Are the numbers ##eπ## and ##e+ π## transcendental?

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  • #1
chwala
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TL;DR Summary
A number is called an algebraic number if it is a solution of a polynomial equation ##a_0 z^n+a_1z^{n-1} + ... a_{n-1}z + a_n =0## where ##a_0,a_1 ...a_n## are integers...otherwise transcendental.
My question is [following the example on the attachment which is apparently clear to me].
1. Are the numbers ##eπ## and ##e+ π## Transcendental?
2. Algebraic numbers can also be rational and not necessarily integers? is that correct?
 

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  • #2
pasmith
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All rational numbers are algebraic: [itex]n/m[/itex] is the solution of [itex]mz - n = 0[/itex].
 
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fresh_42
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TL;DR Summary: A number is called an algebraic number if it is a solution of a polynomial equation ##a_0 z^n+a_1z^{n-1} + ... a_{n-1}z + a_n =0## where ##a_0,a_1 ...a_n## are integers...otherwise transcendental.

My question is [following the example on the attachment which is apparently clear to me].
1. Are the numbers ##eπ## and ##e+ π## Transcendental?
Presumably. The standard theorems about transcendency don't apply to them, but I haven't checked in detail. More interesting is the question: Who cares? These numbers do not occur naturally and I haven't seen any theorem that needed to know whether ##\mathbb{Q}[e,\pi ]## is of transcendental degree one or two. I guess literally nobody will expect it to be one.

2. Algebraic numbers can also be rational and not necessarily integers? is that correct?
##\mathbb{Z}\subsetneq \mathbb{Q} \subsetneq \mathbb{A}\subsetneq \mathbb{C}## if ##\mathbb{A}## is the field of all algebraic numbers over the rationals.
 
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  • #4
chwala
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What i would add on this is that ##e## and ## π## are irrational thus we cannot solve for them in a given equation say,

##z-e- π=0##

as is the case with algebraic terms in any given polynomial.
 
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fresh_42
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What i would add on this is that ##e## and ## π## are irrational thus we cannot solve for them in a given equation say,

##z-e- π=0## as is the case with polynomials.
You must be careful. ##\pi, -\pi## and ##\pi^{-1}## are transcendent, but neither is ##(\pi)+(-\pi)## nor ##(\pi)\cdot (\pi^{-1}).##
 
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pbuk
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What i would add on this is that ##e## and ## π## are irrational
So is ## \sqrt 2 ## (but it is not of course transcendental, by definition).
You must be careful. ##\pi, -\pi## and ##\pi^{-1}## are transcendent, but neither is ##(\pi)+(-\pi)## nor ##(\pi)\cdot (\pi^{-1}).##
And nor is ## e^{i \pi} ##.
 
  • #7
nuuskur
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I believe it is an open problem: are ##\frac{\pi}{e}, e\pi, e+\pi## irrational? Transcendental?
See no.22

It is known that either ##\pi+e## or ##e\pi## is transcendental.
If both are algebraic, then ##(\pi+e)^2 - 4e\pi## is algebraic. So, ##\pi-e## is algebraic, which implies
[tex]
\frac{1}{2}((\pi+e) - (\pi-e)) = e
[/tex]
is algebraic, a contradiction.
 
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