From this, as a layman I wonder if the same goes for $\sqrt2+\pi$? How about $\pi+\log2$?
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6[This](http://math.stackexchange.com/questions/1197927/isthesumofanalgebraicandtranscendentalcomplexnumbertranscendental) 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

1I don't know ($\ln 2$ is transcendental). – David Mitra Jul 26 '16 at 10:43

1Algebraic 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 $x^22$), as is every rational number ($a/b$ is a root of $bxa$). So if $\sqrt2+\pi=r$ then $\pi=r\sqrt2$ is algebraic, as by closure of addition. – user1729 Jul 26 '16 at 10:45
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Suppose the sum is rational, say $\;r\;$, but then
$$r=\sqrt2+\pi\implies r^22r\pi+\pi^2=2\implies \pi\;\;\text{is a root of the polynomial}$$
$$p(x)=x^22rx+r^22\in\Bbb Q[x]\;,\;\;\text{which of course is absurd}$$
DonAntonio
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3With $\log 2$, it pretty well may be an open problem, and an immensely hard one at that. – Ivan Neretin Jul 26 '16 at 14:39