Inference from Multiplication and Inequality

Theorem

Let $a,b,c\in \mathbb{R}$.

• $(a>b\wedge b>c)\Rightarrow a>c$

• $a>b\Rightarrow (a+c>b+c)$

• $(a>b\wedge c>0)\Rightarrow ca>cb$ $(a>b\wedge c<0)\Rightarrow ca<cb$

Proof

Let $ab>0$. Then $a\ne 0$ and $b\ne 0$. (If either $a=0$ or $b=0$, then from 2.1.3 (2) we have $ab=0$, which violates tricothomy property)

Because $a\in \mathbb{R}$, and $a\ne 0$, from Tricothomy Property, $a\in \mathbb{P}$ or $-a\in \mathbb{P}$.

Consider all cases for $a$:

• If $a\in \mathbb{P}$, then $\frac{1}{a}\in \mathbb{P}$, so $b=\frac{1}{a}(ab)\in \mathbb{P}$.
• If $-a\in \mathbb{P}$, then $-\frac{1}{a}\in \mathbb{P}$, so $-b=-(\frac{1}{a})(ab)\in \mathbb{P}$.

Both cases guarantees there could only be 2 possibilities:

1. $a>0$ and $b>0$, or
2. $a<0$ and $b<0$