Are all transcendental numbers irrational? I know that not all irrationals are transcendental (for example, $\sqrt{2}$); but I only know of a few transcendental numbers and they are all irrational.
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3Yes, of course: a rational number is a root of a polynomial with integer coefficients; for $a/b$ you can use $bx-a$. – egreg Jan 11 '14 at 00:57
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@egreg If you turn that into an answer (and tie into the definition of transcendental), I'll upvote... – apnorton Jan 11 '14 at 01:13
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I am surprised that this question had not been asked before here. – Lost1 Jan 11 '14 at 01:15
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Perhaps because it usually pops up in algebraic number theory or in fields extensions and then it is really trivial @Lost1? And who knows: perhaps the question's been asked before... – DonAntonio Jan 11 '14 at 01:20
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@DonAntonio for me, google is better resource for this type of thing, which, of course, often returns a math.se link anyway? – Lost1 Jan 11 '14 at 01:22
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Strangely, this question HAS been asked before, but the cabal of moderators here keep shifting things around. – Marcos Oct 22 '15 at 18:42
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To compile what has been said above, any transcendental number must be irrational. As egreg mentions, it suffices to use the polynomial $p(x) = bx-a$ for any rational number $a/b$. Note the definition of "transcendental:" A number is transcendental if it is not the root of any nonzero polynomial with rational coefficients. Hence we have shown that indeed any rational number is algebraic, that is, not transcendental.
