Ordered Field: $|x|\le y$ iff $-y\le x\le y$ I had a question regarding this part of a theorem that describes the inequalities of the absolute value function for order field $\mathbb{F}.$ Here is the theorem:
Theorem: Let $\mathbb{F}$ be an ordered field and $x,y\in\mathbb{F}$. Then
(a)$|x|\geq0$ and $|x|=0\Leftrightarrow{x}=0;$
...
(d) $|x|\leq{y}\Leftrightarrow-y\leq{x}\leq{y}.$
I am trying to prove statement "(d)," as shown above. Here is my proof:
Proof: We prove that (A) $|x|\leq{y}\Rightarrow-y\leq{x}\leq{y}$ and (B) $-y\leq{x}\leq{y}\Rightarrow|x|\leq{y}.$
(A) We know that $|x|=x.$ Thus, $|x|\leq{y}=x\leq{y}.$ Knowing also that $|x|=|-x|,$ then $|-x|\geq-y=x\geq-y.$ So, we know now that ($x\leq{y}$ and $x\geq-y$)$=-y\leq{x}\leq{y}.$  $\therefore |x|\leq{y}\Rightarrow-y\leq{x}\leq{y}.$
(B) $-y\leq{x}\leq{y}=$ ($x\leq{y}$ and $x\geq-y$). If $x\leq{y}$, then $|x|=y$ is also true by (a). It is the same for $x\leq{y}.$ Thus, this shows that $|x|\leq{y}.$ $\therefore -y\leq{x}\leq{y}\Rightarrow|x|\leq{y}.$ $\blacksquare$  
I need to know if I had written this proof right.
 A: Your proof, at least as written, is wrong.


*

*We do not know that $|x| = x$. Since most of what you wrote for (A) depends on that claim, it's not useful.

*You seem to be using $=$ to equate inequalities, which is incorrect: $|x| \leq y = x \leq y$ means that all three of the statements $|x| \leq y$, $y = x$, and $x \leq y$ are true. Try using words to correctly express your intent, or if you must, write $|x| \leq y \Longleftrightarrow x \leq y$ to mean "$|x| \leq y$ if and only if $x \leq y$". You might use $\equiv$ instead of $\Longleftrightarrow$, but in my opinion that doesn't read well.

*I think at the end of your proof of A, you use the fact that you have proven $x \geq -y$. However, nowhere in your proof have you claimed to have proven that statement: the closest you got was to claim it is equivalent to $|-x| \geq y$.

*In your proof of (B), you say you are using (a) to conclude $|x| = y$; however, nothing in (a) directly resembles either that claim, or the hypothesis you are invoking.

*It is not clear what you mean by "it is the same for $x \leq y$".

*Nor is it clear how you are therefore inferring $|x| \leq y$.

I think the proof form you were trying to invoke for part (A) was


*

*Suppose that $|x| = x$. Insert argument here. Therefore $-y \leq x \leq y$.

*Suppose that $|x| = -x$. Insert argument here. Therefore $-y \leq x \leq y$.

*Since $|x| = x$ or $|x| = -x$, we conclude $-y \leq x \leq y$.


although I would normally use "$x \geq 0$ or $x \leq 0$" rather than "$|x| = x$ or $|x| = -x$".
