True statement with a false contrapositive? Statement: $\forall m, n\in \mathbb{Z}$, $m$ is even and $n$ is even $\implies m\cdot n$ is even (true)
Contrapositive: $\forall m, n\in \mathbb{Z}$, $m\cdot n$ is odd $\implies m$ is odd or $n$ is odd  (false)
???
it should be true where $m\cdot n$ is odd $\implies m$ is odd and $n$ is odd, but that's not the contrapositive.
I'm confused
can anyone help me?
 A: The first statement is true, but overly strong.  We don't need both $m,n$ even to get an even product.
As a result, the contrapositive is true, but overly weak.  If $mn$ is odd, then is is true that one, or the other, or both are odd.  In actuality we know that both must be odd but the weaker statement $m$ or $n$ is odd is certainly true.
A weak statement isn't false (unless the statement is claiming an if and only if claim); it's just weak.  
.....
If I had a lot of data and statements and I did a bunch of logical and correct manipulations, an arrived to the conclusion:  New York City has a population of at least 17 people---  that's not a false conclusion;  it's an utterly ridiculous and comically weak conclusion.  But it is true.  The population of New York City is at least 17 people.
This is similar.  If $mn$ is odd the $m$ or $n$ is odd.  Yes, in fact they are both odd.  
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Consider the four cases:
$m$ odd; $n$ odd.  $mn$ odd.  The output is counter to the positive so this will be considered by the contrapositive.
$m$ odd; $n$ even. $mn$ even.  This case was ignored by the positive but doesn't counter the output.  As it was ignored by the positive, it will not be ruled out by the contrapositive.
$m$ even; $n$ odd. $mn$ even.  Same as above.  This was ignored by the positive and so will not be ruled out by the contrapositive.
$m$ even; $n$ even. $mn$ even.  This is the only case considered by the positive that gives a definite result.  Therefore this is the only case that will be ruled out by the contrapositive when we consider the other result.
Positive:  If $m$ and $n$ are both even; then $mn$ is even.
Literal but lazy Contrapositive:  If $mn$ is not even; then $m$ and $n$ are not both $even$.
Using more natural language and the knowledge that "not even = odd" and "not both A and B = (not A) or (not B)" we have
Contrapositive: If $mn$ is odd; then $m$ is odd or $n$ is odd.
.....
Our impression is that the contrapositive is too weak, but that's because the positive was too strong.
We we really want is:
Positive: $m$ or $n$ is even; then $mn$ is even.
Contrapositive: If $mn$ is not even; then it is not true that at least one $m$ or $n$ is even.
Or: If $mn$ is odd; then $m$ and $n$ are odd.
That's a positive that was broad enough to be necessarily complete, which results in a contrapositive that is strong enough to cover every thing.
BUT... that wasn't the positive we were given, overly strong positives result in weak contrapositives and vice versa.
A: You have written the correct contrapositive, and it is true. If $m \times n$ is odd, then $m$ is odd or $n$ is odd.
You are also correct in noticing that in fact, if $m \times n$ is odd, then $m$ and $n$ are both odd. This is a stronger statement, since notice that
$$ \mbox{$m$ and $n$ are both odd } \implies \mbox{ $m$ is odd or $n$ is odd}$$
This stronger statement is the contrapositive of
$$ \mbox{$m$ is even or $n$ is even } \implies \mbox{ $m \times n$ is even}$$
A: The contrapositive is true. If m×n is odd, at least one of m, n is odd, since both are odd.
A: Conditional Statement: If both the two numbers are even then their product is even.
Contrapositive statement: If the product of two numbers is not even then both numbers can't be even.
Observe that your observation doesn't contradict anything.
