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To prove the required zero-existence statement directly, let $S=\{x\in [a,b]:f(x)\le 0\}$ and $c=\sup S$. This set is nonempty and bounded. Continuity and the strict signs at the endpoints give a right neighbourhood of $a$ with negative values and a left neighbourhood of $b$ with positive values; consequently $a<c<b$. If $f(c)<0$, continuity would give points of $S$ to the right of $c$, contradicting its upper-bound property. If $f(c)>0$, continuity would give an interval immediately to the left of $c$ containing no points of $S$, contradicting the approximation property of its supremum. Hence $f(c)=0$ for some $c\in (a,b)$, proving this form of the intermediate value theorem.

For the half-interval claim, put $H(x)=g(x+1/2)-g(x)$ on $[0,1/2]$. It is a continuous function andIf either endpoint value is zero, it supplies the answer. Otherwise the signs are opposite, so the intermediate value theorem, applied also to $-H$ if necessary, gives an interior zero. Thus $g(c+1/2)=g(c)$ for some $c\in [0,1/2]$.

For general positive integer $n$, define $H_n(x)=g(x+1/n)-g(x)$ on $[0,(n-1)/n]$. At its grid points,If a sampled value is zero, it is the desired point. If none is zero, not all sampled values can have the same sign, so two have opposite signs; continuity gives a zero between their grid points. This proves the horizontal chord of reciprocal-integer length:For $n=1$ the interval is the single point zero, which works by the endpoint equality itself. The proof does not require a differentiable function or a periodic extension.

For a chord of length $1/2$, consider the continuous function $h(x)=f(x+1/2)-f(x)$ on $[0,1/2]$. Since $f(0)=f(1)$,An endpoint zero already supplies the chord. Otherwise the endpoint values have opposite signs, so the intermediate value theorem supplies $h(\alpha )=0$ at some $\alpha \in (0,1/2)$. Thus a horizontal chord of length $1/2$ exists.

For each integer $n>1$, define $h_n(x)=f(x+1/n)-f(x)$ on $[0,1-1/n]$. The sampled differences telescope:If any sampled value is zero, it gives the required chord. Otherwise there must be both a positive and a negative sampled value. The intermediate value theorem, applied between their sample points, then gives $h_n(\alpha )=0$. Hence a horizontal chord of length $1/n$ exists for every positive integer $n$; for $n=1$, the endpoint chord itself suffices. This proves the horizontal chord of reciprocal-integer length result, without restricting $n$ to powers of two.

An explicit piecewise linear function isThe adjacent formulas agree at $1/3$ and $2/3$, so the function is continuous, and $f(0)=f(1)=0$. Any pair of points at distance $2/3$ can be written $t,t+2/3$ with $0\le t\le 1/3$. For every such $t$,Thus their values are never equal. The horizontal chord of reciprocal-integer length conclusion does not extend to this distance.

For $n=1$, choose $x=0$, $y=1$. For $n\ge 2$, define the continuous functionIts values at the grid points have the telescoping sumIf one summand is zero, we already have a zero of $g$. Otherwise the summands cannot all have the same sign, so two of these values have opposite signs. The intermediate value theorem between their arguments gives $t\in [0,1-1/n]$ with $g(t)=0$. Set $x=t$, $y=t+1/n$. ThenThis is the horizontal chord of reciprocal-integer length property; the telescoping sum explains why reciprocal integers appear.