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Definition 2. Let $K$ and $L$ be ordered fields. An order-preserving ($f$ is order-preserving means that $x\leq y \implies f(x)\leq f(y)$) ringhomomorphism is called an embedding.

Note that any embedding must be injective since the kernel of a ringhomomorphism is and ideal and $0 \lt 1$.

Definition 2. Let $K$ and $L$ be ordered fields. An order-preserving ($f$ is order-preserving means that $x\leq y \implies f(x)\leq f(y)$) ring homomorphism is called an embedding.

Note that any embedding must be injective since the kernel of a ring homomorphism is and ideal and $0 \lt 1$.

In a bit of an abuse of language, the elements of $I$ can be said to be "infinitely small". For an example of an ordered field which is not archimedean, consider $\mathbb{Q}(X)$, the set of rational functions in one variable over $\mathbb{Q}$ (see [Wikipedia] for how the order is defined) - $0 \lt \frac{1}{X} \lt q$ for all positive rationals $q$.

In a bit of an abuse of language, the elements of $I$ can be said to be "infinitely small". For an example of an ordered field which is not archimedean, consider $\mathbb{Q}(X)$, the set of rational functions in one variable over $\mathbb{Q}$ (see [Wikipedia] for how the order is defined) - $0 \lt \frac{1}{X} \lt q$ for all positive rationals $q$.

Exercise 14. Fill out the details of the following sketch of a proof that any Cauchy sequence in $R$ is convergent:
Proof. Starting with an arbitrary Cauchy sequence $([{\bf q}]^i)$ of equivalence classes, Proposition 11(4) allows us to assume WLOG that for all $i$, $|[{\bf q}]^i - [{\bf q}]^j]| \lt 4^{-i}$. Note that convergence means showing some inequalities hold, so we only need to show those for some representatives, so we are free to pick nice ones. Pick representatives, $(q^i_n)$, such that $|q^i_n - q^i_m| \lt 4^{-n}$. Take ${\bf x} = (q^1_1,q^2_2,\dotsc)$. This is obviously a Cauchy sequence (why?), and we claim that $[{\bf q}]^i \to [{\bf x}]$, so let $\varepsilon \in \mathbb{Q}^+$ be given. Convince yourself that $[{\bf q}]^i \to [{\bf x}]$ means $$\exists K,N : k \geq K, n \geq N \implies \left\lvert q^k_n - q^n_n \right\rvert \lt \varepsilon. $$ Note that the requirement of the existence of the positive $\alpha$ dropped out since the above has to hold for every $\varepsilon > 0$. Now use the standard trick that $|q^k_n - q^n_n| \leq |q^k_n - q^k_k| + |q^k_k - q^n_n|$ and use that the both the sequence and the representatives have that "$4^{-m}$"-property to conclude that the RHS is eventually smaller than $\varepsilon$.

Update 3

Fixed up the proof sketch in the last exercise. Also I keep finding these small annoying errors... I'm sorry that I keep up bumping this thread, I just hope that you don't mind too much.

Exercise 14. Fill out the details of the following sketch of a proof that any Cauchy sequence in $R$ is convergent:
Proof. Starting with an arbitrary Cauchy sequence $([{\bf q}]^i)$ of equivalence classes, Proposition 11(4) allows us to assume WLOG that for all $i,j$ with $j \geq i$ that $\left\lvert[{\bf q}]^i - [{\bf q}]^j]\right\rvert \lt 4^{-i}$. Note that convergence means showing some inequalities hold, so we only need to show those for some representatives, so we are free to pick nice ones. Pick representatives, $(q^i_n)$, such that $|q^i_n - q^i_m| \lt 4^{-n}$. Take ${\bf x} = (q^1_1,q^2_2,\dotsc)$. This is obviously a Cauchy sequence (why?), and we claim that $[{\bf q}]^i \to [{\bf x}]$. Convince yourself that this is the same as this: $$\forall \varepsilon \in \mathbb{Q}^+ \exists K,N : k \geq K, n \geq N \implies \left\lvert q^k_n - q^n_n \right\rvert \lt \varepsilon. $$ Note that the requirement of the existence of the positive $\alpha$ dropped out since the above has to hold for every $\varepsilon > 0$. So let $\varepsilon \gt 0$ be given. We use the standard "trick" of inserting a clever element into the inequality: $|q^k_n - q^n_n| \leq |q^k_n - q^k_k| + |q^k_k - q^n_n|$ and use that the both the sequence and the representatives have that "$4^{-m}$"-property to conclude that if we choose $N,K$ such that $N=K$ and $2^{-N} \lt \varepsilon$ then RHS is eventually smaller than $\varepsilon$.