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 integern , there is some a , such that f(a)=f(a+1n) .

For any other non-zero realr (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 .

You want to use the intermediate value theorem, but not applied tof 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.

How do we prove

In fact, for every positive integer

For any other non-zero real

**ANSWER:-**You want to use the intermediate value theorem, but not applied to

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