Proving that there exists a horizontal chord with length $1/n$ for a continuous function $f: [0,1] \to \mathbb R$ Given a continuous function $f: [0,1] \to \mathbb R$ and that the chord which connects $A(0, f(0)), B(1, f(1))$ is horizontal then prove that there exists a horizontal chord $CD$ to the graph $C_f$  with length $(CD) = 1 / n$ where $n$ is a nonzero natural number.
I have proven the statement but (by methods of  reductio ad absurdum and signum) but I was wondering if there is a proof by induction, namely proving that there exists some $x_n \in [0,1]$ such that $f(x_n) = f(x_n + 1/n)$. My logic is the following:


*

*Proving the statement for $n=2$ (pretty easy, just apply Bolzano's theorem to f(x) - f(x + 1/2) ). Note that $f(0)=f(1)$

*For $n = k$ we let $h_k : A_k \to \mathbb R$ be a continuous function where $$A_k = \bigg [0, \frac  {k-1} k \bigg ] \\ h_k(x) = f(x) - f \bigg (x + \frac 1 k \bigg )$$ which satisfies our hypothesis (the existence of at least one $x_k: h_k(x_k) = 0$)

*Proving the statement for $n = k + 1$. I was thinking of applying Bolzano's theorem to some interval but I have not found such an interval. Is there one? 


Intervals with length $\frac{1}{k}$ and $\frac{1}{k+1}$ overlap pretty bad, so I do not think it is a good idea to use induction on $k$. Instead, just take $h_n:[0,1-1/n]\to\mathbb{R}$:
$$ h_n(x) = f\left(x+\frac{1}{n}\right)-f(x) $$
that is obviously continuous. If $h_n$ is zero somewhere in $I=[0,1-1/n]$ we have nothing to prove, and the same holds if $h_n$ change its sign over $I$, by continuity. However, $h_n$ cannot have always the same sign, because in such a case:
$$ 0 = f(1)-f(0) = \sum_{k=0}^{n-1}h_n\left(\frac{k}{n}\right)\neq 0$$
we get a contradiction.
