# 1974 AHSME Problems/Problem 6

## Problem

For positive real numbers $x$ and $y$ define $x*y=\frac{x\cdot y}{x+y}$' then $\mathrm{(A)\ } \text{*" is commutative but not associative} \qquad$ $\mathrm{(B) \ }\text{*" is associative but not commutative} \qquad$ $\mathrm{(C) \ } \text{*" is neither commutative nor associative} \qquad$ $\mathrm{(D) \ } \text{*" is commutative and associative} \qquad$ $\mathrm{(E) \ }\text{none of these} \qquad$

## Solution

First, let's check for commutivity. We have $$y*x=\frac{y\cdot x}{y+x}=\frac{x\cdot y}{x+y}=x*y$$, so $*$ is commutative.

Now we check for associativity. We have $$(x*y)*z=\left(\frac{x\cdot y}{x+y}\right)*z=\frac{\frac{x\cdot y}{x+y}\cdot z}{\frac{x\cdot y}{x+y}+z}=\frac{x\cdot y\cdot z}{x\cdot y+z(x+y)}=\frac{x\cdot y\cdot z}{x\cdot y+y\cdot z+x\cdot z}.$$ Also, $$x*(y*z)=x*\left(\frac{y\cdot z}{y+z}\right)=\frac{x\cdot\frac{y\cdot z}{y+z}}{x+\frac{y\cdot z}{y+z}}=\frac{x\cdot y\cdot z}{x(y+z)+y\cdot z}=\frac{x\cdot y\cdot z}{x\cdot y+y\cdot z+x\cdot z}=(x*y)*z ,$$ and so $*$ is also associative. $\boxed{\text{D}}$

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