Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In my homework, we are given the following set $M = \{ (x, y) \in \mathbb{R}^2\, |\, x^2 + y^2 \leq 1 \}$.

Obviously, this represents the set of all points $(x, y)$ that lie within a circle of radius $1$.

However, I'm confused about the $\mathbb{R}^2$, I know that is usually means "all positive real numbers", but could it in this case mean $\mathbb{R}\times\mathbb{R}$ (Cartesian product) since we have a two dimensional set?

share|cite|improve this question
It usually means the cartesian product. Could it be that you confuse it with $\mathbb{R}^{+}$? – t.b. May 8 '11 at 9:10
I suspect the OP thought $\mathbb{R}^2$ meant the set $\{ x^2 : x \in \mathbb{R} \}$, which in all fairness is not completely unjustified, since we can write e.g. $\mathfrak{m}^2$... – Zhen Lin May 8 '11 at 10:26
Yes, that's exactly why I thought that. However $\mathbb{R}\times\mathbb{R}$ makes more sense now. – Hannesh May 8 '11 at 14:30
up vote 8 down vote accepted

No. $\mathbb{R}^2$ is not the set of positive real numbers. I do not know of any such convection. $\mathbb{R}^2$ is $\mathbb{R} \times \mathbb{R}$.

share|cite|improve this answer
on your last statement, you should correct the cross product to remove the 'squared' term. – mixedmath May 8 '11 at 9:13
I must've misheard something along the way then. Thanks! – Hannesh May 8 '11 at 9:16
@mixedmath Thanks, but I'm new to latex. @Theo Thanks for editing. – Dinesh May 8 '11 at 9:16
Another small LaTeX-thing: It's better to write \mathbb{R}^2 instead of \mathbb{R^2}. Compare $\mathbb{R^n}$ (\mathbb{R^n}) to $\mathbb{R}^n$ (\mathbb{R}^n). – t.b. May 8 '11 at 9:20
@Theo. Yeah, good point. Thanks – Dinesh May 8 '11 at 9:23

It means 2d co-ordinate space.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.