# logic question/ contrapositive

• Sep 6th 2009, 10:29 AM
p00ndawg
logic question/ contrapositive
1. If pq is odd, then p is odd and q is odd.

for this one, I had already proved that if p is odd and q is odd then pq is odd.
So in relation to this question i proved that q => p, so with a truth table I used that since p&q are the same(odd) then the reverse(p=>q) is also true.

does that sound alright?

2. if p^2 is even, then p is even.

This one im not quite sure how to prove ive tried the contrapositive:

so if p^2 is not even, the p is not even.

I substituted for p 2k+1, and tried setting it equal to 2k + 1 but my answer doesnt really make sense to me.

any help?
• Sep 6th 2009, 11:33 AM
Plato
Please explain what you are to do with these.
It is not clear what is going on.
What does contrapositive have to do with these.
• Sep 6th 2009, 12:43 PM
Sampras
Quote:

Originally Posted by p00ndawg
1. If pq is odd, then p is odd and q is odd.

for this one, I had already proved that if p is odd and q is odd then pq is odd.
So in relation to this question i proved that q => p, so with a truth table I used that since p&q are the same(odd) then the reverse(p=>q) is also true.

does that sound alright?

2. if p^2 is even, then p is even.

This one im not quite sure how to prove ive tried the contrapositive:

so if p^2 is not even, the p is not even.

I substituted for p 2k+1, and tried setting it equal to 2k + 1 but my answer doesnt really make sense to me.

any help?

2. Suppose p is odd. Then $p^2 = (2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1$ is odd. This is the contrapositive.
• Sep 6th 2009, 12:59 PM
p00ndawg
Quote:

Originally Posted by Plato
Please explain what you are to do with these.
It is not clear what is going on.
What does contrapositive have to do with these.

oops im sorry. the contrapositive had to do with a hint the book gave me to try and prove the statement.

Suppose p and q are integers. prove the following:

Quote:

Originally Posted by Sampras
2. Suppose p is odd. Then $p^2 = (2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1$ is odd. This is the contrapositive.

i think this is right.

so in the end since it is odd, that makes the opposite true?
• Sep 6th 2009, 02:12 PM
HallsofIvy
Quote:

Originally Posted by p00ndawg
1. If pq is odd, then p is odd and q is odd.

for this one, I had already proved that if p is odd and q is odd then pq is odd.
So in relation to this question i proved that q => p, so with a truth table I used that since p&q are the same(odd) then the reverse(p=>q) is also true.

Please don't use the same letters to mean the same thing! You start by talking about p and q being odd, implying that they are integers. You then talk about p&q and p=>q implying that they are statements. Which is it?

And you give a statement, "If pq is odd, then p is odd and q is odd", but don't say what you want to do with it! Prove it? Find the contrapositive of it?

Quote:

does that sound alright?

2. if p^2 is even, then p is even.

This one im not quite sure how to prove ive tried the contrapositive:

so if p^2 is not even, the p is not even.

I substituted for p 2k+1, and tried setting it equal to 2k + 1 but my answer doesnt really make sense to me.

any help?
Proof by contradiction. Suppose p is odd. Then p= 2k+1 as you say. What is its square?