Binary operations

Step 1: The sum of two natural numbers is again a natural number. Step 2: Thus, if (a,b \in N), then (a+b \in N). Step 3: To check closure, see whether the result stays in the same set.

Step 1: Closure requires (a-b \in N) for all (a,b \in N). Step 2: (2*5=2-5=-3), which is not a natural number. Step 3: One counterexample is enough to show that closure fails.

Step 1: The square of an integer is an integer. Step 2: The sum of two integers is also an integer. Hence (a^2+b^2) remains an integer. Step 3: In closure questions, identify the type of the result carefully.

Step 1: (a*b=\frac{a+b}{2}). Step 2: (b*a=\frac{b+a}{2}), and (a+b=b+a). Step 3: The commutativity of addition makes this operation easy to check.

Step 1: Put (a*e=a). Step 2: (\frac{a+e}{2}=a) gives (e=a). This is not one fixed element. Step 3: An identity element must be the same for every (a).

Step 1: Put (a*e=a). Step 2: (a+e-1=a) gives (e=1). Step 3: When a constant is subtracted, the identity shifts accordingly.

Step 1: The identity element is (1). Step 2: Put (7*x=1). Then (7+x-1=1), so (x=-5). Step 3: Do not directly use the usual additive inverse; use the given operation.

Step 1: First find (ab). Here (2\cdot2=4). Step 2: The remainder when (4) is divided by (3) is (1). Step 3: In remainder-based operations, the final answer is the remainder, not the original product.

Step 1: Multiplying by (1) keeps the number unchanged. Step 2: In (a*1), (a\cdot1=a), and its remainder modulo (3) is the same (a) for elements of (A). Step 3: For modulo multiplication, check (1) first as identity.

Step 1: Expanding ((a+1)(b+1)) gives (ab+a+b+1). Step 2: Subtracting (1) gives (ab+a+b). Step 3: Such rewritten forms help in associativity and inverse questions.

Step 1: ((1-a)(1-b)=1-a-b+ab). Step 2: (1-(1-a)(1-b)=a+b-ab). Step 3: Rewriting an operation often makes checking its properties easier.

Step 1: Substitute (a=2) and (b=3). Step 2: (2*3=2+3+2\cdot3=11). Step 3: Do not treat the symbol as ordinary multiplication; use the given definition.

Step 1: (2*4=2+3\cdot4=14). Step 2: (4*2=4+3\cdot2=10). Step 3: If changing the order changes the value, the operation is not commutative.

Step 1: (a+b) can range from (2) to (8). Step 2: (\min(a+b,4)) is always one of (2,3,4), all of which belong to (A). Step 3: For closure, match every possible result with the given set.

Step 1: For the (\max) operation, the identity element is (1). Step 2: We need (2*x=1), but (\max(2,x)) can never be (1). Step 3: Elements larger than the identity often do not have inverses in this type of operation.

Step 1: From (a*e=a), (a+e+4=a), so (e=-4). Step 2: For inverse of (10), put (10*x=-4). Then (10+x+4=-4), so (x=-18). Step 3: Keep the identity value correctly while solving.

Step 1: First find the identity (e). From (a+e+4=a), (e=-4). Step 2: Put (10*x=-4). Then (10+x+4=-4), so (x=-18). Step 3: Finding the identity first prevents mistakes in inverse questions.

Step 1: For identity (e), we need (a*e=a). Step 2: (a+e+4=a) gives (e=-4). Step 3: In an operation of the form (a+b+k), the identity is usually (-k).

Step 1: Solve the bracket first. (2*1=2+1+4=7). Step 2: Now (3*7=3+7+4=14). Step 3: In operation questions, always follow the brackets.

Step 1: First find (3*2=3+2+4=9). Step 2: Now (9*1=9+1+4=14). Step 3: Here both bracketings give the same value because this operation is associative.