Let $ \alpha$ be a new number. If $ \alpha<0$ one can consider $ -\alpha$ which is greater than $ 0$. So assume that $ \alpha>0$. If for each positive real number $ r$, $ \alpha>r$ then, $ \frac 1{\alpha}$ is the number we are after, because for each positive number $ r$, $ \alpha>\frac 1{r}$ which gives $ \frac 1{\alpha}<r$. But $ \alpha>0$ which shows that $ \frac 1{\alpha}>0$. So assume that there is a positive real number $ R$ such that $ \alpha<R$. Let $ r$ be the superimum of all the real numbers less than $ \alpha$. (This exists because $ \alpha<R$ and $ 0$ is less than $ \alpha$) Let $ \epsilon=\alpha-r$. Then $ \epsilon\neq0$ because $ \alpha$ is not a real number. If $ \epsilon>0$, there is no positive real number $ a$, less than $ \epsilon$. Because if $ a$ is such a number, $ r+a<\alpha$ which contradicts $ r$ being the superimum. But if $ \epsilon<0$ then there is no negative real number $ a$ greater than $ \epsilon$. Because if $ a$ is such a number, then $ \alpha<r+a$ which shows that the superimum should have been less than or equal to $ r+a$. So in this case, $ -\epsilon$ is the required number. For the second part of the problem, assume $ P$ is a polynomial with real coefficients. $ P$ can be factored into linear and quadratic terms, each with real coefficients. So for $ \alpha$ to be a root of this polynomial, it has to be a root of one of the terms. (because the new system of numbers is a field) If $ ax+b$ is a linear term ($ a\neq 0$), then $ a\alpha+b=0$ gives $ \alpha=\frac{-b}{a}$ which shows that $ \alpha$ is a real number. If $ ax^2+bx+c$ is a quadratic term, ($ a\neq 0$ and $ b^2<4ac$ so that this term cannot be factored into linear ones) then $ a\alpha^2+b\alpha+c=a\left((\alpha+\frac{b}{2a})^2+\frac{4ac-b^2}{4a^2}\right)$ But each term of the last expression is positive, because if $ c$ is a number in the new system, either $ c\leq 0$ which shows that $ c\times c\geq0\times 0=0$, or $ c\geq 0$ which shows that $ c\times c\geq0\times0=0$. So $ c^2\geq 0$. So $ (\alpha+\frac{b}{2a})^2\geq0$. And $ 4ac-b^2>0$ which gives $ \frac{4ac-b^2}{4a^2}>0$. Hence $ \alpha$ can not be a root of this term, which shows that $ \alpha$ is not a root of the original polynomial.

Omid Hatami wrote: $ P$ can be factored into linear and quadratic terms, each with real coefficients. So for $ \alpha$ to be a root of this polynomial, it has to be a root of one of the terms. (because the new system of numbers is a field) Sorry, but how did deduce that $ P$ can be factored into such terms?

It is a classical result of Fundamental theorem of algebra. $ P$ can be factored to linear terms in $ \mathbb C[x]$ and for each non-real complex root $ \alpha$ of $ P$, $ \bar\alpha$ is also another root. So it can be easily deduced that $ P$ can be factored into linear and quadratic terms, each with real coefficients.

Ah yes. Thank you.

Additional problem: c) Construct such numbers. A more abstract proof for b): For a field to be able to carry an ordering (coherent with the field's operations) is equivalent for $ -1$ not being a sum of squares. Thus in total, an ordered field cannot have a square root of $ -1$. As $ \mathbb C$ is an algebraic extension of degree $ 2$ of $ \mathbb R$, we just need to show that $ -1$ doesn't have a square root in this new numbers, so we are done.

part b) is just the fundamental theorem of Algebra, since the polynomial only has roots in C[x], and the only non-complex numbers in this set are the real numbers, so you're done...

You have to show that there are no complex numbers there (Not that this is really difficult...)

Like since no complex numbers are between any 2 real numbers, they cannot be complex?

A more mathematical approach (reasoning from given properties) would be usefull. There is a square root of $ -1$ in appropriate completions of $ \mathbb Q$, thus there are complex numbers even between rational ones

OK, but proving that $ F$ can be ordered if $ -1$ is not square is not trivial.

If there is a non-real root of a polynomial, it must have the form a+bi, with $ b\ne 0$. Let this number be right after the real number c. Then 2(a+bi) will be right after 2c, since otherwise if there was another number in between, then there would be a number between a+bi and c. Similarly, a+c+bi will be right after c, since otherwise if there was a number in between, then there would have been a number between a+bi and c. Therefore, 2a+bi=a+c+bi so b=0, contradiction!

Omid Hatami wrote: OK, but proving that $ F$ can be ordered if $ - 1$ is not square is not trivial. You mean: not sum of squares. But here you just need the the trivial direction (ordered => $ - 1 < 0$ => not a square). rofler wrote: Let this number be right after the real number c. There is no reason why there should be a "next" number. You didn't even use any property of complex numbers.