Converses and negations

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One of our first-year students asked (on Piazza) whether converse and negation were the same thing. One of my colleagues explained the differences in terms of propositional logic. I added some comments afterward to see if some specific examples might help. I don’t know whether this helped or not! Here is what I said.

Dr Feinstein adds:

Sometimes specific examples can help here.

So consider the following statement, which we will call statement \(P\):

  (\(P\))                          \(6\) is divisible by \(3\)

Then the negation of \(P\) is the statement \(\neg P\):

 (\(\neg P\))                   \(6\) is not divisible by \(3\)

Notice here that \(P\) is true and \(\neg P\) is false. 

Consider instead the statement \(Q\):

 (\(Q\))                            \(3\) is divisible by \(6\)

and its negation

 (\(\neg Q\))                    \(3\) is not divisible by \(6\)

This time \(Q\) is false and \(\neg Q\) is true.

This is no coincidence, because for any (correctly parsed) mathematical statement \(S\), exactly one of the statements \(S\) and \(\neg S\) will be true. If this doesn't happen, then either you have negated incorrectly, or maybe you are in the middle of a proof by contradiction (in which case you expect to arrive at a contradiction).

Notice that these simple statements \(P\) and \(Q\) (and their negations) don't have any implications in them, so taking the converse doesn't even make sense for these.

Once you do have some form of implication (or if) in your statement you can look at taking the converse or the contrapositive. In this case, the contrapositive always has the same truth value as the original statement. But all bets are off when it comes to the converse. For example, consider the following six statements about integers \(n\). (Three "mutually converse" pairs.)

 (\(S_1\))                           If \(n\) is divisible by \(3\), then \(n\) is divisible by \(6\)

 (\(S_2\))                           If \(n\) is divisible by \(6\), then \(n\) is divisible by \(3\)

 (\(S_3\))                           If \(n\) is divisible by \(3\), then \(n\) is divisible by \(5\)

 (\(S_4\))                           If \(n\) is divisible by \(5\) then \(n\) is divisible by \(3\)

 (\(S_5\))                           If \(n\) is odd then \(n^2\) is odd

 (\(S_6\))                           If \(n^2\) is odd, then \(n\) is odd

I won't give all of the details here, because it would be good for you to check that you agree with me! But here, \(S_1\) and \(S_2\) are converses of each other, and one is true but the other is false.

We also have that \(S_3\) and \(S_4\) are converses of each other, but this time they are both false.

Finally, \(S_5\) and \(S_6\) are converses of each other, but both are true. 

Note that the proofs of the (mutually converse) statements \(S_5\) and \(S_6\) are different, and neither statement follows directly from the other!

(Exercise: check my claims above. Hint for \(S_6\): I think the contrapositive of \(S_6\) is easier to see.)

 


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