Let

*s*:

**R**->

**R** be an

automorphism. By considering

*s*(1+1+...+1)=

*s*(1)+

*s*(1)+...+

*s*(1) we see that

*s(n)*=

*n* for all

integers

*n*; since

*s* preserves

division,

*s(r)*=

*r* for any

rational number *r*.

So much is standard. Now, if we could prove that *s* is continuous, we would be done (since it is the identity function on a dense set). But what properties does *s* have that we could use? **NONE!**

Instead, we'll use some properties of **R**. Call a number *x* "nonnegative" if *x=y*^{2} for some *y*, in **R**. Note that this corresponds exactly with our notion of this concept. Note also that *s(x)=s(y)*^{2}, so *s* preserves nonnegativity. But this means that *s* preserves the order on **R** (since *a*<*b* iff *a+c*=*b* for some nonnegative *c*).

Now, to every real number corresponds a unique "section" of the rationals (a partition of the rationals into those "smaller than" the real number and those "greater than" (or equal, if it's rational) it. But *s* maps all elements of the section to themselves, and preserves order. So the section of every real number *x* is transformed into the section of *s(x)* (since *s* preserves order), and yet remains constant (since *s* is the identity on the rationals). Thus it must be that *s(x)=x*.