Let $(\mathbb F,\mathbb F^{+})$ an ordered field. Are the following statements true? Prove (or disprove).
(i) For every element $x \in \mathbb F$ there are two other elements in $\mathbb F$ which are smaller and larger than $x$.
(ii) For every $x,y \in \mathbb F$ there exists $z\in \mathbb F$ such that $x=0$ or $xz=y$.
Sorry that I'm completely lost. I think both statements are true. The first one is true since ordered fields can not be finite. The rational numbers are the smallest ordered field so my intuition is that the second statement is true as well. Can anyone help me to give a more formal proof of the two statements?
(i) Your intuition is correct. You can simply take the numbers to be $x - 1$ and $x + 1$. These two are distinct from each other and from $x$ because $1 + 1 \neq 0 \neq 1$ in an ordered field (a consequence of $1 > 0$). $1 > 0$ also gives tells us that $x - 1 < x < x + 1$.
(ii) If $x = 0$, then you're done. Else, you may take $z = x^{-1}y.$ ($x^{-1}$ exists because every non-zero element in a field has a multiplicative inverse.)