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A real number $r>0$ is written. If there is a number $a$ written, we are allowed to write down $a+1$. If there are numbers $a,b$ written (possibly $a=b$ but they must be written twice in that case), we are allowed to write down the (0, 1, or 2) real roots of $x^2+ax+b$. Is it true that we can always eventually write down the number $\sqrt{r}$?

A particular case where this is certainly possible is if $\sqrt{r}$ is an integer. Starting with $r$, we can write down numbers $2s$ and $s^2$ for some integer $s>0$. The root of $x^2+2sx+s^2$ is $x=-s$, so we can now get all integers at least $-s$. This includes all positive integers.

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  • $\begingroup$ It also works for $r=2$: Take $a = 4$ and $b = 2$ and note that $(\sqrt 2 - 2)^2 + a (\sqrt 2 -2 ) + b = 0$. Hence you can 'write down' $\sqrt 2 -2 $ and consequently $\sqrt r = \sqrt 2$. $\endgroup$
    – Jan Bohr
    Mar 27, 2018 at 9:53
  • $\begingroup$ (I think the question would be easier to understand if phrased in terms of sets: Let $r > 0$ a real number and denote $S\subset \mathbb{R}$ the set inductively defined by: (1) $r\in S$, (2) $x\in S \Rightarrow x +1 \in S$, (3) $a,b\in S \Rightarrow$ reals roots of $x^2+a x + b$ lie in $S$. Question: Is $\sqrt{r}\in S$?) $\endgroup$
    – Jan Bohr
    Mar 27, 2018 at 9:59

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Given $r>0$, we have $r+1$.

So, we have the roots of $x^2+(r+1)x+r=(x+r)(x+1)$, which are $-r$ and $-1$.

Adding $1$ to $-1$, we have $0$, and thus we have the roots of $x^2+0x-r$, so we have $\sqrt{r}$.

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    $\begingroup$ Nice. You need a few $ symbols on the last line. $\endgroup$
    – quasi
    Mar 27, 2018 at 10:50
  • $\begingroup$ The key was getting 0. Nice. $\endgroup$ Feb 3, 2019 at 6:20

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