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 $B > C$?

  • 2
    $\begingroup$ $A>B\geq C \implies A>C$ $\endgroup$
    – Lion Heart
    Aug 13 at 11:14

2 Answers 2


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 $A=2, B=0, C=1$.

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


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