In classical logic, why is $(p\Rightarrow q)$ True if both $p$ and $q$ are False?

I am studying entailment in classical first-order logic.

The Truth Table we have been presented with for the statement $(p \Rightarrow q)\;$ (a.k.a. '$p$ implies $q$') is: $$\begin{array}{|c|c|c|} \hline p&q&p\Rightarrow q\\ \hline T&T&T\\ T&F&F\\ F&T&T\\ F&F&T\\\hline \end{array}$$

I 'get' lines 1, 2, and 3, but I do not understand line 4.

Why is the statement $(p \Rightarrow q)$ True if both p and q are False?

We have also been told that $(p \Rightarrow q)$ is logically equivalent to $(~p || q)$ (that is $\lnot p \lor q$).

Stemming from my lack of understanding of line 4 of the Truth Table, I do not understand why this equivalence is accurate.

Administrative note. You may experience being directed here even though your question was actually about line 3 of the truth table instead. In that case, see the companion question In classical logic, why is $(p\Rightarrow q)$ True if $p$ is False and $q$ is True? And even if your original worry was about line 4, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines.

Here is an example. Mathematicians claim that this is true:

If $x$ is a rational number, then $x^2$ is a rational number

But let's consider some cases. Let $P$ be "$x$ is a rational number". Let $Q$ be "$x^2$ is a rational number".
When $x=3/2$ we have $P, Q$ both true, and $P \rightarrow Q$ of the form $T \rightarrow T$ is also true.
When $x=\pi$ we have $P,Q$ both false, and $P \rightarrow Q$ of the form $F \rightarrow F$ is true.
When $x=\sqrt{2}$ we have $P$ false and $Q$ true, so $P \rightarrow Q$ of the form $F \rightarrow T$ is again true.

But the assertion in bold I made above means that we never ever get the case $T \rightarrow F$, no matter what number we put in for $x$.

Here are two explanations from the books on my shelf followed by my attempt. The first one is probably the easiest justification to agree with. The second one provides a different way to think about it.

From Robert Stoll's “Set Theory and Logic” page 165:

To understand the 4th line, consider the statement $(P \land Q) \to P$. We expect this to be true regardless of the choice of $P$ and $Q$. But if $P$ and $Q$ are both false, then $P \land Q$ is false, and we are led to the conclusion that if both antecedent and consequent are false, a conditional is true.

From Herbert Enderton's “A Mathematical Introduction to Logic” page 21:

For example, we might translate the English sentence, ”If you're telling the truth then I'm a monkey's uncle,” by the formula $(V \to M)$. We assign this formula the value $T$ whenever you are fibbing. In assigning the value $T$, we are certainly not assigning any causal connection between your veracity and any simian features of my nephews or nieces. The sentence in question is a conditional statement. It makes an assertion about my relatives provided a certain condition — that you are telling the truth — is met. Whenever that condition fails, the statement is vacuously true.

Very roughly, we can think of a conditional formula $(p \to q)$ as expressing a promise that if a certain condition is met (viz., that $p$ is true), then $q$ is true. If the condition $p$ turns out not to be met, then the promise stands unbroken, regardless of $q$.

That's why it's said to be “vacuously true”. That $(p \to q)$ is True when both $p$, $q$ are False is different from saying the conclusion $q$ is True (which would be a contradiction). Rather, this is more like saying “we cannot show $p \to q)$ to be false here” and Not False is True.

Here's a slightly different answer from the ones given.

The last line of the truth table is indeed counter-intuitive and this is exploited by the Wason Selection task. In the test, subjects are asked to solve this following puzzle:

There are 4 cards placed on a table. One side of the card has a number, while the opposite side is only coloured. The visible faces of the cards show 3, 8, red and blue. Which card(s) should you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is red?

In Wason's original experiment, only 10 percent responded correctly. Most failed to list the blue card as the card that also must be turned over (apart from the card with number 8). Now, suppose we turn the blue card. Only if the other side fails to have an even number, would the proposition be true. Why? Because if the other side was even, then you'd have a card with even number on one face whose opposite face was not red. You can look at this as the intuition behind the last line of the truth-table. To correctly test the verity of the proposition, we must check that when the consequent is False, then the antecedent must be False too.