Metrics (Distances) on $\mathbb{F}$ Theorem Proof

Question

I had a question regarding a Theorem I had come across that described metrics (distances) on ordered field $\mathbb{F}.$ Here it is:

Theorem: If $\mathbb{F}$ is an ordered field, then $d(x,y)=|x-y|$ is a metric on $\mathbb{F}.$

I was told to use this other theorem to support the proof:

Theorem: Let $\mathbb{F}$ be an ordered field and $x,y\in\mathbb{F}.$ Then

(a) $|x|\geq0$ and $|x|=0\Leftrightarrow{x}=0;$

(b) $|x|=|-x|;$

(c) $|x+y|\leq|x|+|y|.$

I would like to know how I could approach this or you could possibly show me how you would prove it.

Answer 1

Hint: Suppose $\mathbb{F}$ is a field and $x,y \in \mathbb{F}$, then $x-y\in \mathbb{F}.$

Here is the first condition: Using (a), we have $|x−y|≥0$ and $|x-y|=0⇔x-y=0$, hence $d(x,y)≥0$ and $d(x,y)=0⇔x=y$

Here is the second condition: Using (b), we have $|x−y|=|−(x-y)|=|y−x|$, hence $d(x,y)=d(y,x).$

And the third condition: Using (e), we have $|x−y|=|(x−z)+(z-y)|≤|x-z|+|z-y|$, hence $d(x,y)\leq d(x,z) + d(z,y).$