A Real Number Puzzle

Arithmetic, algebra, number theory, sequence and series, analysis, ...
Post Reply
K Sengupta
Posts: 35
Joined: Thu Oct 26, 2006 4:32 pm

A Real Number Puzzle

Post by K Sengupta » Tue Jan 30, 2007 9:33 am

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

ThomasH
Posts: 117
Joined: Sun Mar 26, 2006 7:41 am
Location: Berlin, Germany

Post by ThomasH » Tue Jan 30, 2007 4:43 pm

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.

rmillika
Posts: 39
Joined: Thu Apr 05, 2007 3:37 pm

Post by rmillika » Thu Apr 05, 2007 4:49 pm

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.

User avatar
Rainy Monday
Posts: 18
Joined: Sun Sep 25, 2011 2:55 pm

Re: A Real Number Puzzle

Post by Rainy Monday » Mon Jan 26, 2015 12:30 pm

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.
Old topic, but that's an interesting puzzle.

- 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(x-n) = f(x) - n = - f(n-x),
n - f(x) = f(n-x) = f(n),
n = f(x) + f(n-x).

- Based on III, f(1/x) definitely equals 1/x if x is an integer. Combined with n = f(x) + f(n-x), 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(m-x) - (m-x) = -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.

User avatar
ggoyo
Posts: 25
Joined: Sun Jun 22, 2014 9:45 am
Location: Paris
Contact:

Re: A Real Number Puzzle

Post by ggoyo » Fri Feb 13, 2015 6:35 pm

Rainy Monday wrote: Let the x with the lowest absolute value satisfying the condition that f(x) - x = y =! 0. Then:
What if the lower bound of $\{ |x| : f(x)-x \neq 0\}$ is 0 ?

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 : $\varphi = 1+1/\varphi$

$f(\varphi) = f(1+1/\varphi) = 1+f(\varphi)/\varphi^2$

$f(\varphi)(1-1/\varphi^2) = 1$

$f(\varphi) = \varphi$


But what if $x$ is transcendental ? Or even just not periodic when expressed in continued fraction form ?
Image

User avatar
Rainy Monday
Posts: 18
Joined: Sun Sep 25, 2011 2:55 pm

Re: A Real Number Puzzle

Post by Rainy Monday » Sat Feb 14, 2015 3:37 pm

ggoyo wrote:What if the lower bound of $\{ |x| : f(x)-x \neq 0\}$ is 0 ?
Why? It can't be 0, because f(0)=0.

User avatar
ggoyo
Posts: 25
Joined: Sun Jun 22, 2014 9:45 am
Location: Paris
Contact:

Re: A Real Number Puzzle

Post by ggoyo » Mon Feb 16, 2015 11:06 am

So the lower bound of $A = \{1/n$ $|$ $n \in \mathbb{N}\}$ is not 0 because 0 is not in it ?
Image

User avatar
Rainy Monday
Posts: 18
Joined: Sun Sep 25, 2011 2:55 pm

Re: A Real Number Puzzle

Post by Rainy Monday » Tue Feb 17, 2015 9:48 am

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.

User avatar
ggoyo
Posts: 25
Joined: Sun Jun 22, 2014 9:45 am
Location: Paris
Contact:

Re: A Real Number Puzzle

Post by ggoyo » Tue Feb 17, 2015 1:08 pm

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 ?
Image

Pengolodh
Posts: 6
Joined: Wed Nov 21, 2012 9:03 am

Re: A Real Number Puzzle

Post by Pengolodh » Fri Apr 17, 2015 6:54 pm

We don't need any continuity, here. Substituting $x\rightarrow-x$ in (ii), we get $f(1-x)=1-f(x)$, and from that, using (iii), $f\left(1/(1-x)\right)=(1-f(x))/(1-x)^2$. The substitution $x\rightarrow 1/(1-x)$ gives $f((x-1)/x)=(x^2-2x+f(x))/x^2$, and once more setting $x\rightarrow 1/(1-x)$ gives $f(x)=2x-f(x)$, i.e. $f(x)=x$.

Post Reply