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text: '[summary(Technical): A [4bc real] or [4zw complex] number $z$ is said to be *transcendental* if there is no (nonzero) [-48l]-coefficient [-polynomial] which has $z$ as a [root_of_polynomial root].]\n\n[summary: A *transcendental* number $z$ is one such that there is no (nonzero) [-polynomial] function which outputs $0$ when given $z$ as input. $\\frac{1}{2}$, $\\sqrt{6}$, $i$ and $e^{i \\pi/2}$ are not transcendental; $\\pi$ and $e$ are both transcendental.]\n\nA [4bc real] or [4zw complex] number is said to be *transcendental* if it is not the root of any (nonzero) [-48l]-coefficient [-polynomial].\n("Transcendental" means "not [algebraic_number algebraic]".)\n\n# Examples and non-examples\n\nMany of the most interesting numbers are *not* transcendental.\n\n- Every integer is *not* transcendental (i.e. is algebraic): the integer $n$ is the root of the integer-coefficient polynomial $x-n$.\n- Every [4zq rational] is algebraic: the rational $\\frac{p}{q}$ is the root of the integer-coefficient polynomial $qx - p$.\n- $\\sqrt{2}$ is algebraic: it is a root of $x^2-2$.\n- $i$ is algebraic: it is a root of $x^2+1$.\n- $e^{i \\pi/2}$ (or $\\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{2}}{2}i$) is algebraic: it is a root of $x^4+1$.\n\nHowever, $\\pi$ and $e$ are both transcendental. (Both of these are difficult to prove.)\n\n# Proof that there is a transcendental number\n\nThere is a very sneaky proof that there is some transcendental real number, though this proof doesn't give us an example.\nIn fact, the proof will tell us that "[almost_every almost all]" real numbers are transcendental.\n(The same proof can be used to demonstrate the existence of [54z irrational numbers].)\n\nIt is a fairly easy fact that the *non*-transcendental numbers (that is, the algebraic numbers) form a [-countable] subset of the real numbers.\nIndeed, the [fundamental_theorem_of_algebra Fundamental Theorem of Algebra] states that every polynomial of degree $n$ has exactly $n$ complex roots (if we count them with [multiplicity multiplicity], so that $x^2+2x+1$ has the "two" roots $x=-1$ and $x=-1$).\nThere are only countably many integer-coefficient polynomials [todo: spell out why], and each has only finitely many complex roots (and therefore only finitely many—possibly $0$—*real* roots), so there can only be countably many numbers which are roots of *any* integer-coefficient polynomial.\n\nBut there are uncountably many reals ([reals_are_uncountable proof]), so there must be some real (indeed, uncountably many!) which is not algebraic.\nThat is, there are uncountably many transcendental numbers.\n\n# Explicit construction of a transcendental number\n[todo: Liouville's constant]',
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