I don't really know a lot about this subject but I was wondering if the product of a transcendental number by an integer is transcendental?
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1$e\cdot \frac1{e} = 1$. – zarathustra Apr 11 '15 at 19:56
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Ok, but what about e*5? – Apr 11 '15 at 19:57
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Thus he spaketh! – gary Apr 11 '15 at 19:58
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zarathustra's example answers your question completely ... maybe you want to ask "which pairs of trancendental numbers and real numbers have products that are also transcendental?" – Zubin Mukerjee Apr 11 '15 at 19:58
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Yes, that's what I meant. Sorry my question wasn't clear enough ! – Apr 11 '15 at 19:59
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@Astroman Consider editing your question to broaden its scope. Extra details provided in comments are not 'canonical' to the question per se. – Myridium Apr 11 '15 at 20:01
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@Zubin Mukerjee: Still, Astroman did ask for multiplication by integers , so Zarathustra's is not a counterexample? – gary Apr 11 '15 at 20:11
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1@gary not to the newly edited post it isn't but it was to the original.:) – DRF Apr 11 '15 at 20:32
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The product of a transcendental number by a nonzero algebraic number is necessarily transcendental (as the ratio of two algebraic numbers is algebraic). To prove the case that an algebraic number divided by a nonzero integer is algebraic (as in the revised question), if $x$ satisfies $\alpha_kx^k + \ldots + a_0 = 0$, then $y = x/n$ satisfies $n^k\alpha_k y^k + n^{k-1}\alpha_{k-1}y^{k-1} + \ldots + n\alpha_1 y + \alpha_0 = 0$.
The product of two transcendental numbers can easily be algebraic, however. As suggested by zarathustra in the comments, $e \times 1/e= 1$.
Rolf Hoyer
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Perhaps you could improve this by providing a proof of your first statement, and an example demonstrating your second. – Myridium Apr 11 '15 at 20:02
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A transcendental number is a real or complex number that is not algebraic—that is, it is not a root of a non-zero polynomial equation with rational coefficients. The most prominent examples of transcendental numbers are $\pi$ and $e$. Though only a few classes of transcendental numbers are known (in part because it can be extremely difficult to show that a given number is transcendental), transcendental numbers are not rare. Indeed, almost all real and complex numbers are transcendental, since the algebraic numbers are countable while the sets of real and complex numbers are both uncountable. All real transcendental numbers are irrational, since all rational numbers are algebraic. The converse is not true: not all irrational numbers are transcendental; e.g., the square root of 2 is irrational but not a transcendental number, since it is a solution of the polynomial equation $x^2 − 2 = 0$. No rational number is transcendental and all real transcendental numbers are irrational. A rational number can be written as $p/q$, where p and q are integers. Thus, $p/q$ is the root of $qx − p = 0$. However, some irrational numbers are not transcendental. For example, the square root of 2 is irrational and not transcendental (because it is a solution of the polynomial equation $x^2 − 2 = 0$). The same is true for the square root of other non-perfect squares.
$sin(a), cos(a)$ and $tan(a)$, and their multiplicative inverses $csc(a), sec(a)$ and $cot(a)$, for any nonzero algebraic number a (by the Lindemann–Weierstrass theorem).
$ln(a)$ if a is algebraic and not equal to 0 or 1, for any branch of the logarithm function (by the Lindemann–Weierstrass theorem).
Look for more to ''Baker, Alan (1975). Transcendental Number Theory. Cambridge University Press. ISBN 0-521-20461-5. Zbl 0297.10013''.
ksantas
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