# Obtaining $\sqrt{r}$ from $r$ with quadratics

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.

• 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 $\sqrt r = \sqrt 2$. Mar 27, 2018 at 9:53
• (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 $\sqrt{r}\in S$?) Mar 27, 2018 at 9:59

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 $$\sqrt{r}$$.

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