This seemingly easy proof is giving me some trouble..

For every number n $\in$ $Z$, if $n>n^2+1$ then $n\leq0$.

I find that proving the conditional statement P implies Q is false. For example $n=-2$, it is false that $-2>(-2)^2+1$.

But proving the contrapositive is true (not Q implies not P). If $n>0$ then $n\leq n^2 +1$

Can someone point out where I might be going wrong? Thanks!

  • $\begingroup$ Have you tried sketching a graph of $n$ and $n^2+1$? $\endgroup$ – Mark S. Jan 13 '16 at 17:46
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    $\begingroup$ Mark - Thank you, I didn't think of that. The statement seems true after all! P implies Q is a False implying anything, so the statement should be true which which is consistent now with the contrapositive being true. Thanks! $\endgroup$ – Justin Jan 13 '16 at 17:50

The statement given is false. Note that $n^2+1$ is strictly positive so if $n>n^2+1$ then $$n>n^2+1>0$$

  • $\begingroup$ Ok - but why does it seem the contrapositive is true? Did I state it correctly? For the contrapositive I would have the statement: $0<n<n^2+1$, which is true. $\endgroup$ – Justin Jan 13 '16 at 17:38
  • $\begingroup$ Daniel, you may want to reread the whole statement. @Justin, that's not quite the contrapositive. $\endgroup$ – Mark S. Jan 13 '16 at 17:47
  • $\begingroup$ The statement given in the problem is true, not false. Justin has written the reason why in a comment in his question. $\endgroup$ – Mark S. Jan 13 '16 at 17:52
  • $\begingroup$ But $n^2+1>0$... @MarkS. whatt am I missing here? $\endgroup$ – fosho Jan 13 '16 at 17:54
  • $\begingroup$ Certainly, $A>B>0$ always implies $A>0$. But does that mean it can't also imply $A\le0$? $\endgroup$ – Mark S. Jan 13 '16 at 18:03

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