Let $K$ be an ordered field that satisfies the least upper bound property. Prove that there is a unique function $T:\mathbb{Q} \rightarrow \mathbb{Q}_K$ that is order isomorphic. That is, show that it satisfies the following properties: For $a, b \in \mathbb{Q}$, $T(a) + T(b) = T(a + b)$, $T(a) \cdot T(b) = T(ab)$, and $a < b \implies T(a) < T(b)$
Note: Problem above follows from proof about Peano axioms of set of natural numbers in an ordered field. So when $\mathbb{N}_K$ is mentioned, it refers to the set of natural numbers in $K$. Similarly for $\mathbb{Z}_K$. Also, $\mathbb{Q}_K$ is the set of equivalence classes defined as $\mathbb{Q}_K = \{\frac{a}{b} | \frac{a}{b}$ is the equivalence class of a pair $(a, b) \in \mathbb{Z}$ x $(\mathbb{Z}-0)\}$
My attempt: I constructed the functions as follows:
Begin by taking $T(1) = 1$. Then, consider $\mathbb{N}$:
We define $T(n) = 1 + 1 + ... + 1 = N \in \mathbb{N}_K$.
Then, extend it to $\mathbb{Z}$:
We define $-T(n) = T(-n)$.
Finally, extend it to $\mathbb{Q}$:
Define $T(n) = T(\frac{a}{b}) = \frac{T(a)}{T(b)}$, where $n \in \mathbb{Q}$, and $a, 0 \neq b \in \mathbb{Z}$.
Now, we check each property. Assume $a, b, c, d \in \mathbb{Z}$ with $b, d \neq 0$.
$T(\frac{a}{b} + \frac{c}{d}) = T(\frac{ad + bc}{bd}) = \frac{T(ad + bc)}{T(bd)} = T(\frac{ad}{bd}) + T(\frac{bc}{bd}) = T(\frac{a}{b}) + T(\frac{c}{d})$
$T(\frac{a}{b} \cdot \frac{c}{d}) = T(\frac{a \cdot c}{bd}) = T(\frac{a}{b}) \cdot T(\frac{c}{d})$
I am not sure how to proceed.
Uniqueness (TBD).
I was wondering if my construction and proof thus far are correct? I am confused with how to do the inequality part so any help there is appreciated.