A Good Example of an Argument That Cannot Be Reversed My students are having a hard time with assuming what they must prove.  A good example of what they are doing is when one wishes to show that $\lim_{x\to 3}(2x+1)=7$.  They will assume that $|(2x+1)-7|<\epsilon$, and then derive $|x-3|<\frac{\epsilon}{2}$.  There is no real harm here, since each step of the argument can be reversed.  But, technically, one should assume $|x-3|<\frac{\epsilon}{2}$ and derive the other inequality.  
I would like a good example of an "If $P$ then $Q$" statement where if one tries to assume $Q$, derive $P$, and then work backwards there is a step along the way that cannot be reversed.
What do I mean by good example?  The "if-then" should be FALSE.  And the statement and steps of the argument should be very easy to understand.  I tried to do one in class using $\epsilon-\delta$, but I think the concept of a limit is a little over their heads right now.  So, I'd prefer the math not get in their way.
EDIT: It is amazing what one bad word can do.  I have edited it to say "The "if-then" should be FALSE".  Thus what @Dan Shved said in his answer where $Q\Rightarrow P$ is true.  I'd prefer one that is a little more subtle than the examples where you divide by $0$ by mistake to prove $1=0$.  The students know to avoid dividing by variables.
 A: Well, maybe this isn't exactly what you want, but here is an algebraic example. Let $P$ say "numbers $a$ and $b$ have the same sign and $a^2 = b^2$". Let $Q$ say "a = b".
Now, if you start from $Q$, you immediately conclude that since $a=b$, their signs are the same, and also $a^2 = b^2$. So $Q$ implies $P$ obviously.
None of the two individual steps above ("$a=b \to a^2=b^2$" and "$a = b \to \operatorname{sign} a = \operatorname{sign} b$") can be reversed. Implication $P \to Q$ is true, but it cannot be proved "working backwards" from the argument above.
I think that maybe a more instructive example would be an implication $P \to Q$ that is false, but $Q \to P$ is true. Then you demonstrate a derivation of $P$ starting from $Q$, and if someone thinks that you have thus proved $P\to Q$, you simply show that $P \to Q$ is not true...
UPDATE: again, I feel that my example above, even though it seems valid, will not drive it home for the students who make this kind of mistake. Maybe you should come up with seemingly correct arguments that sort of prove $4=5$ or something because of sloppy implication reversals. $4=5$ in the conclusion should make a sane person think that something was wrong in the argument. If they see a bad argument that still proves a correct statement - I doubt it will disturb them that much.
A: The question is somewhat unclear. Is this an example? 

If $n$ is an odd prime then $\operatorname{gcd}(n,2) = 1$.



*

*Suppose $m > 1$ divides $n$ and $2$.

*Then $m =2$, because the only divisors of $2$ are $1$ and $2$.

*But, because $n$ is an odd prime, $2$ does not divide $n$.

*Thus the only divisor of $2$ that also divides $n$ is $1$.

*Therefore $\operatorname{gcd}(n,2) = 1$
Step (3) clearly does not reverse. 
