A Real Number Puzzle

 Posts: 35
 Joined: Thu Oct 26, 2006 5:32 pm
A Real Number Puzzle
f:R>R is a function satisfying the following properties:
(i) f(x) =  f(x)
(ii) f(x+1) = f(x) + 1
(iii) f(1/x) = f(x)/(x^2) for x not equal to 0.
Determine whether or not f(x) = x for all real values of x.
NOTE: R denotes the set of all real numbers
(i) f(x) =  f(x)
(ii) f(x+1) = f(x) + 1
(iii) f(1/x) = f(x)/(x^2) for x not equal to 0.
Determine whether or not f(x) = x for all real values of x.
NOTE: R denotes the set of all real numbers
a try
f(x+1) = f(x) + 1 implies, that f(x+2) = f((x+1)+1) = f(x+1) + 1 = f(x) + 2
and further f(x+y)=f(x)+y for all y.
Thus f(x) is a linear function with a slope of 1.
In f(x) = x + a , a must be zero because f(x)=  f(x).
So I see no other possibity than f(x) = x .
Edit:
There is an other posssibility: a periodic function with period 1 and antisymmetrie can be added to x ( i.e. g(x)= sin(2*pi*x) or h(x)= sum(all n) (a_n sin(n*2*pi*x) )and the first two conditions are not violated. But the third condition will not be fulfilled with this periodic function.
f(x+1) = f(x) + 1 implies, that f(x+2) = f((x+1)+1) = f(x+1) + 1 = f(x) + 2
and further f(x+y)=f(x)+y for all y.
Thus f(x) is a linear function with a slope of 1.
In f(x) = x + a , a must be zero because f(x)=  f(x).
So I see no other possibity than f(x) = x .
Edit:
There is an other posssibility: a periodic function with period 1 and antisymmetrie can be added to x ( i.e. g(x)= sin(2*pi*x) or h(x)= sum(all n) (a_n sin(n*2*pi*x) )and the first two conditions are not violated. But the third condition will not be fulfilled with this periodic function.
You have indeed shown that f(x)=x for all integers. I believe that by using the continued fraction representation you can show f(x)=x for all the rationals (as the continued fraction terminates.) If we were given that f is continuous that would be sufficient. But I can't see how to exclude that f(pi)=pi+delta. We then get constraints on many other numbers, as f(pi + any rational)=pi+that rational+delta, f(1/pi)=(pi+delta)/pi^2, etc.
 Rainy Monday
 Posts: 18
 Joined: Sun Sep 25, 2011 3:55 pm
Re: A Real Number Puzzle
Old topic, but that's an interesting puzzle.K Sengupta wrote:(i) f(x) =  f(x)
(ii) f(x+1) = f(x) + 1
(iii) f(1/x) = f(x)/(x^2) for x not equal to 0.
 According to I, f(0) =  f(0), so f(0) must be 0.
 Based on I and II, f(x) definitely equals x if x is an integer. Letting x be any real number and n an integer greater than x, we can also obtain:
f(xn) = f(x)  n =  f(nx),
n  f(x) = f(nx) = f(n),
n = f(x) + f(nx).
 Based on III, f(1/x) definitely equals 1/x if x is an integer. Combined with n = f(x) + f(nx), it leads to:
n = f[n  (1/x)] + f(1/x)
Let the x with the lowest absolute value satisfying the condition that f(x)  x = y =! 0. Then:
f(x+n)  (x+n) = y (with n being an integer),
f(mx)  (mx) = y (with m being an integer)
f[1/(x+n)] = f(x+n)/(x+n)^2
f[1/(x+n)] = x+y+n / (x+n)^2
By giving n sufficiently large values, f[1/(x+n)] whose absolute values are even lower than f(x) and do not equal 1/(x+n) can be obtained, which leaves us with a contradiction.
Re: A Real Number Puzzle
What if the lower bound of $\{ x : f(x)x \neq 0\}$ is 0 ?Rainy Monday wrote: Let the x with the lowest absolute value satisfying the condition that f(x)  x = y =! 0. Then:
Here is an inductive proof for $f_{\mathbb{Q}}$ :
let $x \in \mathbb{Q}$. Because of (I) and (II), we can suppose $0<x<1$.
 If $x \in \mathbb{N}$ (in other words if x is a rational with denominator 1) then $f(x)=x$.
 Let $q$ be a fixed integer and suppose that for any rationals $p/k$ with $p$ and $k$ coprimes and $k \leq q$, we have $f(p/k) = p/k$.
 Now take $x=p'/(q+1)$ with $p'$ coprime with $q+1$, since $x<1$, we have $p' \leq q$ and we have exactly what we want because $1/x$ can be written $n+y$ where $n \in \mathbb{N}$ and $y \in \mathbb{Q}$ has denominator $p'$ and so $f(1/x) = 1/x$ and then (III) gives $f(x) = x$.
You probably can prove it for any periodic continued fraction using the same technique as : $\varphi = 1+1/\varphi$
$f(\varphi) = f(1+1/\varphi) = 1+f(\varphi)/\varphi^2$
$f(\varphi)(11/\varphi^2) = 1$
$f(\varphi) = \varphi$
But what if $x$ is transcendental ? Or even just not periodic when expressed in continued fraction form ?
 Rainy Monday
 Posts: 18
 Joined: Sun Sep 25, 2011 3:55 pm
Re: A Real Number Puzzle
Why? It can't be 0, because f(0)=0.ggoyo wrote:What if the lower bound of $\{ x : f(x)x \neq 0\}$ is 0 ?
Re: A Real Number Puzzle
So the lower bound of $A = \{1/n$ $$ $n \in \mathbb{N}\}$ is not 0 because 0 is not in it ?
 Rainy Monday
 Posts: 18
 Joined: Sun Sep 25, 2011 3:55 pm
Re: A Real Number Puzzle
For any existing lower bound, an even lower one can be found and it should converge to 0, which gives a contradiction. Don't see what's wrong with this kind of logic.
Re: A Real Number Puzzle
What's wrong is that you mistake lower bound for minimum. A bounded subset of $\mathbb{R}$ can have no minimum.
Look, I'm going to apply your reasoning to my example $A$ :
Let x be the smallest element of $A$.
Then, exists $n$ such that $x=1/n$.
But wait, 1/(n+1) is lower than $x$.
That contradicts the fact that $x$ is the smallest. Then we must have $A = \emptyset$. So writing 1/2 is a mathematical mistake.
Still does not see what's wrong with that kind of logic ?
Look, I'm going to apply your reasoning to my example $A$ :
Let x be the smallest element of $A$.
Then, exists $n$ such that $x=1/n$.
But wait, 1/(n+1) is lower than $x$.
That contradicts the fact that $x$ is the smallest. Then we must have $A = \emptyset$. So writing 1/2 is a mathematical mistake.
Still does not see what's wrong with that kind of logic ?
Re: A Real Number Puzzle
We don't need any continuity, here. Substituting $x\rightarrowx$ in (ii), we get $f(1x)=1f(x)$, and from that, using (iii), $f\left(1/(1x)\right)=(1f(x))/(1x)^2$. The substitution $x\rightarrow 1/(1x)$ gives $f((x1)/x)=(x^22x+f(x))/x^2$, and once more setting $x\rightarrow 1/(1x)$ gives $f(x)=2xf(x)$, i.e. $f(x)=x$.