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Transcendental number
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==Properties== A transcendental number is a (possibly complex) number that is not the root of any integer polynomial. Every real transcendental number must also be [[irrational number|irrational]], since every [[rational number]] is the root of some integer polynomial of [[degree of a polynomial|degree]] one.<ref>{{harvnb|Hardy|1979}}</ref> The set of transcendental numbers is [[uncountable|uncountably infinite]]. Since the polynomials with rational coefficients are [[countable]], and since each such polynomial has a finite number of [[zero of a function|zeroes]], the [[algebraic number]]s must also be countable. However, [[Cantor's diagonal argument]] proves that the real numbers (and therefore also the [[complex number]]s) are uncountable. Since the real numbers are the union of algebraic and transcendental numbers, it is impossible for both [[subset]]s to be countable. This makes the transcendental numbers uncountable. No [[rational number]] is transcendental and all real transcendental numbers are irrational. The [[irrational number]]s contain all the real transcendental numbers and a subset of the algebraic numbers, including the [[quadratic irrational]]s and other forms of algebraic irrationals. Applying any non-constant single-variable [[algebraic function]] to a transcendental argument yields a transcendental value. For example, from knowing that {{mvar|Ο}} is transcendental, it can be immediately deduced that numbers such as <math>5\pi</math>, <math>\tfrac{\pi - 3}{\sqrt{2}}</math>, <math>(\sqrt{\pi}-\sqrt{3})^8</math>, and <math>\sqrt[4]{\pi^5+7}</math> are transcendental as well. However, an [[algebraic function]] of several variables may yield an algebraic number when applied to transcendental numbers if these numbers are not [[algebraically independent]]. For example, {{mvar|Ο}} and {{math|(1 β ''Ο'')}} are both transcendental, but {{math|''Ο'' + (1 β ''Ο'') {{=}} 1}} is obviously not. It is unknown whether {{math|''e'' + ''Ο''}}, for example, is transcendental, though at least one of {{math|''e'' + ''Ο''}} and {{mvar|eΟ}} must be transcendental. More generally, for any two transcendental numbers {{mvar|a}} and {{mvar|b}}, at least one of {{math|''a'' + ''b''}} and {{mvar|ab}} must be transcendental. To see this, consider the polynomial {{math|(''x'' β ''a'')(''x'' β ''b'') {{=}} ''x''<sup>2</sup> β (''a'' + ''b'') ''x'' + ''a b''}} . If {{math| (''a'' + ''b'')}} and {{mvar|a b}} were both algebraic, then this would be a polynomial with algebraic coefficients. Because algebraic numbers form an [[algebraically closed field]], this would imply that the roots of the polynomial, {{mvar|a}} and {{mvar|b}}, must be algebraic. But this is a contradiction, and thus it must be the case that at least one of the coefficients is transcendental. The [[non-computable numbers]] are a strict subset of the transcendental numbers. All [[Liouville number]]s are transcendental, but not vice versa. Any Liouville number must have unbounded partial quotients in its [[simple continued fraction]] expansion. Using a [[Cantor's diagonal argument|counting argument]] one can show that there exist transcendental numbers which have bounded partial quotients and hence are not Liouville numbers. Using the explicit continued fraction expansion of {{mvar|e}}, one can show that {{mvar|e}} is not a Liouville number (although the partial quotients in its continued fraction expansion are unbounded). [[Kurt Mahler]] showed in 1953 that {{mvar|Ο}} is also not a Liouville number. It is conjectured that all infinite continued fractions with bounded terms, that have a "simple" structure, and that are not eventually periodic are transcendental<ref>{{harvnb|Adamczewski|Bugeaud|2005}}</ref> (in other words, algebraic irrational roots of at least third degree polynomials do not have apparent pattern in their continued fraction expansions, since eventually periodic continued fractions correspond to quadratic irrationals, see [[Hermite's problem]]).
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