Rules of Inference Have a couple of questions...
How do I show invalid arguments?

If $x$ is a real number such that $x > 1$, then $x^2 > 1$. 

Suppose that $x^2$ is $> 1$, then $x > 1$. Okay, I know that this is wrong due to 
$P \rightarrow Q$,
$ Q$,
therefore $P$... is invalid, but how do I show or verify this? 
 A: From Propositional Logic, $P \rightarrow Q \vdash \lnot Q \rightarrow \lnot P$, but the inference from "$P \rightarrow Q$" to "$Q \rightarrow P$" is NOT correct.
"The fallacy of affirming the consequent" is the following WRONG proof : assuming the fallacious $( P \rightarrow Q ) \rightarrow ( Q \rightarrow P )$, you can have :


*

*"$P \rightarrow Q$" --- assumption

*$Q$ --- assumption

*$( P \rightarrow Q ) \rightarrow ( Q \rightarrow P )$ --- fallacious !!

*$P$ --- from 3 with 1&2 and two applications of ($\rightarrow$-elimination)
A: The formal definition of an invalid argument ( or, equivalently, of an invalidation of an argument) is that there is an assignment( example) where the premises are true, but the conclusion is false. So, in case you want, e.g., to show that $x^2>1 \rightarrow  x>1$ is false, then you just find a case where $x^2>1$ but $x \leq 1$. An invalid argument is one where the premises are true, but the conclusion is false. 
So, if you wanted to show your argument is valid, you must show that, whenever $x^2>1$, it must be the case that $x>1$. If this does not hold, then you can show that the argument is not valid by giving an example of a number $x$ , where $x^2>1$, but $x \leq 1$ (notice that $x\leq 1$ is the negation of $x>1$).
EDIT: the converse of the statement $x>1$ , then $x^2>1$ is the statement: if $x^2 >1$ , then $x>1$ is false; take the example of $x=-2$. Then $x^2=4>1$ , but $-2 \leq -1$ . So you showed the converse is invalid , by showing there is a case where the premise $x^2>1$is true, but the conclusion is false.   
