... and awesome.
You added a (?) to "known" and the answer is, 'sort of'. There's infinity, and then there's infinity. 
You've got your aleph_nought infinity of sets that can be mapped bijectively to the natural counting numbers, which includes integers and rationals and the algebraic numbers on the real line. Then you've got your aleph_1 least upper bound of the reals, which includes the uncountable set of transcendental numbers. We think of aleph_1 being infinitely larger than aleph_0, even though aleph_0 is infinitely large. Right? In fact, formally aleph_1 := 2^{aleph_0}, whatever that means. Heheh... So yeah, most of the reals are transcendentals... the larger infinitude, whatever sense that makes.
For which set of q_i's in the rationals does the equation q(pi) = q0 + q1*pi + q2*pi^2 + ... qn*pi^n = 0? The answer is there is no such set. You're right it's some heavy math (abstract algebra), but the proof calls for induction with some long division of polynomials... ie., q(pi) / (q1*pi + q0) searching for irreducible polynomials.
But to get an intuitive sense of 'transcendental', just plug pi in to a polynomial instead of the unknown "x". The result is the same -- "x" isn't a variable, it's an abstract algebraic placeholder called an "unknown" and algebraically, plugging pi in will give you the same result: irreducible polynomials.
Ex., x^2 - 4 = (x + 2)(x - 2) --> x = +-2 are roots in Q. (x + 2) and (x - 2) are irreducible polynomials in Q because x^2 - 2 = (x + sqrt(2))(x - sqrt(2)) and sqrt(2) is not in Q. So sqrt(2) is irrational. But sqrt(2) isn't transcendental (it is algebraic) for this very reason: sqrt(2) is the root of some polynomial with coefficients in Q (x^2 - 2). The same cannot be said for pi.
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