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That was a misprint. He meant A^2B^2. The point is (AB)^2= ABAB is NOT equal to A^2B^2 because matrix multiplication is not, generally, "commutative". For most matrices BA is not equal to AB.

I simply cannot bring myself to say "factorized" instead of the simpler "factored"!

Would you like to see my tan line?

What do you mean by "least common multiple" of non-integer numbers?

And I treat bourbon as water. Look where that got me.

"English is the most ambiguous language in the world." I don't know what you mean by that. Could you say it in some other language?

(Moderator's Note: Post moved from this thread.) It could be worse! I remember a colleague showing me an essay turned in by one of his students. In the middle of one page it said "see figure 12.2". Of course there was no figure!

And you thought it would be amusing not to tell us what the problem really asked you to do?

How can you say what the person who set the problem intended? As I said before, this problem can be done without any Calculus at all.

What do you get when you divide the circumference of a pumpkin by its diameter? Pumpkin \pi!

Oh, I think it was mildly challenging! I did it using "Venn Diagrams".

For many months now (possibly as soon as I started coming here) I have been getting messages saying "only secure content is displayed" and have to click on "show all content" to get the full board. But recently I am also getting "There is a problem with this website's security certificate" and...

And what do we learn from this?

Strictly speaking, critical points are points and so are neither values of x or f(x) but (x, f(x))

Find two sets, P and Q, in R^2, such that 1) P and Q are both contained in the square with vertices at (1, 1), (1, -1), (-1, -1) and (-1, 1). 2) P contains the diagonally opposite points (1, 1) and (-1, -1) and Q contains (1, -1) and (-1, 1). 3) P and Q are disjoint. 4) P and Q are both...

Here's a problem I came up with many years ago: Give two sets P and Q such that 1) P and Q are completely contained in the square -1\le x\le 1, -1\le y\le 1. 2) P contains the diagonally opposite points (-1, -1) and (1, 1). 3) Q contains the diagonally opposite points (-1, 1) and (1, -1)...

Isn't it wonderful that rectangles only have integer length and width.