1
$\begingroup$

I understand that there are infinitely many more irrational numbers than there are rationals - there are many ways in which one can intuitively understand this. However, consider the following:

LEMMA: For every pair of distinct irrational numbers, there exists a rational number between them.

PROOF: Let $a,b$ be distinct irrational numbers such that $b>a$ and let $\delta = b-a$. Then for some large enough $N$, we have $N\delta >1$ so there exists an integer in the interval $[Na,Nb]$ and as such there exists a rational number in the interval $[a,b]$.

By this logic, we can find a rational number for every pair of distinct irrational numbers - why does this not imply there are as many rationals as irrationals?

$\endgroup$
4
$\begingroup$

By this logic, we can find a rational number for every pair of distinct irrational numbers

Not really. Or rather, we can indeed do so, but not injectively - the same rational works for many different pairs of irrationals.

In order to get an appropriate bijection, we need to assign to a pair of irrationals $(\alpha,\beta)$ a rational $q$ which is unique, in the sense that a different pair $(\alpha',\beta')$ will always get a different rational $q'$. Intuitively, there are a lot of tasks to be done (= pairs of irrationals to separate) and few workers (= rationals), but each worker can do a lot of tasks (= separate continuum many pairs).

$\endgroup$

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.