Originally Posted by

**swtdelicaterose** $\displaystyle

\forall q \in \textbf{Q}, \exists r \in\textbf{Q}$ so that $\displaystyle q + r \notin \textbf{Z}$ and $\displaystyle qr \in \textbf{Z}

$

(where Q is set of all rationals, and Z is set of all integers)

So my approach was to write out the contrapositive, which is the same statement logically. Which is (I think):

$\displaystyle

\exists q \in \textbf{Q}, \forall r \in\textbf{Q}$ so that either $\displaystyle q + r \in \textbf{Z}$ or $\displaystyle qr \notin \textbf{Z}

$

I'm not sure about the and statement. Is this contrapositive correct?

(if it is correct, then)

So now I just need to let q = a rational number and find a counter example r so that one of the statements in the or fails. So for example, let q = 0 which is rational, and r = 1, q + r is an integer, but qr is also an integer, however since q+r is an integer, the or statement is true, which should mean the original statement is false?

Is my logic correct? I'm not asking for how to prove this, I'm just asking if I can form a proof from what I've established so far.

Thanks!