Same amount of irrational numbers in equal size intervals?

Question

Let's say you have the intervals $(q_1, q_2),\ q_i,\in \mathbb{Q}$ and $(q_1+a,q_2+a),\ a\in\mathbb{Q}$.

Is the cardinality of $(q_1,q_2)\cap \mathbb{I}$ equal to the cardinality of $(q_1+a,q_2+a)\cap \mathbb{I}$?

Any interval contains the same "amount" of rational numbers, where I assume that "amount" means "cardinality". Specifically, every interval contains a countably infinite number of rational numbers. — Xander Henderson, Sep 20 '17 at 18:35
If you mean cardinality, then yes, but that doesn't say much: every interval has the cardinality of the continuum, the countable many rational numbers don't change that. A more sensible notion than "amount" would be "measure", and if you choose Lebesgue measure, the answer is "yes", again. — , Sep 20 '17 at 18:39

score 2 · Answer 1 · answered Sep 20 '17 at 18:48

2

The set of irrational numbers between $q_1$ and $q_2$ has Lebesgue measure $q_2-q_1$ and cardinality $\mathfrak c$.

answered Sep 20 '17 at 18:48

Chris Culter

Where can I find a demonstration? – Jorge Martín Pérez Sep 20 '17 at 18:56
1

An upper-division undergrad textbook on real analysis ought to cover both concepts. Personally, I like Pugh. – Chris Culter Sep 20 '17 at 19:03

score 1 · Answer 2 · answered Sep 20 '17 at 18:47

For your specific example, there is a simple 1-1 correspondence between irrationals in these intervals.

Specifically, the map $x \mapsto x+a$ does the trick. That's because if $a$ is rational, then $x$ is irrational if and only if $x+a$ is irrational.

So, yes, if by "amount" you mean cardinality.

2 Answers2