Another interesting property of unit fractions (numbers of the form $\dfrac{1}{n}$ for positive integer $n$).
Part A)
Suppose $f:[0,1] \to \mathbb{R}$ is any continuous function such that $f(0) = f(1)$.
Show that for every natural number $n \gt 0$, there is some $c \in [0,1]$ ($c$ dependent on $n$ and $f$) such that $f(c) = f\left(c + \dfrac{1}{n}\right)$
Basically the graph of $f$ has a chord of length $\dfrac{1}{n}$.
Part B)
Suppose $r \in (0,1)$ is a real number such that $r$ is not a unit fraction.
Show that there is some continuous function $f_r$ (dependent on $r$) such that
$f_r:[0,1] \to \mathbb{R}$, $f_r(0) = f_r(1)$ and for any $c \in [0,1-r]$, $f_r(c) \ne f_r(c+r)$.
i.e the graph of $f_r$ has no chord of length $r$.
[Solution]
Part A)
Suppose $f:[0,1] \to \mathbb{R}$ is any continuous function such that $f(0) = f(1)$.
Show that for every natural number $n \gt 0$, there is some $c \in [0,1]$ ($c$ dependent on $n$ and $f$) such that $f(c) = f\left(c + \dfrac{1}{n}\right)$
Basically the graph of $f$ has a chord of length $\dfrac{1}{n}$.
Part B)
Suppose $r \in (0,1)$ is a real number such that $r$ is not a unit fraction.
Show that there is some continuous function $f_r$ (dependent on $r$) such that
$f_r:[0,1] \to \mathbb{R}$, $f_r(0) = f_r(1)$ and for any $c \in [0,1-r]$, $f_r(c) \ne f_r(c+r)$.
i.e the graph of $f_r$ has no chord of length $r$.
[Solution]
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