# Are there infinitely many irrationals between every 2 rationals?

Is this a correct proof that there are infinitely many irrationals between two rationals?

Take $n\in \mathbb N$, $q=\cfrac 1n \in \mathbb Q$.

I chose this because $1$ is the smallest positive integer and we want to take $q \in \mathbb Q$ as small as possible. As $n \to \infty, q \to 0$ so for any $\epsilon \gt 0$, we can find $q$ with $\epsilon \gt q \gt 0$.

Now take an irrational number $1 \gt \alpha\gt \epsilon$.

Consider $\frac 1 n \alpha$. This is obviously an irrational number always less than $q$. Since there are infinitely many irrational numbers $\alpha \in (\epsilon, 1)$ there are infinitely many irrationals between $0$ and $\epsilon$ in $\mathbb Q$.

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In case the question in the title is the intended question, then: Yes. If $a,b$ are rationals (or reals for that matter) with $a<b$, then there are uncountably many irrationals $\alpha$ with $a<\alpha<b$ (and countably many rationals $q$ with $a<q<b$) –  Hagen von Eitzen Jun 8 at 12:23
Short answer: Yes, as there are infinitely many irrational numbers by definition. –  JohnWO Jun 8 at 12:27
I have a textbook that claims there is a rational between every 2 irrationals. –  Daniel Margolis Jun 8 at 12:29
which is true aswell. Thats why you say: $\mathbb Q$ is dense in $\mathbb R$ –  CBenni Jun 8 at 12:33
@DanielMargolis Let the difference between two irrationals be $\gamma \gt 0$ then we can choose a natural number $N \gt \frac 1{\gamma}$ and there will be a rational number of the form $\frac {M+1}N$ in the interval between - indeed let $M$ be the largest integer such that $\frac MN$ falls below the interval concerned. –  Mark Bennet Jun 8 at 12:45