6
$\begingroup$

I have found out that between every 2 rational numbers there is an irrational number, and between every 2 irrational numbers, there is a rational number.

Does this mean that the rational and irrational numbers are laid out in an alternating pattern along the real line?

$\endgroup$
  • $\begingroup$ It doesn't make sense to talk about the "next" real number under the usual ordering of $\mathbb{R}$. $\endgroup$ – user61527 Apr 19 '14 at 4:01
  • $\begingroup$ Not really, any interval of the real line contains a dense subset of both rational and irrational numbers. So in fact you can find any number of rational or irrational numbers between any two numbers. $\endgroup$ – BoZenKhaa Apr 19 '14 at 4:03
  • $\begingroup$ But doesn't the link imply that you will never find two "successive" irrational or rational numbers? I realize I am probably wrong, but I am at a loss as to why. $\endgroup$ – Superbest Apr 19 '14 at 4:04
  • $\begingroup$ Also keep in mind that there are "more" irrational numbers than rational numbers. Imagining them as laid out in an alternating pattern may give you wrong impressions about that. $\endgroup$ – user142299 Apr 19 '14 at 4:05
  • 1
    $\begingroup$ @Superbest Keep in mind that there is no such thing as two "successive" numbers. For example, what if I asked you what number comes after $1$? That doesn't have a good answer. $\endgroup$ – user142299 Apr 19 '14 at 4:06
3
$\begingroup$

No. Your statements are correct, but it is incorrect to think of the real numbers like this. Because the rationals and irrationals are dense in $\mathbb{R}$, for any two distinct numbers in either set you can find another in between them (in either set, as you mentioned), so you can't possibly have a notion of 'adjacent'-ness of rationals and irrationals.

For a deeper answer you may be interested in order theory.

Here is another argument:

To have the rationals and irrationals 'alternate when lined up' in some sense, we would have to take each irrational number, and put it in between two other rational numbers. However, this would imply that for every irrational, there is a distinct rational number that is lined up to it's right, so we could make a 1-to-1 mapping from the irrationals to the rationals.

But the rationals are countable, so that would imply that the irrationals are countable too!

But the irrationals are uncountable!

so it's impossible to line up the rationals and irrationals so that they alternate and each number only appears once.

$\endgroup$
  • $\begingroup$ Off the top of my head (order theory is not my forte) I believe this has to do with the fact that $\mathbb{R}$ with a usual inequality relation isn't well-ordered en.wikipedia.org/wiki/Well-order#Reals $\endgroup$ – enthdegree Apr 19 '14 at 4:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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