The book 'How to Read and Do Proofs' by Daniel Solow has the following exercise:
Find the contrapositive of the proposition, “If $n$ is an integer for which $n^2$ is even, then $n$ is even.”
I thought the answer would be: "If $n$ is not even, $n$ is not an integer or $n^2$ is not even"
(I reasoned that the condition in the exercise statement is an 'and' statement and hence it's negation involving 'or', according to De Morgan's law)
But the solution to this is given as: "If $n$ is an odd integer, then $n^2$ is odd"
I have these doubts:
- Why is my answer wrong, if it is?
- Is my negation of the condition, and the reasoning behind it, right?
- How is not("$n$ is even") equivalent to "$n$ is odd". Aren't there numbers which are neither odd nor even?
- Let $A$ be any statement. Can any statement $B$ which is logically equivalent to $\mathrm{not}(A)$ be called the negation of $A$?