Binary operations

Step 1: Put (a\ast e=a). Step 2: From (\frac{ae}{2}=a), we get (e=2). The same value works for (e\ast a). Step 3: In fractional operations, do not assume (1) is always the identity.

Step 1: For identity (e), (\frac{a+e}{2}=a) must hold. Step 2: This gives (e=a), which changes with (a). No fixed (e) exists. Step 3: An identity element must be the same for the whole set.

Step 1: To be binary on (R), the result must lie in (R). Step 2: (a^2) and (b^2) are real, so their sum is also real. Step 3: Check closure with respect to the set; being positive is not a problem here.

Step 1: The result must always be an integer. Step 2: (1\ast2=\frac{1+2}{2}=\frac{3}{2}), which is not an integer. Step 3: An average operation is not always closed on integers.

Step 1: The sum of two even integers is again an even integer. Step 2: Therefore, for every (a,b), (a+b) remains in the same set. Step 3: In even-odd questions, check closure using parity rules.

Step 1: The sum of two odd integers is an even integer. Step 2: For example, (3+5=8), which is not odd. Step 3: While checking closure, the result must return to the same set.

Step 1: The product of two even integers is again an even integer. Step 2: Therefore, (ab) remains in the same set. Step 3: Use even-odd product rules to check closure quickly.

Step 1: The product of two odd integers is an odd integer. Step 2: For example, (3\cdot5=15), which is odd. Step 3: The odd-number product property proves closure here.

Step 1: In non-zero real numbers, (b\neq0), so division is defined. Step 2: (\frac{a}{b}) is also non-zero because (a\neq0). Step 3: In division questions, carefully check the role of zero.

Step 1: (R) includes (0). Step 2: If (b=0), then (\frac{a}{0}) is not defined. So the operation is not available for every pair. Step 3: A binary operation must be defined for every ordered pair.

Step 1: (a\ast b=a), while (b\ast a=b). Step 2: These are equal only when (a=b), not for every pair. Hence, the operation is generally not commutative. Step 3: For commutativity, compare both orders.

Step 1: The rule says the result of the operation is the second element. Step 2: In (x\ast y), the second element is (y), so the answer is (y). Step 3: In definition-based questions, apply the rule exactly as given.

Step 1: Put (a=7) and (b=3) in the rule. Step 2: (7\ast3=7+3-5=5). Step 3: Write each term separately to avoid sign mistakes.

Step 1: Write (a\ast e=a). Step 2: From (a+e-5=a), we get (e=5). Also, (e+a-5=a) gives the same. Step 3: In an operation like (a+b-c), the number (c) often becomes the identity.

Step 1: The identity of this operation is (5). Step 2: Let the inverse of (2) be (x). Then (2+x-5=5), so (x=8). Step 3: While finding an inverse, set the result equal to the identity, not always zero.

Step 1: Put (a\ast e=a). Step 2: (2ae=a) and (a\neq0), so (2e=1) and (e=\frac{1}{2}). Step 3: If multiplication has a coefficient, the identity can change.

Step 1: For inverse (x), (4\ast x=\frac{1}{2}) must hold. Step 2: (2\cdot4\cdot x=\frac{1}{2}), so (8x=\frac{1}{2}), giving (x=\frac{1}{16}). Step 3: Solve fractional equations slowly to avoid calculation errors.

Step 1: First find the identity: (2ae=a), so (e=\frac{1}{2}). Step 2: Let the inverse of (4) be (x). Then (2\cdot4\cdot x=\frac{1}{2}), giving (x=\frac{1}{16}). Step 3: The target of an inverse is always the identity.

Step 1: First solve inside the bracket: (3\ast4=3+4+1=8). Step 2: Now (2\ast8=2+8+1=11). Step 3: In composite operation values, solve the inner part first.

Step 1: First, (2\ast3=2+3+1=6). Step 2: Then, (6\ast4=6+4+1=11). Step 3: The value is the same here, but in exams always apply the rule step by step.