# Bijection

• Feb 21st 2006, 09:10 AM
juef
Bijection
Hey all,

I know there's a bijection between \$\displaystyle \mathbb{R}\$ and \$\displaystyle \rbrack0,1\lbrack\$, but I was wondering if there was one between \$\displaystyle \mathbb{R}\$\\$\displaystyle \mathbb{Q}\$ and \$\displaystyle \mathbb{R}\$\\$\displaystyle \mathbb{Q}\bigcap\$\$\displaystyle \rbrack0,1\lbrack\$.

Can anyone help me? Thank you!
• Feb 21st 2006, 02:51 PM
ThePerfectHacker
What do you mean by \$\displaystyle \mathbb{R}/\mathbb{Q}\$ because that is used to represent a factor ring but the problem is that \$\displaystyle \mathbb{Q}\$ is not an indeal in \$\displaystyle \mathbb{R}\$. Thus you mean something else by the symbol \$\displaystyle /\$ but I do not know what you mean.
• Feb 21st 2006, 03:29 PM
juef
Sorry for the confusion, the backslash would mean, in this case, \$\displaystyle \mathbb{R}\$ without \$\displaystyle \mathbb{Q}\$, or in other words the irrationals. Indeed, I am not trying to represent a factor ring. :)
• Feb 22nd 2006, 02:42 PM
ThePerfectHacker
Quote:

Originally Posted by juef
Sorry for the confusion, the backslash would mean, in this case, \$\displaystyle \mathbb{R}\$ without \$\displaystyle \mathbb{Q}\$, or in other words the irrationals. Indeed, I am not trying to represent a factor ring. :)

But the cardinality of the irrationals must be the countinuum. Because the cardinality of the rationals is \$\displaystyle \aleph_0\$.