Arrow lab

P is sufficient for Q when every case of P is a case of Q: the set P sits inside Q. P is necessary for Q when Q cannot happen without P: Q sits inside P. Build two sets and watch which way the arrow points.

Try

$P = (3, \infty)$ $Q = (2, \infty)$

P QDrag a bracket, or use the editor below. $[$ includes the end, $($ leaves it out.

Sufficient, not necessary

$P \Rightarrow Q, \quad Q \not\Rightarrow P$

Read it aloud

$x > 3$ is sufficient for $x > 2$ : whenever $x > 3$ , also $x > 2$ .
$x > 3$ is not necessary for $x > 2$ : $x = \tfrac{5}{2}$ has $x > 2$ without $x > 3$ .

P \subset Q

P

(3, \infty)

Q

(2, \infty)