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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 !

Mahidevran
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  • In your last equation, you should have $\sigma(q+b)-\sigma(b)$, not $\sigma(q+b)-\sigma(-b)$. – Eric Wofsey Dec 06 '15 at 05:30
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    I really don’t understand the difficulty. Knowing that $\sigma$ takes squares to squares, you know that if $a\ge0$ then $\sigma(a)\ge0$. Consequently, if $a\le b$ (so that $b-a\ge0$), it follows that $\sigma(a)\le\sigma(b)$ (because $\sigma(b)-\sigma(a)=\sigma(b-a)\ge0$). – Lubin Dec 06 '15 at 05:42

2 Answers2

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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 $\sigma(b-a)>0$.)

Eric Wofsey
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  • So if we don't use $q$ , then we have $\sigma(b - a) > 0$ because we assume $a < b$. So $\sigma(b) > \sigma (a)$. ? – Mahidevran Dec 06 '15 at 05:37
  • if we show it without the $q$ , then just by that we can conclude $\sigma a < \sigma b$. Which is much faster than introducing $q$. – Mahidevran Dec 06 '15 at 05:38
  • Yep! And if you think about it, what your argument with $q$ is doing is just making that same argument twice to separately show that $\sigma(a)<\sigma(q)$ and $\sigma(q)<\sigma(b)$. You don't actually need the fact that $q=\sigma(q)$. – Eric Wofsey Dec 06 '15 at 05:42
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You are right for the part $1$ of the problem.

Now note that any positive real $r$ is square of some real number say $s$ i.e. $r= s^2$and by part $1$ $\sigma$ carries squares to squares so $\sigma (r)=(\sigma (s ))^2$ which will be a positive real number.

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