# Difference between revisions of "1983 IMO Problems/Problem 1"

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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>. | ||

− | Let <math>x=y=1</math> and we have <math>f(f(1))=f(1)</math>. Now, let <math>y=f(1)</math> and we have <math>f( | + | 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>. |

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

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

## Revision as of 09:56, 20 August 2008

Find all functions defined on the set of positive reals which take positive real values and satisfy: for all ; and as .

Let and we have . Now, let and we have since we have .

Plug in and we have . If is the only solution to then we have . We prove that this is the only function by showing that there does not exist any other :

Suppose there did exist such an . Then, letting in the functional equation yields . Then, letting yields . Notice that since , one of is greater than . Let equal the one that is greater than . Then, we find similarly (since ) that . Putting into the equation, yields . Repeating this process we find that for all natural . But, since , as , we have that which contradicts the fact that as .