The truth table shows the following statement is a tautology, but it doesn't make sense. the truth table of the sentence $$(p \rightarrow q) \vee (q \rightarrow p)$$ 
is 
\begin{array}{ c  c  l }
  p & q & (p \rightarrow q) \vee (q \rightarrow p) \\ \hline 
  T & T & \, \, T \; T \> T \> \>  \mathbf{T} \> \> T \; T \> T\\
  T & F & \, \, T \; F \> F \> \> \mathbf{T} \> \> F \; T \> T\\
  F & T & \, \, F \; T \> T \> \> \mathbf{T} \> \> T \; F \> F\\
  F & F & \, \, F \; T \> F \> \> \mathbf{T} \> \> F \; T \> F\\
\end{array}
where the italic truth values are of subclauses and the boldface truth values are of the whole statement. From the truth table it is seen that the given statement is a tautology. So far so good. For me the problem arises when I verbally think of the statement. It can be translated into metalanguage as,
the statement "either $p$ implies $q$, or $q$ implies $p$", is a tautology. 
This tautology means that, given two arbitrary statements $p$ and $q$, if $p$ does not imply $q$, then $q$ must imply $p$, and the other way  around. Or in other words, it is not the case that none of them implies the other.  This doesn't make too much sense to me.  Why it is always the case that, given two arbitrary statements, one should be implying the other for sure? Why two random statements should bound in such way? 
 A: In classical logic, there is nothing wrong in the above "un-natural" tautology.
You can see this recent post (with many links) about the "correct" reading of the truth functional aspects of $\rightarrow$.
You can check with truth table that :

$p \rightarrow q$ and $\lnot p \lor q$

are equivalent.
Thus, we can rewrite :

$(p→q)∨(q→p)$ (called also : Dumemtt's law)

as :

$(\lnot p \lor q) \lor (\lnot q \lor p)$.

Now we can rearrange the disjuncts to get :


$(\lnot p \lor p) \lor (\lnot q \lor q)$.


Now, the tautology "sounds" much more ... tautological.

As per comment above, you can avoid this counter-intuitive aspects leaving classical logic and adopting, for example, the intuitionsitic one; but you have to remember that the truth functional properties of the connectives (i.e. the truth table) are nor more valid in it.
A: Use the equivalences $$[A\implies B] \equiv \neg[A\land \neg B]\equiv [\neg A\lor B]$$ to show that $$[[p\implies q]\lor [q\implies p]]\equiv[[p\land \neg q]\implies[\neg q \lor p]] $$
