0
$\begingroup$

It's a quite strange task.

We have a 5 long worm on [0,5] interval. After some "twisting", now it is on the interval [0,1] only. Prove that the worm has a point, which didn't change position during this process.

I'm not sure if you can understand, but I can't really prove this funny, but interesting task! (Maybe it has to do something with the continuum amount of numbers on this intervates)? :)

$\endgroup$
  • 5
    $\begingroup$ This is what's known as a "fixed-point theorem." This might be useful:en.wikipedia.org/wiki/Banach_fixed-point_theorem $\endgroup$ – Mnifldz Mar 2 '15 at 17:31
  • 1
    $\begingroup$ You might want to define "wimple." $\endgroup$ – Newb Mar 2 '15 at 17:32
  • $\begingroup$ The point 0 did not change its position in this case. Hence, proved? $\endgroup$ – shauryachats Mar 2 '15 at 17:32
  • $\begingroup$ If the point 0 is the case, isn't the task too easy? :D $\endgroup$ – Atvin Mar 2 '15 at 17:36
  • $\begingroup$ The transformation function, or method of twisting, was left implicit, so technically speaking any point on [0,1] could be a fixed point. $\endgroup$ – Zach466920 Mar 2 '15 at 19:27
4
$\begingroup$

I shall try to tackle this interesting question using a graph, though I suppose this proof is not quite formal.

First, I shall only concern the "tail" of the worm that is originally located at $[0, 1]$, because this is the part where a point exists without changing its position (if any).

Next, I create a graph as follows:

enter image description here
The $x$-axis denotes the original position of the worm, $y$-axis denotes the new (twisted) position. Hence, the tail of the worm will be shown as a red dot, while the part of the worm $1$ unit from the tail is marked as the purple dot. Note that the red dot must have a coordinate of $(0, p)$, while the purple dot must have a coordinate of $(1, q)$, where $0 \leq p \leq 1$ and $0 \leq q \leq 1$.

Now, if $p = 0$ or $q = 1$, it means the red dot or the purple dot will lie on the blue line. This blue line is actually $y = x$. In other words, if the red dot or purple dot lies on the blue line, at that point, the original position equals the new position, so there exists a point that the worm has not changed its position.

If $p \neq 0$ and $q \neq 1$, then, just like the figure above, the red dot must be above the blue line (area A), while the purple dot must be below the blue line (area B). The body of the worm is continuous, meaning a continuous curve has to be drawn to connect the red dot and the purple dot. So, it must intersect the blue line, meaning there must be a point that the worm has not changed its position.

$\endgroup$
1
$\begingroup$

(Remark: The following is a streamlined version of LaBird's answer. Give the credit to LaBird above.)

First, I shall only deal with the part of the worm that originally takes up the interval $[0,1]$, because only for this part a fixed point can occur.

Consider an atom originally at the point $x\in[0,1]$, and denote its final position by $f(x)$. In this way we obtain a continuous function $f:\>[0,1]\to[0,1]$ which we can plot in the $(x,y)$-plane. The figure shows in red the point $(0,p)$ and in purple the point $(1,q)$ of the resulting graph.

If it so happens that $p=0$, i.e., $f(0)=0$, then $0$ is a fixed point, and if $q=1$ then in the same way $1$ is a fixed point. If $p>0$ and $q<1$ then the auxiliary function $$g(x):=f(x)-x$$ is $>0$ at $x=0$ and $<0$ at $x=1$. Therefore, by the intermediate value theorem, there is a point $\xi\in\ ]0,1[\ $ with $g(\xi)=0$, or $f(\xi)=\xi$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.