Is $\sqrt2+\pi$ irrational?

Question

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

@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
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

score 11 · Answer 1 · answered Jul 26 '16 at 10:35

11

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

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

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

answered Jul 26 '16 at 10:35

DonAntonio

Thank you, can you elaborate about $\pi+\log 2$? – Ahmbak Jul 26 '16 at 10:49
3

With $\log 2$, it pretty well may be an open problem, and an immensely hard one at that. – Ivan Neretin Jul 26 '16 at 14:39

1 Answers1