If $\sigma$ takes squares to squares, conclude $a < b$ implies $\sigma a < \sigma b$. Prove that any $\sigma \in $ Aut$(\mathbb{R/\mathbb{Q}})$ takes squares to squares and takes positive reals to positive reals. Conclude that $a < b$ implies $\sigma a < \sigma b$ ,$\forall a,b 
\in R$.
proof: Suppose $\sigma: a → b$, such that $a = c^2$ where $b$ is a square. 
Then $\sigma(a) = \sigma(c^2) = \sigma(c *c) = \sigma (c) \sigma(c) = (\sigma(c))^2$ which is a square. So it takes positive reals to positive reals.
Then suppose that if $a < b$ , where $a, b\in R$. 
Then there is a $q \in \mathbb{Q}$ such that $a < q < b$. So $q = \sigma q = \sigma ( q + a - a) =  \sigma (q - a) + \sigma a > \sigma a.$
So $q > \sigma(a)$.
And $q = \sigma (q) = \sigma (q + b - b) = \sigma( q + b ) - \sigma (-b)$
can someone please help  me with $q < \sigma(b)$? and please verify I am on the right track. Anything could help thank you ! 

Instead of splitting $\sigma(q+b-b)$ as $\sigma(q+b)-\sigma(b)$, split it as $\sigma(q-b)+\sigma(b)=-\sigma(b-q)+\sigma(b)$.  Since $b>q$, $\sigma(b-q)>0$, so you get $\sigma(q)<\sigma(b)$.
(Actually, you can do this without introducing $q$ at all!  Just notice that $\sigma(b)=\sigma(a+b-a)=\sigma(a)+\sigma(b-a)$, and $\sigma(b-a)>0$.)
