Suppose that $(F, \preceq)$ is an ordered field, that $H \subseteq F$, and that $a \in F$. Suppose further that $\sup H = t \in F$. Finally, let $H+a = \set{x + a \mid x \in H} \subseteq F$. Prove that $H+a$ has a supremum in $F$, and that $\sup(H+a) = \sup(H)+a$.
Prove the following useful facts.
Suppose that $\emptyset \subsetneq E_1 \subseteq E_2 \subseteq \R$. If $E_2$ is bounded above, then $\sup E_1$ exists (in $\R$) and $\sup E_1 \leq \sup E_2$.
(Note: the symbol "$\subsetneq$" means "is a proper subset of", so $A \subsetneq B$ means $A \subseteq B$ and $A \neq B$. Thus $\emptyset \subsetneq E_1$ means "$E_1$ is non-empty". In contrast, the symbol $\nsubseteq$ means "is not a subset of". The symbol "$\subset$" is inconsistently used, usually meaning $\subsetneq$ but sometimes meaning $\subseteq$, depending on the author.)
If $x, y \in \R$ with $0 \leq x \leq y$ then $x^2 \leq y^2$. Moreover, assuming that $x^{1/2}, y^{1/2}$ exist in $\R$ with $0 \leq x^{1/2}, y^{1/2}$, prove that $x^{1/2}\leq y^{1/2}$.
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment. They are coloured by approximate difficulty: easy/medium/hard.
Suppose that $\F$ is a field. If $p \in \N$ is such that \[0_\F = \sum_{i=1}^p 1_\F = \underbrace{1_\F + 1_\F + \ldots + 1_\F}_{p \text{ times}},\] and $p$ is minimal with this property, then $\F$ is said to be "of characteristic $p$". If $\F$ is not of characteristic $p$ for any $p \in \N$, we say that $\F$ is of characteristic zero.
Now, let us suppose that $(\F, \preceq)$ is an ordered field.
Let $F = \Q[X] / \paren{X^2-2}$; this is the field of polynomials with rational coefficients in a single indeterminate $X$, modulo the ideal generated by $X^2-2$. Every element of $F$ is of the form $s + tX$, with addition defined component-wise and multiplication defined by \[(s_1+t_1X)\cdot_F(s_2+t_2X) = (s_1s_2+2t_1t_2) + (s_1t_2+t_1s_2)X.\]