Can the Set of Rational Numbers Be Counted? And the Irrational Numbers?

[dabdobber]

I need help with this math problem:

Show that the set of rational numbers, Q, is countable.

and

Show that the set of irrational numbers is uncountable.

 

[micromass]

What do you know about cardinality already? Are there any theorems you've seen that might be useful?
What did you try already??

 

[dabdobber]

well as far as the first one goes I know that the set of integers Z is a subset of Q and that Z is countable. now I make B as the set of all non-integral rational numbers.

if B is countable, then Q is countable because a union of two countable sets is countable.
that's my approach, but I'm not much of an example person.

and for the second one I'm thinking that I can show the union of the reals IR with the irrational, thus making them uncountable as well?

 

[lanedance]

1) try and set up a 1:1 map between Q and Z
2) show that no such map exists for R-Q

 

[blue_raver22]

Yes, the set of rational numbers, Q, can be counted. This is because the rational numbers can be expressed as a ratio of two integers, and there is a finite number of possible combinations of integers. This means that each rational number can be assigned a unique position in a counting sequence, and thus the set of rational numbers is countable.

To show that the set of irrational numbers is uncountable, we can use a proof by contradiction. Suppose that the set of irrational numbers is countable. This would mean that there exists a counting sequence that includes all the irrational numbers. However, we know that the set of irrational numbers is infinite and uncountable, meaning that there would always be an irrational number missing from the counting sequence. This contradicts our initial assumption that the set of irrational numbers is countable. Therefore, we can conclude that the set of irrational numbers is uncountable.