Question

think about the commutative property of real-number operations as it applies to addition and subtraction of functions. how do you think this property might extend to multiplication and division of functions?

Explanation:

1. Commutative Property for Real - Number Operations:

• For addition of real numbers, the commutative property states that for any two real numbers \(a\) and \(b\), \(a + b=b + a\). When we extend this to addition of functions, if we have two functions \(f(x)\) and \(g(x)\), the sum \((f + g)(x)=f(x)+g(x)\) and \((g + f)(x)=g(x)+f(x)\). Since addition of real numbers is commutative, \(f(x)+g(x)=g(x)+f(x)\), so \((f + g)(x)=(g + f)(x)\). Thus, addition of functions is commutative.
• For subtraction of real numbers, the commutative property does not hold. That is, in general, \(a - b

eq b - a\) (e.g., \(5 - 3 = 2\) and \(3 - 5=-2\)). When we consider subtraction of functions \((f - g)(x)=f(x)-g(x)\) and \((g - f)(x)=g(x)-f(x)\). Since \(f(x)-g(x)
eq g(x)-f(x)\) (in most cases, similar to real - number subtraction), subtraction of functions is not commutative.

1. Extending to Multiplication and Division of Functions:

• Multiplication of Real Numbers and Functions:
• For multiplication of real numbers, the commutative property holds: for any two real numbers \(a\) and \(b\), \(a\times b = b\times a\) (or \(ab = ba\)). When we extend this to multiplication of functions, if we have two functions \(f(x)\) and \(g(x)\), the product \((f\times g)(x)=f(x)\times g(x)\) and \((g\times f)(x)=g(x)\times f(x)\). Since multiplication of real numbers is commutative, \(f(x)\times g(x)=g(x)\times f(x)\), so \((f\times g)(x)=(g\times f)(x)\). Thus, multiplication of functions is commutative.
• Division of Real Numbers and Functions:
• For division of real numbers, the commutative property does not hold. In general, \(\frac{a}{b}

eq\frac{b}{a}\) (e.g., \(\frac{4}{2}=2\) and \(\frac{2}{4}=\frac{1}{2}\)). When we consider division of functions \((\frac{f}{g})(x)=\frac{f(x)}{g(x)}\) (where \(g(x)
eq0\)) and \((\frac{g}{f})(x)=\frac{g(x)}{f(x)}\) (where \(f(x)
eq0\)). Since \(\frac{f(x)}{g(x)}
eq\frac{g(x)}{f(x)}\) (in most cases, similar to real - number division), division of functions is not commutative.

Answer:

• Addition of Functions: Commutative (follows from commutative property of real - number addition: \((f + g)(x)=(g + f)(x)\) since \(f(x)+g(x)=g(x)+f(x)\) for all \(x\) in the domain).
• Subtraction of Functions: Not Commutative (follows from non - commutative property of real - number subtraction: \((f - g)(x)

eq(g - f)(x)\) in general as \(f(x)-g(x)
eq g(x)-f(x)\) for most functions \(f\) and \(g\)).

• Multiplication of Functions: Commutative (follows from commutative property of real - number multiplication: \((f\times g)(x)=(g\times f)(x)\) since \(f(x)\times g(x)=g(x)\times f(x)\) for all \(x\) in the domain).
• Division of Functions: Not Commutative (follows from non - commutative property of real - number division: \((\frac{f}{g})(x)

eq(\frac{g}{f})(x)\) in general as \(\frac{f(x)}{g(x)}
eq\frac{g(x)}{f(x)}\) for most functions \(f\) and \(g\) with non - zero denominators).