Difference between revisions of "1974 AHSME Problems/Problem 6"

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==See Also==
 
==See Also==
 
{{AHSME box|year=1974|num-b=5|num-a=7}}
 
{{AHSME box|year=1974|num-b=5|num-a=7}}
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[[Category:Introductory Algebra Problems]]
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Latest revision as of 11:42, 5 July 2013

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. $\boxed{\text{D}}$

See Also

1974 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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