Help with the contrapositive, converse and inverse of statements.

What is the contrapositive, converse and inverse of the below statements? I am a little confused by these concepts and would like a bit of help pls.

I am doing a uni degree with discreet maths as module and this an exercise in the one of the units. I kinda understand what they are but it all seems a bit muddled so if someone could help me understand these examples that would be great.

I think i understand the words contrapositive, converse and inverse. Am I right in thinking contrapositive is the opposite? For example...

Contrapositive for If P then Q would be If not P then not Q Inverse for If P then Q would be If not Q then not P Converse for If P then Q would be If Q then P.

I know that using "if P then Q" is a super simple example and I understand it this simple but I struggle when the statement gets more complicated like the below.

i) ∀x ∈ ℝ, if x > 3 then x² > 9

ii) ∀x ∈ ℝ, if x(x + 1) > 0 then x > 0 or x < -1

If anyone could help out that'd be fantastic.

1 Answer

If P then Q
Contrapositive: If not P then not Q

Inverse: If not Q then not P

Converse: If Q then P.

Your suggested converse is correct, whereas you have switched around the contrapositive and the inverse.

Note that

• a statement's converse and inverse are logically equivalent to each other,
• a statement is logically equivalent to its contrapositive.

(When statements are logically equivalent, they are either all true or all false.)

i) $∀x ∈ ℝ,$ if $x > 3$ then $x² > 9$

contrapositive: $∀x ∈ ℝ,$ if $x² \leq 9,$ then $x \leq 3$

converse: $∀x ∈ ℝ,$ if $x² > 9,$ then $x > 3$

ii) $∀x ∈ ℝ,$ if $x(x + 1) > 0$ then $(x > 0$ or $x < -1)$

contrapositive: $∀x ∈ ℝ,$ if $-1\leq x\leq0,$ then $x(x + 1) \leq 0$

converse: $∀x ∈ ℝ,$ if $(x > 0$ or $x < -1),$ then $x(x + 1) > 0$

In case you are wondering why $$\text{not }(x > 0 \:\:\text{or }\; x < -1)$$ is equivalent to $$-1\leq x\leq0,$$ just illustrate the $(x > 0 \:\:\text{or }\; x < -1)$ on a number line (the $x$-axis) and notice that its complementary region corresponds to $-1\leq x\leq0.$