How can we measure how "irrational" a number is? I wonder, can we, for instance, say that $\pi$ is more irrational than $e$? Or that $e$ is more irrational than $\sqrt{2}$? What kind of irrationality measurement can we use to say that $A$ is (much) more irrational than $B$? (in particularly rigorous sense)
Well, there is a sort of extension of the idea of irrationality called "transcendental". You can think of "irrational" as meaning "There is no way to relate this number to $1$ by thinking only about addition and subtraction". That is, a number $q$ is rational if we can write something like:
$$q+q+q-1-1-1-1=0$$
or some other similar equation involving only $q,\,+,$ and $1$. The above represents the number $\frac{4}3$ and more generally, you can, for $q=\frac{a}b$ write:
$$\underbrace{q+\ldots+q}_{b \text{ repetitions}}-\underbrace{1-\ldots-1}_{a \text{ repetitions}}=0.$$
or appropriately change signs for negative $a$.

The next logical step would be to consider the idea of allowing multiplication too - and this extends the rational numbers to algebraic numbers, which are (potentially) irrational, but can be related to the rationals via multiplication and addition. For instance, $x=\sqrt{2}$ satisfies
$$x^2-2=0$$
and we can come up with similar expressions (the set of such equations would be polynomials to be precise) for any expression built with radicals and addition and such - and we could say that the degree of an algebraic number (i.e. how many multiplications we need) represents how irrational it is - that is $\sqrt{2}$ can be related to integers by squaring it, but $\sqrt[3]{2}$ requires cubing to get to an irrational - and numbers like $\sqrt{2}+\sqrt{3}$ require fourth powers. We might say that the degree represents a sort of "distance" from the rational numbers.

However, this only extends to a narrow class of numbers - $\pi$ and $e$ are both transcendental, meaning "not algebraic". We have much less understanding of these, since we can't relate them to the rationals through arithmetic - so we could be justified in saying that they are less well behaved then algebraic numbers. One way we can talk about their
 irrationality measure which essentially tells us how far from rational numbers a given irrational is, relating the growth of the denominator of the best rational approximations to how close they are - this represents how these numbers are comprehended by looking at sequences of rational approximations, rather than algebraic properties. However, this isn't terribly useful to compare actual numbers, because we hardly know how to calculate any. We do know that, from its infinite series, that $e$ acts very similarly to an algebraic number (looking at its series definition) and its irrationality measure is $2$. We don't know the irrationality measure of $\pi$, though we might suspect that it greater than $2$ - but it's hard to compare numbers this way, given our limited knowledge thereof, and all it means to have a small irrationality measure is that it's "far" from any rationals with small denominators.
No rational number is irrational, but in the same way we can measure how &quot;black&quot; a shade of grey is, given that every irrational number has an infinite continued fraction, a fair measure of how irrational a rational number is, is by the length of its continued fraction.
As part of my studies of the Collatz conjecture I discovered (proof thanks to Torsten Schoeneberg) that every rational number has precisely one finitely-long continued fraction of the form $1-\dfrac1{2^{m_0}-\dfrac1{2^{m_1}-\dfrac1{2^{m_2-\frac1{\ldots}}}}}$
provided the terms are measured by letting $\lvert x\rvert_2$ be the 2-adic value of $x$ and iterate $1-\dfrac1{x\lvert x\rvert_2}$ until you get to zero.  The number of steps taken is the length of the continued fraction and the measure.
What is more, if you wish more resolution, you can assign every rational number (up to bijection) an ordinal less than $\omega^\omega$.  To do this, simply identify $m_0, m_1,\ldots$ by counting the number of divisions by $2$ required at each step. Then use any bijection $f$ from $\Bbb Z\to \Bbb N$ to send $m_i\to n_i$ and:
$1-\dfrac1{2^{m_0}-\dfrac1{2^{m_1}-\dfrac1{2^{m_2-\frac1{\ldots}}}}}\mapsto \displaystyle\sum_{i=0}^\infty\omega^i\cdot n_i=\displaystyle\sum_{i=0}^\infty\omega^i\cdot f(m_i)$
is a bijection from the rationals into the ordinal $\omega^\omega$.