1973 AHSME Problems/Problem 5

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Problem

Of the following five statements, I to V, about the binary operation of averaging (arithmetic mean),

$\text{I. Averaging is associative }$

$\text{II. Averaging is commutative }$

$\text{III. Averaging distributes over addition }$

$\text{IV. Addition distributes over averaging }$

$\text{V. Averaging has an identity element }$

those which are always true are

$\textbf{(A)}\ \text{All}\qquad\textbf{(B)}\ \text{I and II only}\qquad\textbf{(C)}\ \text{II and III only}$

$\textbf{(D)}\ \text{II and IV only}\qquad\textbf{(E)}\ \text{II and V only}$

Solution

We can consider each statement independently and see which ones are true. The average of two numbers $x$ and $y$ is $avg(x, y) = \frac{x + y}{2}$.

Statement I $$avg(avg(x, y), z) = avg(x, avg(y, z))$$ $$\frac{\frac{x + y}{2} + z}{2} = \frac{x + \frac{y + z}{2}}{2}$$ $$\frac{x + y + 2z}{4} = \frac{2x + y + z}{4}$$ The above statement is not true for all $(x, y, z)$, so Statement I is false.

Statement II $$avg(x, y) = avg(y, x)$$ $$\frac{x + y}{2} = \frac{y + x}{2}$$ Since addition is commutative, the above is true for all $(x,y)$, so Statement II is true.

Statement III $$avg(x, y + z) = avg(x, y) + avg(x, z)$$ $$\frac{x + y + z}{2} = \frac{x + y}{2} + \frac{x + z}{2}$$ $$\frac{x + y + z}{2} = \frac{2x + y + z}{2}$$ The above statement is not true for all $(x, y, z)$, so Statement III is false.

Statement IV $$x + avg(y, z) = avg(x + y, x + z)$$ $$x + \frac{y + z}{2} = \frac{2x + y + z}{2}$$ $$\frac{2x + y + z}{2} = \frac{2x + y + z}{2}$$ The left side and the right side are equivalent expressions, so Statement IV is true.

Statement V Let $i$ be the identity element. If such $i$ exists, then the following is true: $$avg(x, i) = x$$ $$\frac{x + i}{2} = x$$ $$x + i = 2x$$ $$i = x$$ However, since $x$ is a variable, but $i$ must be a constant, fixed number, there is no identity element for the binary operation of averaging. Therefore, Statement V is false.

Statements II and IV are true, so the answer is $\boxed{\textbf{(D)} \text{II and IV only}}$.