Difference between revisions of "1974 AHSME Problems/Problem 6"
m (added category) |
|||
| Line 20: | Line 20: | ||
{{AHSME box|year=1974|num-b=5|num-a=7}} | {{AHSME box|year=1974|num-b=5|num-a=7}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:42, 5 July 2013
Problem
For positive real numbers 
and 
define 
' then
Solution
First, let's check for commutivity. We have ![\[y*x=\frac{y\cdot x}{y+x}=\frac{x\cdot y}{x+y}=x*y\]](http://latex.artofproblemsolving.com/1/4/4/144ffa2f41a12855644c53fbb76cd0cbb710824b.png)
, so 
is commutative.
Now we check for associativity. We have ![\[(x*y)*z=\left(\frac{x\cdot y}{x+y}\right)*z=\frac{\frac{x\cdot y}{x+y}\cdot z}{\frac{x\cdot y}{x+y}+z}=\frac{x\cdot y\cdot z}{x\cdot y+z(x+y)}=\frac{x\cdot y\cdot z}{x\cdot y+y\cdot z+x\cdot z}.\]](http://latex.artofproblemsolving.com/8/0/5/8058038396e45c0c82addc82b1b6292de87a03b9.png)
Also, ![\[x*(y*z)=x*\left(\frac{y\cdot z}{y+z}\right)=\frac{x\cdot\frac{y\cdot z}{y+z}}{x+\frac{y\cdot z}{y+z}}=\frac{x\cdot y\cdot z}{x(y+z)+y\cdot z}=\frac{x\cdot y\cdot z}{x\cdot y+y\cdot z+x\cdot z}=(x*y)*z ,\]](http://latex.artofproblemsolving.com/f/8/b/f8be0e79766a4e840b681f9a676912da4c868d1e.png)
and so is also associative. 
See Also
| 1974 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
