3

From this, as a layman I wonder if the same goes for $\sqrt2+\pi$? How about $\pi+\log2$?

Ahmbak
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    [This](http://math.stackexchange.com/questions/1197927/is-the-sum-of-an-algebraic-and-transcendental-complex-number-transcendental) may be of interest. – David Mitra Jul 26 '16 at 10:30
  • @DavidMitra Thank you, so the first one is obvious. Can I ask you further if $\log 2$ is algebraic? – Ahmbak Jul 26 '16 at 10:40
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    I don't know ($\ln 2$ is transcendental). – David Mitra Jul 26 '16 at 10:43
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    Algebraic numbers are closed under addition and multiplication (in fact, they form a field). That is, if $\alpha$ and $\beta$ are roots of some polynomial then so are $\alpha+\beta$ and $\alpha\beta$. This is what is going on in Don Antonio's answer. Clearly $\sqrt2$ is algebraic (root of $x^2-2$), as is every rational number ($a/b$ is a root of $bx-a$). So if $\sqrt2+\pi=r$ then $\pi=r-\sqrt2$ is algebraic, as by closure of addition. – user1729 Jul 26 '16 at 10:45

1 Answers1

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Suppose the sum is rational, say $\;r\;$, but then

$$r=\sqrt2+\pi\implies r^2-2r\pi+\pi^2=2\implies \pi\;\;\text{is a root of the polynomial}$$

$$p(x)=x^2-2rx+r^2-2\in\Bbb Q[x]\;,\;\;\text{which of course is absurd}$$

DonAntonio
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