Need clarification on what makes an argument invalid or valid In the argument below the hypotheses and conclusion are all true:
$5$ is not an even number.
If $5$ is an even number, then $7$ is an even number.
$\therefore 7$ is not an even number.
The argument is invalid. But I was told that to show an argument is invalid the hypotheses must be true while the conclusion is false. And in this case hypotheses and the conclusion are true which is what makes an argument valid. I don't understand why its invalid if:
True.
False $\rightarrow$ False.
$\therefore $True
 A: Generally speaking for statements $A,B$ from $A\Rightarrow B$ you may not conclude, that $\neg A \Rightarrow \neg B$, which is what is being done in the argument.
A: If $A$ means "5 is an even number" and $B$ means "7 is an even number", then formally the argument is:

$1.\; \lnot A$
$2.\; A \implies B$
$C.\; \lnot B$

However, notice that (2) is equivalent to:

$2a. \lnot B \implies \lnot A$

This is called the contrapositive of (2). Assuming that $\lnot A$ is true, means that $\lnot B$ could be either true or false, because $\lnot A$ being true means it doesn't matter what truth value $\lnot B$ takes, (2a) will be true regardless (the only way for it to be false would be if $\lnot B$ was true AND $\lnot A$ was false). Therefore it doesn't follow that $\lnot B$ is necessarily true.
A: 
... i was told that to show an argument is invalid the hypotheses must be true while the conclusion is false.

Please know that an argument is invalid if and only if it is possible for the premises to be true and the conclusion to be false.  Whether it is actually the case that the premises are true and conclusion is false is not relevant when it comes to validity.
Of course, one way to demonstrate that an argument is invalid, i.e. to demonstrate that it is possible for the premises to be true and the conclusion to be false, is to create a possible world or scenario in which the premises are true and the conclusion is false.
So, while in the 'normal' world of arithmetic 5 and 7 are both not even, try to imagine a world where 5 is not even, but 7 is.  Now, I understand that may be difficult to do, but one way to think about this is that ultimately logic is only about symbol structures ... logic has no 'understanding' of what '5', '7', or 'even' mean ... as far as logic is concerned, to say that '7 is even' is not any different from a claim like 'snow is white': it just says that there is some object that happens to be labeled '7', has some property that happens to be labeled 'even'. And if we have some further object labeled '5', and if we assume that that object does not have the property of 'even', then in that imagined scenario or world, both premises are true, but the conclusion is false, meaning that the argument is invalid.
So again, the premises don't have to be true and the conclusion false in our world for an argument to be invalid. You just need to have a logically possible world (which for classical logic means: claims cannot both be true and false at the same time) in which that is so ... so that need not be our world (or, in this case, the classical world of arithmetic).
A: As others have mentioned, the general form of the argument is invalid, except on a technicality. That technicality is that most proof systems allow you to conclude anything that’s already been proven, so assuming “7 is not even” has already been proven, then you can use any assumption to re-prove it. This is the gist behind the common axiom schema $P \to (Q \to P)$. This same general idea can be used as an inference rule which allows you to introduce anything that already has a proof into another proof or derivation.
