- #1

- 12

- 0

For any two rational numbers q1<q2, card ((q1,q2) cap Q) = card Q, right?

How to prove it, if it's true?

How to prove it, if it's true?

Last edited:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter kindlychung
- Start date

- #1

- 12

- 0

How to prove it, if it's true?

Last edited:

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 964

I don't know what the "I" in "(q1, q2)-I". The set of integers? So that "(q1, q2)- I" is the set of all numbers that are NOT integers between q1 and q2? If so then the answer is as before: If (q1, q2)- I is the set of all non-integer

- #3

- 12

- 0

((q1, q2) cap Q) is the set of all non-integer rational numbers between q1 and q2. By intuition I know it has the same cardinality as Q, yet I don't know how to build a 1-1 correspondence between it and Q, please help.

- #4

- 12

- 0

Let m, n be any positive integers, any positive element q of Q can be written as [tex]m/n[/tex], and any negative element q of Q can be written as [tex]-m/n[/tex].

Let's define function f as follows:

[tex]f(q) = \frac{q_{1}+q_{2}}{2} + \frac{q_{2}-q_{1}}{2}\cdot\frac{m}{m+n} (q>0)[/tex]

[tex]f(q) = \frac{q_{1}+q_{2}}{2} - \frac{q_{2}-q_{1}}{2}\cdot\frac{m}{m+n} (q<0)[/tex]

[tex]f(q)=0 (q=0)[/tex]

Share: