# Compatibility in Real numbers?

I'm reading Good Math by Mark Chu-Carroll and it says,

• "$$\le$$" is compatiable with "+" and "*":

1. ...
2. if x $$\le$$ y, then for all z where 0 $$\le$$ z, (x * z) $$\le$$ (y * z).
3. if x $$\le$$ y, then for all z where z $$\le$$ 0, (x * z) $$\le$$ (y * z).

Is that 3. true?

When

x = -2, y = -1, and z = -1

Then

(-2 * -1) $$\le$$ (-1 * -1) ?

• The sign is reversed. Example, $3\leq 5$ but $(3\cdot (-1))\leq (5\cdot (-1))$ is false. Rather, $(3\cdot (-1))\color{red}{\geq}(5\cdot (-1))$. Multiplication by a negative "flips" the inequality sign. – JMoravitz Oct 22 at 14:46

For example, $$-2 \leq 3$$, and $$-5 \leq 0$$, but $$-5 \times -2 = 10 \not\leq -15 = -5 \times 3.$$
1. if x $$\le$$ y, then for all z where z $$\le$$ 0, (y * z) $$\le$$ (x * z). Alternately, you can reverse the $$\le$$ to $$\ge$$ and keep $$x$$ and $$y$$ where they are.