We know there are numbers like $\pi$, $e$, $\phi$ or also $\zeta(3)$ which are transcendental numbers.
I was wondering if combinations of transcendental numbers are still transcendental numbers, like $e + \pi$ or $\frac{1}{\phi}$ or whatever.
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4We don't even know if $e+\pi$ is rational or not, never mind whether it is transcendental. – Dylan Nov 23 '15 at 22:14
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5The golden ratio, $\phi$ is algebraic. It satisfies $\phi^2-\phi-1=0$. – lulu Nov 23 '15 at 22:14
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1It is also not known whether $\zeta(3)$ is transcendental. It has only been proven to be irrational. – Dylan Nov 23 '15 at 22:16
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@lulu Whoops, my bad. I made a mistake! – Enrico M. Nov 23 '15 at 22:17
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@Dylan I guess the answer is then "we don't know" or better: "it depends" :D – Enrico M. Nov 23 '15 at 22:17
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1You might like to look into Schanuel's conjecture. – Rob Arthan Nov 23 '15 at 22:39
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Hint:
$\pi $ and $1-\pi$ are transcendental. What about $\pi+(1-\pi)$?
Emilio Novati
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1That's cool! I guess that then we cannot say it a priori. It depends upon numbers we have.. – Enrico M. Nov 23 '15 at 22:18
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Polynomials over the rationals in one transcendental number are transcendental, almost (but not quite) by definition. $e^5 + 3e^3 - 2e + 2$ is transcendental.
In two transcendental numbers... well of course $\pi + (1-\pi)$ is rational.
And as for $e + \pi$, who knows?
djechlin
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I believe not all such polynomials are transcendental. For instance $e^2 + be^2 - 1$ is not transcendental where $b$ is defined such that $e^2 + be^2 - 1 = 0$. Perhaps you meant all polynomials with rational algebraic linearly independent coefficients by the Lindemann Weierstrass theorem https://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem. – abligh Nov 24 '15 at 07:06
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$e+\pi$ and $e\pi$ both can not be rational, because if both be rational then we have $x^2-(e+\pi)x+e\pi=0$ has roots $\pi$ and $e$ and leads to contradiction. Also if $z$ is transcendental, then $\frac{1}{z}$ is transcendental, if not then we have $\sum_{i=0}^na_i(\frac{1}{z})^i=0$ for some $a_i\in \mathbb{Z},a_n\not = 0$ and this means $\sum_{i=0}^na_{n-i}z^i=0$, if there exists a $j<n$ such that $a_j\not =0$, this leads to contradiction because it shows that $z$ is algebraic and if for all $j<n$ we have $a_j=0$, then we have $\frac{a_n}{z}=0$ and it contradicts by $a_n\not =0$.
Erfan Khaniki
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