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Prove that, $f:\mathbb{R}\rightarrow \mathbb{R} $ is a isomorphism if, only if $f(x)=x $.

For any rational element that is right, but do not know how to prove to the irrational. I would like to prove it using only the properties of complete ordered field.

Henfe
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    Possible duplicate http://math.stackexchange.com/questions/449404/is-an-algebraic-automorphism-of-the-field-of-real-numbers-the-identity-map – Prism Aug 09 '13 at 04:42

1 Answers1

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Using the fact that $f(1) = 1$, it follows that $f(n) = n$ for any integer $n$, and so $f(\frac{n}{m}) = \frac{n}{m}$ for any rational $\frac{n}{m}$.

Now show that $f$ preserves the ordering on the reals:

If $a \geq 0$, then

$$f(a) = f(\sqrt{a}^2) = f(\sqrt{a})^2 \geq 0$$

It follows that if $a < b$, $f(a) < f(b)$.

Finally, choose $r \in \mathbb{R}$ and let $q_i$ be a sequence of rational numbers with

$$q_1 < q_2 < ...$$

and $q_i \to r$ as $i \to \infty$. Use the above point about orderings to show that

$$q_i = f(q_i) < f(r)$$

for every $i$, so that

$$r = \limsup_{i \to \infty} q_i \leq f(r)$$

Do something similar with decreasing sequences to get the reverse inequality.

The reverse direction of the proof should be immediately clear.

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    Great answer. This is what I tried to hint at, but then I got lost in the details. +1 :) – Alex Wertheim Aug 09 '13 at 03:59
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    What properties must this isomorphism preserve? Addition? Multiplication? Both? Ordering? Until you state what your restrictions are, we can't conclude that f(x) = x is the 'only' one. For example f(x) = -x preserves addition; f(x) = 1/x preserves multiplication; and both preserve order although they do reverse it. – Betty Mock Aug 09 '13 at 05:04
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    @BettyMock Based on context, I'm assuming that $f$ is a field automorphism. That is, $f(x + y) = f(x) + f(y)$ and $f(xy) = f(x) f(y)$ for all $x, y \in \mathbb{R}$. –  Aug 09 '13 at 05:10
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    Voted for your nice solution, but went on thinking about this. Can you enlighten me a bit further? I don't think an isomorphism has to preserve ordering, given that various fields don't have any ordering. Is that correct? Also, I don't recall a requirement that the multiplicative identity in the domain must map into the mult. identity in the range, i.e. f(1) = 1; only that there must be such an identity in the range, so it must have some precursor in the domain. Do I remember wrong? (It's very possible). thx. – Betty Mock Aug 10 '13 at 20:12
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    @BettyMock Right, arbitrary fields frequently have no sense of ordering; part of the characterization of $\mathbb{R}$ is that it is the (unique) complete, ordered field containing $\mathbb{Q}$ as a subfield. Further, any injective field homomorphism must map $1_d$ to $1_r$ ($d$ and $r$ stand for domain and range); for a proof, consider $$f(1_d) = f(1_d^2) = f(1_d)^2 \implies f(1_d) (f(1_d) - 1_r) = 0$$ Hence, either $f(1_d) = 0$ (failing 1-1) or $f(1_d) = 1_r$. –  Aug 10 '13 at 20:18
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    Thanks, this is helpful, in pointing out that R is an unusual field with special characteristics. I figured out the part about the multiplicative identity shortly after I posted my comment. There's nothing like writing something down wrong to clear the mind. – Betty Mock Aug 12 '13 at 02:35