Let f(x) be a continuous function on [0, 3] such that f(0) = f(3). Show that there exists an element, c, in [0, 2] such that f(c) = f(c+1).
Consider the continuous function $\displaystyle g(x) = f(x + 1) - f(x)$ on the interval $\displaystyle \left[ {0,2} \right]$.
Consider: $\displaystyle \left. \begin{gathered} g(0) = f(1) - f(0) \hfill \\ g(1) = f(2) - f(1) \hfill \\ g(2) = f(3) - f(2) \hfill \\ \end{gathered} \right\}\; \Rightarrow \;g(0) + g(1) + g(2) = 0$.
If each of $\displaystyle g(0),g(1),g(2)$ is zero then we are done.
Else, this means that some one of the pairs, $\displaystyle \left\{ {g(0),g(1)} \right\},\left\{ {g(2),g(1)} \right\},\left\{ {g(0),g(2)} \right\}$ has one negative and one positive member. WHY?
Now apply the intermediate value theorem to that pair.
The point is we cannot be sure which pair has one positive and one negative member.
We just know one must exist. Say it is the pair $\displaystyle \{g(1),g(2)\}$.
That means that zero is between them so $\displaystyle \left( {\exists c \in (1,2)} \right)\left[ {g(c) = 0} \right]$ which inturn implies that $\displaystyle f(c + 1) = f(c)$.