Prove that if $aProve that if $a<b$, then $-b<-a$
I'm a bit lost in this one, this is what I did:
First case: $0<a<b$
$|a|<|b|$, so  $|-a|<|-b|$
Since both are negative and $|-b|$ is greater than $|-a|$, then $-b<-a$
Second case: $a<0<b$
Since $a$ is negative, then $-a$ is positive.
Since $b$ is positive, then $-b$ is negative.
So clearly $-b<-a$
Third case: $a<b<0$
$|a|>|b|$
$-a$ and $-b$ are positive and $|-a|>|-b|$
So $-b<-a$.
Is this right? It seems clear enough to me, but I'm not sure if it's the right way to prove this.
 A: Assume: $\exists a, b. a<b \wedge -b\geq-a$
Then: $a-a\le a-b<b-b \implies 0\le a-b<0$ but $a\neq b$ so we have derived a contradiction, hence, $\neg[\exists a, b. a<b \wedge -b\geq-a]$ is true, which implies:
$$\forall a,b. a\geq b \vee-b<-a$$
Thus $a<b \implies -b<-a$ is true when $a<b$ is true.
A: In every ordered field 1. if x>0 then -x<0
                       2. if x>0 then y>z ⟹ xy>xz
then x(y-z)>0 implies -(x(y-z))=(-x)(y-z)<0
so that -xy<-xz
A: Let $a,b,c$ be integers.

Lemma 2. $a>b$ if and only if $a-b$ is a positive natural number.
Lemma 3 (Addition preserves order). If $a>b$, then $a+c>b+c$.

You can prove Lemma 3 by Lemma 2.
Proof. We have $a<b$. By Lemma 3, adding $-a$ and $-b$, we have, $-b<-a$.

Maybe you need consider this definition:

Definition 1 (Ordering of the integers). Let $n$ and $m$ be integers. We say that $n$ is greater than or equal to $m$, and write $n\ge m$ or $m\le n$, iff we have $n=m+a$ for some whole number $a$. We say that $n$ is strictly greater than $m$, and write $n>m$ or $m<n$, iff $n\ge m$ and $n\ne m$.

A: Are you allowed to assume the addition property of inequality, i.e.,
if $x>y$, then $x+a>y+a$ for any $a\in\mathbb{R}$?
If yes, then the proof can be made significantly simpler.
A: If you are working in an ordered field $F$, then this is almost a direct consequence of the axiom which states that for all $a,b,c\in F$, if $a<b$, then $a+c<b+c$.
If we assume that $a<b$, then we have
$$
-b=(a-b)-a=a-(b+a)<b-(b+a)=-a
$$
where we add $c=-(b+a)$ to $a$.
