0
$\begingroup$

This question already has an answer here:

I am working on a simple proof involving rational and irrational numbers. Is it safe to assume that if a number is not rational, it is irrational, and that if a number is not irrational, it is rational?

Example: Let $P(x)=\text{x is rational}$ and $Q(x)=\text{x is irrational}$.

Then is it true that $\forall x\ P(x)\text{ xor }Q(x)$?

$\endgroup$

marked as duplicate by Ross Millikan, rschwieb, Watson, Namaste discrete-mathematics Sep 17 '16 at 11:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ A double negation is an affirmation. $\endgroup$ – Pedro Tamaroff Sep 16 '16 at 1:47
  • $\begingroup$ As long as by "number" you mean real number. $\endgroup$ – Joey Zou Sep 16 '16 at 4:56
  • $\begingroup$ I feel that it's important to point out that double negation isn't always an affirmation. See constructive mathematics. $\endgroup$ – James Wood Sep 17 '16 at 20:45
0
$\begingroup$

The real numbers are composed of the rational numbers and the irrational numbers so yes if a number is not one, it is the other.

See Dominik's answer:

Are there real numbers that are neither rational nor irrational?

There he provided this excellent picture: enter image description here

$\endgroup$
2
$\begingroup$

A rational number is defined as a number that can be expressed as the ratio of two integers, i.e. $\frac{p}{q}$, where $q\ne0$. An irrational number is a real number that cannot be expressed as a ratio. So, yes a real number is either rational or irrational, but not both.

$\endgroup$
1
$\begingroup$

An irrational number is a number that is not rational.

Thus a non-irrational number is not (a number that is not rational), thus it is rational.

In set notation irrational numbers are $\mathbb{R}\setminus\mathbb{Q}$, so non-irrational numbers are $\mathbb{R}\setminus(\mathbb{R}\setminus\mathbb{Q})=\mathbb{Q}$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.