Universal Chord Theorem

Let f∈C[0,1] and f(0)=f(1).
How do we prove ∃a∈[0,1/2] such that f(a)=f(a+1/2)?
In fact, for every positive integer n, there is some a, such that f(a)=f(a+1n).
For any other non-zero real r (i.e not of the form 1n), there is a continuous function f∈C[0,1], such that f(0)=f(1) and f(a)≠f(a+r) for any a.


ANSWER:-

You want to use the intermediate value theorem, but not applied to f directly. Rather, let g(x)=f(x)−f(x+1/2) for x∈[0,1/2]. You want to show that g(a)=0 for some a. But g(0)=f(0)−f(1/2)=f(1)−f(1/2)=−(f(1/2)−f(1))=−g(1/2). This gives us the result: g is continuous and changes sign, so it must have a zero.