If $x\leq y$ then $\phi \left( x\right) \leq \phi \left( y\right)$ and $\phi \left( x\right) =x$ for all $x\in\mathbb{R}$.

Question

Let $\phi :\mathbb{R} \rightarrow \mathbb{R}$ be a one to one map such that $\phi \left( x+y\right) =\phi \left( x\right) +\phi \left( y\right)$ and $\phi \left( x^{2}\right) =\phi \left( x\right) ^{2}$ for all $x,y\in\mathbb{R}$, $\phi \left( xy\right) =\phi \left( x\right) \phi \left( y\right)$ for all $x,y\in\mathbb{R}$ and $\phi \left( q\right) =q$ for all $p\in\mathbb{Q}$.

Claim1. If $x\leq y$ then $\phi \left( x\right) \leq \phi \left( y\right)$.

Claim2. $\phi \left( x\right) =x$ for all $x\in\mathbb{R}$.

My proof-trying of Claim1. By the definition, $x\leq y$ $\Leftrightarrow$ $\exists z$ $(z\geq 0 \wedge $x+z=y$)$. Thus, $\phi \left( x\right) +\phi \left( z\right) =\phi \left( y\right)$. Hence, we obtain $\phi \left( x\right) \leq \phi \left( y\right)$. But, I don't sure. Can you check?

Can you give a hint for claim2?

Recall that Show $\phi \left( q\right) =q$ for all $q\in\mathbb{Q}$.

This (and the whole exercise) is proved here. See for "If $a<b$ then $\phi(a)<\phi(b)$" in the proofs. And Claim $2$ says that the identity is the only automorphism. — Dietrich Burde, Nov 30 '16 at 20:58
@Kahler It's not true that $x\le y\Leftrightarrow \exists z, (x+z=y).$ What's true is: $x\le y\Leftrightarrow \exists z, (z\ge 0 ,\land, x+z=y).$ — Mitchell Spector, Nov 30 '16 at 21:08
@DietrichBurde For Claim2, (Field automorphism) i.e., $\phi \left( 1\right) =1$, $\phi \left( x+y\right) =\phi \left( x\right) +\phi \left( y\right)$, $\phi \left( xy\right) =\phi \left( x\right) \phi \left( y\right)$. Thus, take $x=y$. Hence, we have $\phi \left( x\right) =x$. Is it true? — , Nov 30 '16 at 21:15
No, we have a field automorphism because of your first question (your own link, see above). Then you have to show that $\phi(q)=q$, see your first question again. — Dietrich Burde, Nov 30 '16 at 21:20
@DietrichBurde Also, I showed all question of the link:http://math.stackexchange.com/questions/2027980/show-phi-left-q-right-q-for-all-q-in-mathbbq Thus, we can use these. — , Nov 30 '16 at 21:24
@MitchellSpector Actually, we know $\phi \left( z\right) =z$ for all $z$ by the previous proposition in the link: http://math.stackexchange.com/questions/2027980/show-phi-left-q-right-q-for-all-q-in-mathbbq. Thus, is it enough for 'if $z\geq 0$ then $\phi \left( z\right) \geq 0$'. — , Nov 30 '16 at 21:42
Doesn't that link just say that $\phi(z)=z$ for rational $z?$ Unless I'm missing it, I don't see where you show that $z\ge 0\rightarrow \phi(z)\ge 0$ for all real $z.$ — Mitchell Spector, Nov 30 '16 at 21:49

If $x\leq y$ then $\phi \left( x\right) \leq \phi \left( y\right)$ and $\phi \left( x\right) =x$ for all $x\in\mathbb{R}$.

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