0
$\begingroup$

If $xy$ is odd, then $x$ is odd and $y$ is odd

I was just wondering if the correct contrapositive would involve proving these three cases:

1) $x$ is even or $y$ is odd

2) $x$ is odd or $y$ is even

3) $x$ is even or $y$ is even

I'm not too sure about if the last case is necessary as in the answer to this question only the first two cases were shown. I guess what i'm trying to ask is why do we not check the last case, since it is a possible negation of $x$ is odd and $y$ is odd?

$\endgroup$
2
  • $\begingroup$ No, @user170039 It is the statement of the form $P \land Q$, where $P$ means x is odd, and $Q$ means $y$ is odd, the negation of which is $\lnot (P\land Q) \equiv (\lnot P \lor \lnot Q)$ $\endgroup$
    – amWhy
    Mar 10, 2020 at 19:56
  • $\begingroup$ @amWhy: Indeed. Nice catch. I have removed my comment. $\endgroup$
    – user170039
    Mar 11, 2020 at 5:29

1 Answer 1

Reset to default
2
$\begingroup$

The counter-positive of the assertion

if $xy$ is odd, then $x$ and $y$ are odd

is

if $x$ is even or $y$ is even, then $xy$ is even.

And asserting that $x$ is even or $y$ is even is equivalent to asserting that we have one of the following possibilities:

  1. $x$ is even and $y$ is odd;
  2. $x$ is odd and $y$ is even;
  3. both $x$ and $y$ are even.
$\endgroup$
1
  • 2
    $\begingroup$ though we need only two cases: 1. $x$ is even (and $y$ arbitrary); 2. $y$ is even (and $x$ arbitrary) $\endgroup$
    – Hagen von Eitzen
    Mar 10, 2020 at 13:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .