Help, Negation and Contrapositive of the following statement?

[Tek1Atom]

Homework Statement

Consider the statement "if x is odd and x is a multiple of 3, then x ≥ 6." write down the contrapositive and negation of this statement?

2. The attempt at a solution

This is what I worked our as an answer but I am pretty sure it's wrong.

Contrapositive

"if x ≤ 6, then x is even OR x is not a multiple of 3"

Negation

"if x is even OR x is not a multiple of 3, then x ≤ 6"

 

[berkeman]

Homework Statement

Consider the statement "if x is odd and x is a multiple of 3, then x ≥ 6." write down the contrapositive and negation of this statement?

2. The attempt at a solution

This is what I worked our as an answer but I am pretty sure it's wrong.

Contrapositive

"if x ≤ 6, then x is even OR x is not a multiple of 3"

Negation

"if x is even OR x is not a multiple of 3, then x ≤ 6"

Welcome to the PF.

Can you please define the two terms in the context of this problem. What are your book's definition of Negation and Copntrapositive?

 

[Mark44]

Tek1Atom, you are close on the contrapositive. For the implication p ==> q, the contrapositive is ~q ==> ~p.

In your problem, q is the statement "x <= 6" What statement is represented by ~q?
Edit: Fixed error.
For the negation of an implication p ==> q, it might be helpful to note that (p ==> q) <==> (~p v q). So the negation of p ==> q (that is, ~(p ==> q)) is equivalent to the negation of (~p v q) (that is, ~(~p v q)). This last expression can be simplified by using deMorgan's Laws.

 

[Tek1Atom]

Thanks Mark44, as you are aware, there are two variables in this statement so it is quite tricky for me to answer at this point in time, I am pretty new to logic, what would the answer to this question be written down in English?

 

[Mark44]

p: x is odd and x is a multiple of 3
q: x >= 6
Edit: Fixed error.
For the negation of p ==> q, you can instead work with the negation of ~p v q; i.e., ~(~p v q). You'll want to simplify this last expression using deMorgan's Law.

You're going to have to work this out yourself, since the rules of this forum expressly prohibit giving the answers to questions. I have defined statements p and q above, based on your problem, so work with the symbols and plug in the appropriate statements to get your answer in English.

 

[vela]

For the negation of an implication p ==> q, it might be helpful to note that (p ==> q) <==> (p v ~q).

Shouldn't that be (~p v q)? The implication is only false when p is true and q is false.

 

[Mark44]

Yes, you're right. I edited the two posts with this error.

 

[Tek1Atom]

Would this be correct?

p: x is odd and is a multiple of 3
q: x ≥ 6
~p: x is even and x is not a multiple of 3
~q: x ≤ 6

so

Contrapositive

if x ≤ 6, then x is even and x is not a multiple of 3

Negation

x is even and x is not a multiple of 3, therefore x ≤ 6

 

[Mark44]

Would this be correct?

p: x is odd and is a multiple of 3
q: x ≥ 6
~p: x is even and x is not a multiple of 3

Almost. ~(Q ^ R) <==> ~Q V ~R
Here ^ means "and", V means "or".

~q: x ≤ 6

If x = 6, then you have both p and ~p being true, which can't possibly happen.

so

Contrapositive

if x ≤ 6, then x is even and x is not a multiple of 3

Negation

x is even and x is not a multiple of 3, therefore x ≤ 6