- Commutativity: The operation is commutative because \(x \circ y = x^2 + y^2 = y^2 + x^2 = y \circ x\).
- Associativity: The operation is not associative, as shown by the counterexample \((1 \circ 2) \circ 3 = 34
eq 170 = 1 \circ (2 \circ 3)\).
- Conclusion: The operation is commutative and not associative. \(\boxed{\text{Commutative: Yes, Associative: No}}\).
Explanation
- Checking Commutativity
To determine if the operation \(\circ\) defined by \(x \circ y = x^2 + y^2\) on the set of positive real numbers is commutative, we compare \(x \circ y\) with \(y \circ x\). By the definition of the operation, \(x \circ y = x^2 + y^2\). Similarly, \(y \circ x = y^2 + x^2\). Since the addition of real numbers is commutative, \(x^2 + y^2 = y^2 + x^2\), which implies \(x \circ y = y \circ x\) for all positive real numbers \(x\) and \(y\). Thus, the operation is commutative.
- Checking Associativity
To determine if the operation is associative, we must check if \((x \circ y) \circ z = x \circ (y \circ z)\) for all \(x, y, z \in A\).
First, calculate the left-hand side (LHS):
\((x \circ y) \circ z = (x^2 + y^2) \circ z = (x^2 + y^2)^2 + z^2 = x^4 + 2x^2y^2 + y^4 + z^2\).
Next, calculate the right-hand side (RHS):
\(x \circ (y \circ z) = x \circ (y^2 + z^2) = x^2 + (y^2 + z^2)^2 = x^2 + y^4 + 2y^2z^2 + z^4\).
To check if these are identical for all positive real numbers, we can test with specific values. Let \(x=1\), \(y=2\), and \(z=3\).
LHS: \((1 \circ 2) \circ 3 = (1^2 + 2^2) \circ 3 = 5 \circ 3 = 5^2 + 3^2 = 25 + 9 = 34\).
RHS: \(1 \circ (2 \circ 3) = 1 \circ (2^2 + 3^2) = 1 \circ 13 = 1^2 + 13^2 = 1 + 169 = 170\).
Since \(34
eq 170\), the operation is not associative.
- Conclusion
We have determined that the operation \(\circ\) satisfies the commutative property because \(x \circ y = y \circ x\). We have also shown that the operation does not satisfy the associative property because we found a counterexample where \((1 \circ 2) \circ 3
eq 1 \circ (2 \circ 3)\). Therefore, the operation is commutative but not associative. Final conclusion: Commutative: Yes; Associative: No.
Examples
Consider a system where you combine two inputs into a single output by squaring each and adding them together, such as when calculating the combined power contribution in two parallel electrical circuits, where each circuit contributes the square of its current. If the system is commutative, the order in which you consider the circuits doesn't change the total power. However, if it's not associative, the order in which you combine multiple circuits matters significantly for the final outcome. This shows how understanding algebraic properties can help in modeling physical systems or decision-making processes.