$ x > y \implies -x < -y$, why? Following property, perhaps, is something that everyone learns in introductory math courses:

If $x > y$ then $-x < -y$.

While it may seem obvious, but well, why exactly is it true?
In other words, suppose someone asserts $\exists x \exists y(x> y \land -x ≥ -y)$, how do you disprove it?
 A: I would add $-x-y$ to both  sides of $x >y$ to get $-y >-x$ which is equivalent to $-x <-y$. 
A: To disprove the hypothetical assertion, I would add $x+y$ to both sides of the inequality $-x\ge-y,$ to obtain $y\ge x,$ contradicting $x>y.$ A similar approach lets you prove $-x<-y$ rigorously from $x>y.$
As for the intuition behind it, $x>y$ means that $x$ is to the right of $y$ on the number line (under the convention of increasing from left to right). $-x$ is the number on the number line that is the same distance from $0$ as $x$ is, but on the opposite side; likewise with $-y.$ Put another way, we can consider $-x$ and $-y$ to be the reflections of $x$ and $y$ (respectively) across $0.$ This reflection reverses the orientations, so that $-x$ is to the left of $-y,$ meaning $-x<-y.$
A: Note: $(-x),(-y)$ are the additive inverses of x,y resp., i.e. 
$(-x)+x=0$; and $(-y)+y=0$.
Translational invariance:
$(-x)+x >(-x)+y;$
$0>(-x)+y$;
Translational invariance::
$0+(-y) >((-x)+y)+(-y);$
$(-y) >(-x)+ (y+(-y))$;(associativity)
$ (-y) > (-x) +0=(-x);$
A: Recall that $x>y$ means that 
$$
x-y>0.
$$ 
But then
$$
-(x-y)=y-x=(-x)-(-y)<0,
$$
i.e. $-y>-x$.
A: See $3 >2$ but $-3 <-2$ when you multiply or divide an inequation by a negative quantity the sign of the inequation changes. If the multiplying quantity is positive the sign of inequation does not change.
