Do $a>2$ and $b<-2$ imply $-ab>4$? My attempt is, first of all, show that $b<-2$ implies $-b>2$. But it is obvious by the multiplicative property. I then make use of the following lemma:

Let $a$, $b$, $c$ and $d$ be real numbers such that $a>b>0$ and $c>d>0$. Then $ac>bd$.

It then follows that $a(-b)>2(2)$, or $-ab>4$.
However, my teacher told me that $a(-b)>2(2)$ is not correct and I am not convinced by her logic that 'you can't multiply two inequalities together'. I would like to seek for comments from all of you. Thanks in advance.
 A: Your teacher is probably trying to highlight that you can't just multiply two inequalities together in general. What I mean is, we can't say in general that from $b < a$ and $d < c$ we can get $bd < ac$.  As an example, $-1 < 1$ and $-5 < 1$, but that doesn't mean $(-1)(-5) < (1)(1)$, since this gives $5 < 1$, which is false.
BUT, we can do this if we have $0 < b < a$ and $0 < d < c$, as you stated in your lemma.  So the original multiplication in your answer is totally fine.  Your teacher should have looked at it more carefully and better articulated to you that she probably meant she just wants you to be careful because multiplying inequalities doesn't always work (but it works in the special case mentioned in your lemma).
A: I agree with you, you can write
$$-ab=a(-b)>2\cdot 2 =4$$
note that $$b<-2 \implies -b>2$$
A: Your reasoning is correct.
First of all, by your first paragraph $ab < -4$ (notice that the sign is flipped). Then, we have to check three things: the lower bound, the upper bound, and the sign.
As $a$ can be made as large as possible, and $b$ can be made as small as possible, $ab$ has no maximum value. Following this, the smallest value of $ab$ can be obtained when $|a|$ and $|b|$ are minimised. In this case, the smallest value is when $a=2$ and $b=-2$, so the lower bound is correct.
Finally, we can give an example to show the sign is correct. When $a=3$ and $b=-3$, $ab=-9$, so $ab$ should be less (but not equal to) than $-4$.
A: Your reasoning is correct, but perhaps you weren't allowed to use that lemma? Teachers can be picky about what you can and cannot use. And if that's not the case, teachers can be wrong (see comment :-)).
I assume you can use the multiplicative property $x > y \implies xz > yz$ for $z>0$. Then you could do it in two steps, after you already established $b <- 2 \implies -b > 2$.
You have (multiplying both sides with $-b>2>0$):
$$a > 2 \implies \color{green}{-ab} > \color{blue}{-2b}$$
and (multiplying both sides with $2$):
$$-b > 2 \implies \color{blue}{-2b} > \color{red}{4}$$
Now by the transitive property $\color{green}{x} > \color{blue}{y} \mbox{ and } \color{blue}{y} > \color{red}{z} \implies \color{green}{x} > \color{red}{z}$, the result follows.
A: When your teacher said that you can't multiply two inequalities together, she was probably urging caution. And while it's true that you can, under certain circumstances, multiply two inequalities (in the same direction) together, you have to do so carefully.
You're given $a > 2$ and $b < -2$. You validly reversed the second inequality to give $-b > 2$ and multiplied this by the first to (validly) give $-ab>4$. The reason this works is that $-b$ is a clearly positive quantity, and multiplying by it will not change the sign of another inequality.
Let's say you tried this another way. Reverse the first inequality to give $-a < -2$. Now you cannot simply multiply this by $b <-2$ to give $-ab < 4$ (this is invalid). However, if you recognise that $-a$ is a clearly negative quantity, then you know you need to reverse the direction of any inequality you multiply by this quantity. So when you do the multiplication, you need to reverse the sign of the result to give $-ab > 4$, as we have previously arrived at.
Note that disregarding these rules can give some wildly incorrect results. For example, if I were to give you $a<2$ and $b<2$, you cannot simply multiply those two to give $ab<4$ (consider $a=b=-3$, which satisfy the first set of inequalities, but the product $ab = 9$, which clearly doesn't satisfy the resultant (incorrect) inequality we got by naive multiplication). The reason this doesn't work is because $a<2$ implicitly includes the possibility that $a$ can be negative (ditto for $b$). However, if I gave you $a>2$ and $b>2$, it is valid to multiply them together to give $ab>4$.
As I said (and as your teacher probably meant), do it carefully.
