Binary operations

Step 1: A binary operation takes two elements. Step 2: If both inputs are from (A) and the result is also in (A), it is an operation on (A). Step 3: In exams, check closure first.

Step 1: Closure means the result does not leave the set after the operation. Step 2: Here (a*b) again lies in (A), so closure holds. Step 3: Closure is the first sign to check in binary operation questions.

Step 1: Take any two numbers from (\mathbb{N}). Step 2: Their sum remains in (\mathbb{N}). Step 3: Since closure holds, usual addition is a binary operation on (\mathbb{N}).

Step 1: For closure, every pair must give a result inside the set. Step 2: (2,5\in\mathbb{N}), but (2-5=-3), which is not in (\mathbb{N}). Step 3: One counterexample is enough to reject a binary operation.

Step 1: The product of two integers is always an integer. Step 2: So the result does not leave (\mathbb{Z}). Step 3: On integers, both addition and multiplication are closed.

Step 1: For a binary operation, every valid pair must give a result in the same set. Step 2: (1) and (2) are integers, but (\frac{1}{2}) is not an integer. Step 3: Division often fails closure.

Step 1: For commutativity, check (a*b=b*a). Step 2: Since (a+b=b+a), we get (a*b=b*a). Step 3: Addition-based operations often remain unchanged when order is changed.

Step 1: (2*1=2-1=1). Step 2: (1*2=1-2=-1), so they are not equal. Step 3: A small numerical example is useful to test commutativity.

Step 1: For identity (e), we need (a*e=a). Step 2: From (a+e+1=a), we get (e=-1). Step 3: To find identity, place (e) and solve the equation.

Step 1: Put (a*e=a). Step 2: From (a+e+ae=a), (e(1+a)=0), and (e=0) works for every (a). Step 3: An identity must work for all elements.

Step 1: For identity (e), we need (a*e=a). Step 2: (ae+1=a) cannot give one fixed (e) for all (a), and for (a=0) it gives (1=0), impossible. Step 3: An identity must be the same for all elements.

Step 1: We need (\max(a,e)=a) for every (a\ge0). Step 2: Taking (e=0) gives (\max(a,0)=a). Step 3: In a maximum operation, the smallest element is often the identity.

Step 1: We need (\min(a,e)=a) for every (a\ge0). Step 2: This requires (e) to be greater than or equal to all (a), but ([0,\infty)) has no greatest element. Step 3: For a minimum operation, the greatest element is the identity if it exists.

Step 1: Put (a=3) and (b=4) in the operation. Step 2: (3*4=3+2(4)=11). Step 3: Do not treat the symbol as usual multiplication; use the given rule.

Step 1: Substitute (a=2) and (b=3) in the rule. Step 2: (2*3=2+3+2\cdot3=11). Step 3: First read the rule, then calculate normally.

Step 1: For associativity, check ((a*b)*c=a*(b*c)). Step 2: In addition, ((a+b)+c=a+(b+c)). Step 3: Usual addition is associative.

Step 1: ((5*2)*1=(5-2)-1=2). Step 2: (5*(2*1)=5-(2-1)=4). Step 3: They are not equal, so associativity fails.

Step 1: (a*b=a+b+ab=(a+1)(b+1)-1). Step 2: This form is linked to multiplication, and multiplication is associative. Step 3: Rewriting the rule helps identify the property.

Step 1: (1*2=1+4=5). Step 2: (2*1=2+2=4), so the two values are different. Step 3: To test commutativity, interchange the positions and compare.

Step 1: For multiplication identity (e), we need (a\cdot e=a). Step 2: Taking (e=1), (a\cdot1=a) for every real (a). Step 3: Remember that the identity for usual multiplication is (1).

Step 1: For addition identity (e), we need (a+e=a). Step 2: Taking (e=0), (a+0=a) for every (a). Step 3: The identity for addition is (0).

Step 1: The identity under addition is (0). Step 2: The inverse of (7) is the number that gives (0) when added, so (7+(-7)=0). Step 3: Under addition, the inverse of a number is its negative.

Step 1: The identity under multiplication is (1). Step 2: The inverse of (5) is the number that gives (1) when multiplied. Step 3: Since (5\cdot\frac{1}{5}=1), the inverse is (\frac{1}{5}).

Step 1: We need (\min(a,e)=a) for every (a). Step 2: Taking (e=3), (\min(a,3)=a) because (3) is the greatest element of the set. Step 3: For a minimum operation, the greatest element is the identity.

Step 1: We need (\max(a,e)=a) for every (a). Step 2: Taking (e=1), (\max(a,1)=a) because (1) is the smallest element. Step 3: For a maximum operation, the smallest element becomes the identity.