# Intersection of Rationals and (0 to Infinity)?

1. Mar 13, 2012

### IntroAnalysis

1. The problem statement, all variables and given/known data
Let A = [Q$\bigcap$(0,$\infty$)] $\bigcup$ {-1} $\bigcup$(-3, -2]

2. Relevant equations
So A = (0,$\infty$) $\bigcup${-1} $\bigcup$(-3,-2]

3. The attempt at a solution
I understand that the Rational numbers are cardinally equivalent to (0,$\infty$),

but why isn't the intersection of Rationals and (0,$\infty$) =>(0,$\infty$)\Irrationals ?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 13, 2012

### sunjin09

Quite the contrary, the Irrationals are cardinally equivalent to (0,$\infty$)

3. Mar 13, 2012

### HallsofIvy

Staff Emeritus
The set of all rational numbers is countable, unlike $[0, \infty)$.

4. Mar 14, 2012

### IntroAnalysis

Then back to my original question why is the intersection of rationals and (0,∞) = (0,∞)

in other words, why don't irrationals come out of this intersection?

5. Mar 14, 2012

### clamtrox

If x belongs to the intersection of A and B, then x belongs to A and x belongs to B. The intersection of rationals and (0,∞) is the set of numbers which are both rationals and positive real numbers.

6. Mar 14, 2012

### HallsofIvy

Staff Emeritus
It isn't. The intersection of the set of all rational numbers with all positive real numbers is all positive rational numbers.