Binary operations

Step 1: The rule (a*b=a) says the result is always the first element. Step 2: In (2*3), the first element is (2). Step 3: In exams, read the rule first and then substitute the given elements.

Step 1: According to the rule, (a*b) is equal to the second element. Step 2: In (4*7), the second element is (7). Step 3: Do not add or multiply unless the rule says so.

Step 1: The operation is defined by (a*b=a+b). Step 2: So (5*6=5+6=11). Step 3: The symbol may be new, but the rule may simply mean addition.

Step 1: The rule (a*b=ab) represents multiplication. Step 2: (3*8=3\times 8=24). Step 3: Do not read (ab) as a two-digit number, it means product.

Step 1: The rule tells us to subtract the second element from the first. Step 2: (9*4=9-4=5). Step 3: Order is very important in subtraction-based operations.

Step 1: Put (a=3) and (b=5) in the rule. Step 2: (3*5=3+2(5)=13). Step 3: Complete multiplication before addition.

Step 1: Substitute (a=4) and (b=3). Step 2: (4*3=2(4)+3=11). Step 3: Correct substitution is the main skill in simple operation questions.

Step 1: The rule requires the square of the first element. Step 2: (3*2=3^2+2=9+2=11). Step 3: Notice that the square applies only to the first element.

Step 1: The rule uses the square of the second element. Step 2: (4*3=4+3^2=4+9=13). Step 3: Check carefully which element has the power.

Step 1: According to the rule, add the numbers and subtract (1). Step 2: (6*4=6+4-1=9). Step 3: Do not forget the constant term in the operation rule.

Step 1: Put (a=0) and (b=5). Step 2: (0*5=0+5+1=6). Step 3: Even when zero appears, use the complete rule.

Step 1: For closure, (a*b\in A) must hold for every (a,b\in A). Step 2: (2*2=2+2=4), but (4\notin A). Step 3: One counterexample is enough to disprove closure.

Step 1: If (a\in A), then (a*b=a) is also in (A). Step 2: So the result never goes outside (A). Step 3: For closure, check whether outputs remain inside the set.

Step 1: The sum of two natural numbers is again a natural number. Step 2: Subtraction and division do not always give natural numbers. Step 3: Closure must hold for every possible pair.

Step 1: Integers are closed under addition, subtraction, and multiplication. Step 2: Division does not always give an integer, for example (\frac{1}{2}\notin Z). Step 3: To test closure, find an output outside the set.

Step 1: For commutativity, (a*b=b*a) must hold. Step 2: Here (a*b=a+b) and (b*a=b+a), which are equal. Step 3: Addition-based operations usually keep the same value when order changes.

Step 1: For commutativity, (a-b=b-a) must always hold. Step 2: (5-2=3) but (2-5=-3), so they are not equal. Step 3: Subtraction usually changes when order changes.

Step 1: In commutativity, we interchange the two elements. Step 2: In multiplication, (ab=ba), so (a*b=b*a). Step 3: Simple rules based on multiplication are generally commutative.

Step 1: (1*2=1+2(2)=5). Step 2: (2*1=2+2(1)=4), so they are not equal. Step 3: To test commutativity, calculate in both orders.

Step 1: (a*b=a+b+ab). Step 2: (b*a=b+a+ba), which equals (a+b+ab). Step 3: Addition and multiplication do not change when the order is reversed.

Step 1: For identity element (e), (a*e=a) and (e*a=a). Step 2: (a+0=a) and (0+a=a). Step 3: For addition-like operations, check (0) first.

Step 1: An identity element (e) must satisfy (ae=a) and (ea=a). Step 2: (a\cdot 1=a) and (1\cdot a=a). Step 3: For multiplication-like operations, (1) is the first candidate.

Step 1: We need (a*e=a). Step 2: From (a+e+1=a), we get (e=-1). Step 3: To find identity, cancel (a) from both sides and solve for (e).

Step 1: For (a*e=a), we need (a+e-1=a). Step 2: This gives (e=1), and (e*a=e+a-1=a) also works. Step 3: It is a good habit to check identity from both sides.

Step 1: (a*0=a+0+a\cdot 0=a). Step 2: (0*a=0+a+0\cdot a=a), so (0) works from both sides. Step 3: The identity element leaves every element unchanged.