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I understand that a subset $Y$ of a metric space $(X,d)$ is dense in $X$ if $\forall x\in X$, $\forall r\gt0$, $\exists y\in Y$ such that $d(x,y)\lt r$.

I'm struggling to understand how to prove that the Irrationals $\mathbb{R}-\mathbb{Q}$ is dense in $\mathbb{R}$.

If $x\in\mathbb{R}-\mathbb{Q}$, then just take $y=x$. Then $d(x,y)=0\lt r$.

But if $x\in\mathbb{Q}$, how do we choose a $y$ that $\in\mathbb{R}-\mathbb{Q}$ and is guaranteed to have $d(x,y)\lt r?$

Henrik supports the community
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mfaczz
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    Can you prove that $\mathbb{Q}$ is dense in $\mathbb{R}$? Then consider ${\pi + r : r \in \mathbb{Q}}$. – Dermot Craddock May 23 '25 at 13:08
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    @DermotCraddock or $\sqrt2 + r$ so you don't have to prove the irrationality of $\pi$ hahahaha – Bruno Andrades May 23 '25 at 13:09
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    An alternative approach: If $x\in\mathbb R$ and $r>0$, then the ball $B(x,r)$ is uncountably infinite. But there are only countably many rational numbers in $B(x,r)$, so there must be exist an irrational number $y\in B(x,r)$. – Joe May 23 '25 at 13:20
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    Another alternative approach, in a similar vein to @Joe 's, show that every countable set has measure $0$, and now conclude that the complement of a set of measure $0$ is dense hahahahaha – Bruno Andrades May 23 '25 at 14:00

2 Answers2

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The proof is actually easier than for the rationals. As you said, if the ball has center an irrational, then you're done, if not, let the ball be $B(x,r)$ for $r>0$ and $x\in\mathbb Q$. And let $n>\sqrt2/r$ be a natural number; which exists by the Archimedean principle. Now $x+\sqrt2/n$ is an irrational number and

$$\left|x-\left(x+\frac{\sqrt2}{n}\right)\right|=\frac{\sqrt2}{n}<r$$

And so $x+\sqrt2/n\in B(x,r)$ is an irrational number

Also, as pointed in the comments, if you already know the rationals are dense, then, choose a rational $y$ in the ball $B(x-\sqrt2,r)$, and now $y+\sqrt2\in B(x,r)$ is irrational (with this proof you don't even need to separate into cases)

Bruno Andrades
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Since we are in Real line it's sufficient to show that $\forall x < y \in \mathbb{R}$ we can find an irrational number $a$ such that $$x<a<y$$ Proof: Let WLOG $x<y\in \mathbb{R^+}$ . Now from Archemidean principle we can find a $N \in \mathbb{N} $ such that $$\frac{y-x}{√2} > \frac{1}{N}\implies y-x > \frac{√2}{N}$$ Suppose $$\frac{√2}{N}<\frac{2√2}{N}... \frac{k√2}{N} \le y-x $$ Since the difference between any two consecutive terms of the above sequence is$ \frac{√2}{N} $ and the difference between $x$ and $y$ is larger than $\frac{√2}{N} $ we can say $$\frac{k√2}{N}\leq y-x \le\frac{(k+1)√2}{N}$$ is not possible, so there exists at least a multiple of $ \frac{√2}{N} $ between $x$ and $y$. Hence $$x< a=\frac{m√2}{N} <y$$ where $\frac{m√2}{N} $ is an irrational number. .

Mathson
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