Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1983 IMO Problems/Problem 1"
Line 1: Line 1:
 +
==Problem==
 
Find all functions <math>f</math> defined on the set of positive reals which take positive real values and satisfy:  <math>f(xf(y))=yf(x)</math> for all <math>x,y</math>; and <math>f(x)\to0</math> as <math>x\to \infty</math>.  
 
Find all functions <math>f</math> defined on the set of positive reals which take positive real values and satisfy:  <math>f(xf(y))=yf(x)</math> for all <math>x,y</math>; and <math>f(x)\to0</math> as <math>x\to \infty</math>.  
  
 +
==Solution==
 
Let <math>x=y=1</math> and we have <math>f(f(1))=f(1)</math>.  Now, let <math>x=1,y=f(1)</math> and we have <math>f(f(f(1)))=f(1)f(1)\Rightarrow f(1)=[f(1)]^2</math> since <math>f(1)>0</math> we have <math>f(1)=1</math>.   
 
Let <math>x=y=1</math> and we have <math>f(f(1))=f(1)</math>.  Now, let <math>x=1,y=f(1)</math> and we have <math>f(f(f(1)))=f(1)f(1)\Rightarrow f(1)=[f(1)]^2</math> since <math>f(1)>0</math> we have <math>f(1)=1</math>.   
  

Revision as of 21:36, 31 January 2016

Problem

Find all functions $f$ defined on the set of positive reals which take positive real values and satisfy: $f(xf(y))=yf(x)$ for all $x,y$; and $f(x)\to0$ as $x\to \infty$.

Solution

Let $x=y=1$ and we have $f(f(1))=f(1)$. Now, let $x=1,y=f(1)$ and we have $f(f(f(1)))=f(1)f(1)\Rightarrow f(1)=[f(1)]^2$ since $f(1)>0$ we have $f(1)=1$.

Plug in $y=x$ and we have $f(xf(x))=xf(x)$. If $a=1$ is the only solution to $f(a)=a$ then we have $xf(x)=1\Rightarrow f(x)=\frac{1}{x}$. We prove that this is the only function by showing that there does not exist any other $a$:

Suppose there did exist such an $a\ne1$. Then, letting $y=a$ in the functional equation yields $f(xa)=af(x)$. Then, letting $x=\frac{1}{a}$ yields $f(\frac{1}{a})=\frac{1}{a}$. Notice that since $a\ne1$, one of $a,\frac{1}{a}$ is greater than $1$. Let $b$ equal the one that is greater than $1$. Then, we find similarly (since $f(b)=b$) that $f(xb)=bf(x)$. Putting $x=b$ into the equation, yields $f(b^2)=b^2$. Repeating this process we find that $f(b^{2^k})=b^{2^k}$ for all natural $k$. But, since $b>1$, as $k\to \infty$, we have that $b^{2^k}\to\infty$ which contradicts the fact that $f(x)\to0$ as $x\to \infty$.

1983 IMO (Problems) • Resources
Preceded by
First question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_IMO_Problems/Problem_1&oldid=75004"