Why is the probability of $\emptyset$ equal to 0?

It's a consequence of the axioms of a probability measure: if $A$ is any event, then $A\cup\emptyset=A$ and $A\cap\emptyset=\emptyset$, hence $$ \mathbb{P}(A)=\mathbb{P}(A\cup\emptyset)=\mathbb{P}(A)+\mathbb{P}(\emptyset)$$ Therefore we must have $\mathbb{P}(\emptyset)=0$.

I think a more rigorous proof than previously given as answers uses more axioms of a probability measure. Of the three axioms, we need:

1)P($\Omega$)=1

2)Given $A_i$ are mutually disjoint events in a sample space $\Omega$, P($\cup_{i=1}^{\infty}A_i$)=$\sum_{i=1}^{\infty}P(A_i)$

So we say $A_i = \emptyset$, because these are mutually disjoint, we can use axiom 2, such that:

P($\cup_{i=1}^\infty A_i$)=$\sum_{i=1}^{\infty}$P($A_i$)=P($\emptyset$)+P($\emptyset$)+...

However we know from axiom one that the probability of a event must be finite from axiom 1. Therefore P($\cup_{i=1}^\infty A_i$)<$\infty$ is true only if P($\emptyset$)=$0$.

Suppose you pick a number between $1$ and $10$ at random. What's the probability that the number is even? $1/2$. What's the probability that you picked $1$ or $2$? $1/5$. Now, what's the probability that the number you picked is neither even nor odd?