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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?

Chris
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1 Answers1

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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).

Noah Schweber
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