= Horizontal chord of reciprocal-integer length
{title2=$f(alpha+1/n)=f(alpha)$}
If $f:[0,1]\to\mathbb R$ is continuous and $f(0)=f(1)$, then for every positive integer $n$ there is $\alpha\in[0,1-1/n]$ with $f(\alpha+1/n)=f(\alpha)$. For $n>1$, the continuous difference $h_n(x)=f(x+1/n)-f(x)$ satisfies $\sum_{j=0}^{n-1}h_n(j/n)=0$. Either one sampled difference is zero or two have opposite signs; the <intermediate value theorem> then supplies a zero between them. For $n=1$, the endpoint chord suffices.
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