# Write inverse, converse, and contrapositive following statement

#### Joystar1977

##### Active member
Write the inverse, converse, and contrapositive of the following statement:

upside down A x E R, if (x + 2) (x - 3) > 0, then x < -2 or x > 3

Indicate which among the statement, its converse, its inverse, and its contrapositive are true and which are false. Give a conterexample for each that is false.

Please help me with this math problem because I am totally lost and don't understand it at all.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
My guess is that you don't know the definitions of these types of statements. Why don't you start by reading about them in your textbook or Wikipedia? Note that in constructing the inverse, converse, and contrapositive you are supposed to leave the universal quantifier alone and just change the implication. For example, the converse of $\forall x\,(P(x)\to Q(x))$ is $\forall x\,(Q(x)\to P(x))$.

In plain text, you can write "for all" for ∀ and "in" for ∈.

#### Joystar1977

##### Active member
Doesn't the converse, contrapositive, and Inverse as follows:

q arrow p is the converse of p arrow q

slash bar q arrow slash bar p is the contrapositive of p arrow q

slash bar p arrow slash bar q is the inverse of p arrow q

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Yes, this is correct. But you need to write inverse, etc., for the concrete statement given in post #1.

I suggest using notations from this post.

#### Joystar1977

##### Active member
Is another way of saying this as follows:

Statement: if p then q

Converse: if q then p

Inverse: if not p then not q

Contrapositive: if not q then not p

Statement:

upside down A x E R, if (x + 2) (x -3) > 0, then x < -2 or x > 3

So

p : if (x + 2) (x - 3) > 0

q: x < -2 V x > 3

Am I suppose to solve this problem like an inequality or algebraic equation?

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Statement:

upside down A x E R, if (x + 2) (x -3) > 0, then x < -2 or x > 3

So

p : if (x + 2) (x - 3) > 0
No "if".

q: x < -2 V x > 3
Correct.

Am I suppose to solve this problem like an inequality or algebraic equation?
No, you are supposed to
Write the inverse, converse, and contrapositive of the following statement:

upside down A x E R, if (x + 2) (x - 3) > 0, then x < -2 or x > 3
Replace p and q in the statements from the beginning of post #5 by the expressions you found later in that post.

And please use the notation suggestion from post #2:
In plain text, you can write "for all" for ∀ and "in" for ∈.

#### Joystar1977

##### Active member
Is this right for the Converse?

q -> p, or "if q then p" translates to

If x < -2 V x > 3 then (x + 2) (x - 3) > 0

Would one of these be the contrapositive or inverse?

If x > -2 V x < 3 then (x +2) (x-3) < 0

If x > -2 V x < 3 then (x + 2) (x-3) > 0

If x > -2 V x > 3 then (x + 2) (x - 3) > 0

If x > -2 V x > 3 then ( x + 2) (x - 3) < 0
I don't quite understand this because what I see in front of my face is an Inequality

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Is this right for the Converse?

q -> p, or "if q then p" translates to

If x < -2 V x > 3 then (x + 2) (x - 3) > 0.
Yes, this is the converse.

Would one of these be the contrapositive or inverse?

If x > -2 V x < 3 then (x +2) (x-3) < 0

If x > -2 V x < 3 then (x + 2) (x-3) > 0

If x > -2 V x > 3 then (x + 2) (x - 3) > 0

If x > -2 V x > 3 then ( x + 2) (x - 3) < 0
No. As a first try, you could keep the negations. As I said, if you replace p with (x + 2)(x - 3) > 0 and q with x < -2 ∨ x > 3 in ~q -> ~p, you get ~(x < -2 ∨ x > 3) -> ~((x + 2)(x - 3) > 0). Simple, isn't it? Then we can apply simplification and remove negations if we want. Note that ~(x < y) is not (x > y), but (x ≥ y). To figure out ~(x < -2 ∨ x > 3) we apply the De Morgan's law: ~(A ∨ B) = ~A ∧ ~B. Therefore, ~(x < -2 ∨ x > 3) is ~(x < -2) ∧ ~(x > 3), which is x ≥ -2 ∧ x ≤ 3. Some people abbreviate this as -2 ≤ x ≤ 3. Altogether, the contrapositive is (x ≥ -2 ∧ x ≤ 3) -> (x + 2)(x - 3) ≤ 0.