Multiplication of real numbers is commutative, so (ab=ba). Step 2: Multiplication is also associative, so ((ab)c=a(bc)). Step 3: Recognizing ordinary multiplication helps simplify many binary-operation questions.