If one has the inequality

$$A> B$$

and also

$$B \geq C$$

does that also imply

$$A > C$$?

Edit: Additionally, if $$A \geq B$$ and $$A > C$$, does that imply $$B > C$$?

• $A>B\geq C \implies A>C$ Aug 13 at 11:14

Yes: $$A>B$$ implies $$A-B > 0$$ and $$B\geq C$$ implies $$B-C \geq 0$$. Therefore we have

$$A-C = (A-B) + (B-C) \geq A-B > 0$$

which means $$A > C$$.

Regarding your edit: The answer is no, consider $$A=2, B=0, C=1$$.

• Why is $B \ge 0$?
– Fred
Aug 13 at 11:16
• Sorry, I mean $B \geq C$. Aug 13 at 11:18

On the contrary, suppose that $$A \leq C$$. Since $$A \leq C$$ and $$C \leq B$$, by transitivity, we have $$A \leq B$$, which contradicts $$A > B$$.