Understanding Binary Operations and Their Properties

A binary operation is a rule that combines two elements of a set to produce another element of the same set. For this question, we are given a binary operation denoted by '*' defined on the set of real numbers (R) as \( \rm p * q = \frac{p+q}{2} \), for all p, q \( \rm \in \) R. We need to examine two important properties of binary operations: commutativity and associativity.

What is a Binary Operation?

• A binary operation takes two inputs from a set and produces a single output that also belongs to that set.
• For example, addition (+) is a binary operation on real numbers because adding two real numbers always gives another real number.

Commutativity Test for the Binary Operation

A binary operation '\( \rm * \)' on a set S is said to be commutative if for any two elements p and q in S, the order of the elements does not affect the result. Mathematically, this means:

\( \rm p * q = q * p \)

Let's apply this definition to our given operation \( \rm p * q = \frac{p+q}{2} \).

• First, calculate \( \rm p * q \):

  \( \rm p * q = \frac{p+q}{2} \)

• Next, calculate \( \rm q * p \):

  \( \rm q * p = \frac{q+p}{2} \)

Since addition of real numbers is commutative (i.e., \( \rm p+q = q+p \)), we can clearly see that:

\( \rm \frac{p+q}{2} = \frac{q+p}{2} \)

Therefore, \( \rm p * q = q * p \). This confirms that the given binary operation '\( \rm * \)' is commutative.

Associativity Test for the Binary Operation

A binary operation '\( \rm * \)' on a set S is said to be associative if for any three elements p, q, and r in S, the grouping of the elements does not affect the result. Mathematically, this means:

\( \rm (p * q) * r = p * (q * r) \)

Let's apply this definition to our given operation \( \rm p * q = \frac{p+q}{2} \).

• First, calculate \( \rm (p * q) * r \):

  We know \( \rm p * q = \frac{p+q}{2} \).

  So, \( \rm (p * q) * r = \left(\frac{p+q}{2}\right) * r \)

  Using the definition \( \rm A * B = \frac{A+B}{2} \), with \( \rm A = \frac{p+q}{2} \) and \( \rm B = r \):

  \( \rm (p * q) * r = \frac{\frac{p+q}{2} + r}{2} = \frac{p+q+2r}{4} \)

• Next, calculate \( \rm p * (q * r) \):

  We know \( \rm q * r = \frac{q+r}{2} \).

  So, \( \rm p * (q * r) = p * \left(\frac{q+r}{2}\right) \)

  Using the definition \( \rm A * B = \frac{A+B}{2} \), with \( \rm A = p \) and \( \rm B = \frac{q+r}{2} \):

  \( \rm p * (q * r) = \frac{p + \frac{q+r}{2}}{2} = \frac{2p+q+r}{4} \)

Now, let's compare the results for \( \rm (p * q) * r \) and \( \rm p * (q * r) \):

• \( \rm (p * q) * r = \frac{p+q+2r}{4} \)
• \( \rm p * (q * r) = \frac{2p+q+r}{4} \)

For the operation to be associative, these two expressions must be equal for all p, q, r \( \rm \in \) R. However, \( \rm \frac{p+q+2r}{4} \) is generally not equal to \( \rm \frac{2p+q+r}{4} \).

Let's take a counterexample to prove it is not associative. Let p = 1, q = 2, r = 3.

• For \( \rm (p * q) * r \):

  \( \rm (1 * 2) * 3 = \left(\frac{1+2}{2}\right) * 3 = \frac{3}{2} * 3 = \frac{\frac{3}{2} + 3}{2} = \frac{\frac{3+6}{2}}{2} = \frac{\frac{9}{2}}{2} = \frac{9}{4} \)

• For \( \rm p * (q * r) \):

  \( \rm 1 * (2 * 3) = 1 * \left(\frac{2+3}{2}\right) = 1 * \frac{5}{2} = \frac{1 + \frac{5}{2}}{2} = \frac{\frac{2+5}{2}}{2} = \frac{\frac{7}{2}}{2} = \frac{7}{4} \)

Since \( \rm \frac{9}{4} \neq \frac{7}{4} \), the operation is not associative.

Conclusion on Binary Operation Properties

Based on our analysis:

• The binary operation \( \rm p * q = \frac{p+q}{2} \) is commutative because \( \rm p * q = q * p \).
• The binary operation \( \rm p * q = \frac{p+q}{2} \) is not associative because \( \rm (p * q) * r \neq p * (q * r) \).

Therefore, the operation is commutative but not associative.