Can there be a string of irrational numbers in a row without a rational number in between?

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For example. Rational, Irrational, Irrational, Rational...

I am not sure whether this is the case. My intuition says no, but I feel that this proof contradicts this.. I am trying to get my head wrapped around this. At which points is the function discontinuous?

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Whenever you have two different real numbers -- rational or irrationals, that doesn't matter -- there will always be an infinity of both rational and irrational numbers between them.

In particular, for your question: It is not possible to have two different irrationals that don't have any rational number between them.

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There are infinitely many rationals between any two distinct irrationals or, in fact, any two distinct reals. (I dont' quite follow the relation to the other question you linked to.)

Let $a \lt b$ with $a,b \in \mathbb{R} \setminus \mathbb{Q}$ be the two irrationals. The following shows how to actually construct a rational number $r \in (a,b) \cap \mathbb{Q}$. (Note: same proof works for $\forall a,b \in \mathbb{R}$, not necessarily irrational.)

Imagine you choose a sequence of rational numbers in arithmetic progression, and extend it to infinity at both ends. If the common difference of the progression is small enough, then there must exist a term of sequence which falls inside $(a,b)$. Intuitively, all that's required is for the common difference to be smaller than the "width" $b-a$ of the interval $(a,b)$.

To formalize this intuition, choose a positive integer $q \gt \frac{1}{b-a}$ and define $n = \lfloor q \cdot a \rfloor$. Then, by the definition of the greatest integer function:

$$\lfloor q \cdot a \rfloor = n \le q \cdot a \lt n+1$$ $$\iff \quad \frac{n}{q} \le a \lt \frac{n+1}{q}$$

But given $q \gt \frac{1}{b-a}$ it follows that $\frac{1}{q} \lt b-a$ then using the inequalities above gives:

$$a \lt \frac{n+1}{q} = \frac{n}{q} + \frac{1}{q} \lt a + (b - a) = b$$

which proves that $\frac{n+1}{q} \in (a,b)$ is a rational number between $a$ and $b$.

The same construction can be repeated for any integers $q \gt \frac{1}{b-a}$ and (after resolving technicalities about possible duplicates) proves that there are in fact infinitely many rationals between $a$ and $b$.