I'm being asked to prove that the set of irrational number is dense in the real numbers. While I do understand the general idea of the proof:

Given an interval $(x,y)$, choose a positive rational number (say) $z=\sqrt{2}$. By density of rationals, there exists a rational number $p$ in the interval $(x/z, y/z)$, which essentially means that $$\frac{x}{\sqrt{2}} < p < \frac{y}{\sqrt{2}}.$$ I find that $pz$ is irrational, since it is the product of a rational and irrational number. However, my instructions besides writing that proof down is to specifically verify that $y=xz$ is irrational. What does this have to do with anything and how does it prove denseness of irrationals? Regardless, assume that $x$ is a nonzero rational number and that $z$ is irrational. For the sake of contradiction, assume that $y=xz$ is rational. This should mean that $y/x$ rational as well, and therefore that $z$ is rational, a contradiction.

  • 1
    $\begingroup$ I am confused too. It seems to me that at the beginning $x$ and $y$ are arbitrary real numbers with $x < y$, but later you say that $y = xz$... $\endgroup$
    – Tunococ
    Commented Sep 18, 2014 at 1:45
  • $\begingroup$ $z = \sqrt{2}$ is not a positive rational number. $\endgroup$
    – Ben123
    Commented Jun 17 at 20:02

9 Answers 9


Another argument:

$\mathbb{Q}$ is dense in $\mathbb{R}$, so $\mathbb{Q} + \sqrt{2}$ is dense in $\mathbb{R} + \sqrt{2} = \mathbb{R}$. Since $\mathbb{Q} + \sqrt{2}$ is a subset of the irrationals, we conclude that the irrationals are also dense in $\mathbb{R}$.

  • 4
    $\begingroup$ This 'proof' does not contain a rigorous for why $\mathbb{Q}+\sqrt{2}$ is dense in $\mathbb{R}+2$. It's intuitive, almost obvious, but does not actually help prove that the irrationals are dense. $\endgroup$ Commented Feb 3, 2023 at 2:27

By the density of rational numbers, there exists a rational number $r \in (x, y)$.

Since $\frac{y - r}{2} > 0$, by the Archimedian Property, there exists $n \in \mathbb{N}$ such that $\frac{y - r}{2} > \frac{1}{n}$.

Then we have $x < r + \frac{\sqrt{2}}{n} < r + \frac{\sqrt{4}}{n} < y$.

Now we check that $s = r + \frac{\sqrt{2}}{n}$ is an irrational number sitting in $(x, y)$.

  • 1
    $\begingroup$ Isn't it easier/better to understand, if you change $\frac{\sqrt{2}}{n}$ to $\frac{1}{n\sqrt{2}}$? $\endgroup$ Commented May 17, 2016 at 11:30
  • 1
    $\begingroup$ @FlorianWendelborn Yes, it is. And with this, you do not need to show $s\in (x,y)$. $\endgroup$
    – Iain
    Commented Oct 13, 2023 at 23:05

The density of the irrationals follows from the density of the rationals and the existence of positive irrational numbers. Indeed, given an interval $(a,b)$, choose any positive irrational number $z$; for instance, choose $z = \sqrt2$. By the density of the rationals there is a rational number $x$ in the interval $(a/z, b/z)$ so that $zx$ lies in the interval $(a, b)$ and $zx$ is irrational since it is the product of an irrational number and a rational number.

Source: ADVANCED CALCULUS by Patrick M. Fitzpatrick

  • $\begingroup$ $zx$ is irrational since it is the product of an irrational number and a nonzero rational number. $\endgroup$
    – jskattt797
    Commented Dec 14, 2020 at 2:13

I find this way really easy to understand so I thought I would share it. I had come across it in the book Introduction to Real Analysis by William Trench.

Proof: Given that we know that rational numbers are dense in $\mathbb{R}$ . So $\exists$ $r_1 , r_2 \in \mathbb{Q}$ such that:

$x < r_1 < r_2 <y$

Remark: To find a number between $r_1$ and $r_2$ that is irrational, we need to come up with a small enough number that can we can add to $r_1$ that serves two purposes:

  1. Make it irrational
  2. Make sure the sum of this irrational number and $r_1$ is less than $r_2$

To undertand it better, think of this inequality: $\frac{r_2-r_1}{2} < \frac{r_2-r_1}{\sqrt2} < r_2 - r_1$

So, $\frac{r_2-r_1}{\sqrt2}$ is a number that we can add to $r_1$ to make it less than $r_2$ (it serves the two points outlined above).

Let $t = r_1 + \frac{r_2-r_1}{\sqrt2}$

Therefore $x < r_1 < t <r_2<y$

Note - you can further prove that $t$ is irrational by using the facts:

  1. The sum of an irrational and a rational number is irrational.
  2. The product of an irrational and a rational number is irrational.

Setep 1: $[\Bbb{Q}$ is dense in $\Bbb{R}]$

  1. Every non empty open interval interval on $\mathbb{ R }$contains at least one $r\in\mathbb{Q }$.

  2. Limit points of $\mathbb{ Q }$is the whole of $\mathbb{R }$.

  3. For any $x\in \mathbb{R } $ there exists a sequence $(r_n) _{n\in\mathbb{N}} \subset \mathbb{Q}$ such that $(r_n) \to \mathbb{R}$.

  4. Give any $x\in \mathbb{R }$ there exists a rational number $r$ such that $d(x, r)<\epsilon $ for all $\epsilon >0$

  5. $\mathbb{Q }$ is dense in $\mathbb{R }$

All the above statements are equivalent.

It is not difficult to prove $Q$ is dense in $\mathbb{R}$.

I am going to prove using property $1)$ i.e any non-empty open interval contains at least one rational number. (In fact it contains infinitely many of them).

Choose an open interval $(a, b) \subset\mathbb{R } ,a<b $

Then, $\ell(a, b) =b-a>0$

Then by Archimedean Property, $\exists n\in {\mathbb{N}}$ such that $$\frac{1}{n}<(b-a) $$

$$\implies n(b-a) >1$$

Consider the open interval $(na, nb) $

And $$\ell(na, nb) >1$$

$\implies (na, nb) $ contains an integer. Because,

Consider, $$k=[na]=floor(na) $$

Then, $k\le na <k+1$ , $k\in \mathbb{Z}$

Claim: $k+1<nb$

Suppose, $k+1\ge nb$

Then, $[na, nb]\subseteq [k, k+1]$

$\implies \ell[na, nb]\le \ell[k, k+1]$

$\implies n(b-a) \le 1$

But this contradict that $n(b-a) >1$.

Hence, $k+1<nb$ and set, $k+1=m$

And that implies $na<m<nb$

$\implies a< \frac{m}{n}<b$

Hence, $\frac{m}{n} \in (a, b) $ and $\frac{m}{n} \in \mathbb{Q}$

So we proved any non empty open interval contains a rational number.

Hence, $\mathbb{Q}$ is dense in $\mathbb{R}$.

One can start from Step 2

Step 2:$[\Bbb{R\setminus Q}$ is dense in $\mathbb{R}]$

Let, $a, b\in \Bbb{R}$ with $a<b$

Then, the open interval$$(a-\sqrt2 , b-\sqrt2) $$ contains at least one rational number $(\text{ by density of } \Bbb{Q})$, say $r$.

So, \begin{align}& r\in (a-\sqrt2 , b-\sqrt2) \\ &\implies r+\sqrt2\in(a, b) \end{align}

And, $r+\sqrt2 \in {\Bbb{R\setminus Q}}$

Since, $(a, b) $ is any arbitrary open interval, hence every open intervals contains at least one irrational, implies irrationals are dense in Real.

This is a standard textbook proof. You can find it any Real analysis book.

  • $\begingroup$ This implicitly uses the fact that $(\mathbb{N}, \le)$ is well-ordered. $\endgroup$
    – Tran Khanh
    Commented Dec 31, 2023 at 4:47

For any real number $x$, the sequence of irrational numbers: $$ x_n = \frac{\lfloor nx\rfloor}{n}+\frac{1}{\sqrt{n^2+1}} $$ converges towards $x$, since: $$ |\,x_n - x\,|\leq \frac{2}{n}.$$


It sufficient to show that there is a irrational between $a<b\in \mathbb{Q}$. Since $1/\sqrt{2}\in (0,1)$ consider the map $(0,1)\to (a,b)$, $\,t\mapsto(b-a)t+a$, what can you say about $(b-a)/\sqrt{2}+a$?


If $(a,b) = (a,b) \cap\mathbb{Q}$, then $|(a,b)| = \aleph_0$. But $f:(a,b) \to \mathbb{R}$ given by $$f(x) = \tan\left(\frac{\pi}{b-a}\left(x - \frac{b+a}{2}\right)\right)$$

is a bijection.


I guess this is what you're going for.

We first prove that $u/\sqrt{2}$ is irrational whenever $u$ is a non-zero rational number. Suppose, for a contradiction, that $\gamma=u/\sqrt{2}$ is rational. If $\gamma=0$ then we must have $u=0$, contradiction. If $\gamma\neq0$ then we can divide by $\gamma$, whence $\sqrt{2}=u/\gamma$. Since the quotient of two rational numbers is also rational, it then follows that $\sqrt{2}$ is rational, another contradiction. It must follows that $\gamma=u/\sqrt{2}$ must be irrational whenever $x$ is rational.

Pick any two non-zero rational numbers $x$ and $y$, and assume without loss of generality that $x<y$. Then $\alpha = x/\sqrt{2}$ and $\beta = y/\sqrt{2}$ are both irrational. You know that the rationals are dense in the reals. In particular that means that if I take two irrational numbers $\alpha$ and $\beta$ then you can find a rational number $r$ with $\alpha<r<\beta$. Hence you can find a rational number $r$ such that $\lvert x/\sqrt{2}\rvert <r<\lvert y/\sqrt{2}\rvert$, whence $$\lvert x\rvert <\frac{r}{\sqrt{2}}<\lvert y\rvert.$$

  • If $x$ and $y$ are both positive, then we have found an irrational between them.
  • If $x$ and $y$ are both negative, then $-r/\sqrt{2}$ is an irrational between them.
  • If $x$ is negative and $y$ is positive then note that $x<\lvert x \rvert$ and so $r/\sqrt{2}$ is again an irrational between them.

This proves that between any two rational numbers there exists an irrational number. You already know that between any two numbers $a$ and $b$ there exists a rational number $r_{1}$, and then between $r_{1}$ and $b$ you can find another rational number $r_{2}$. Hence between any two numbers $a$ and $b$ there are two rational numbers, and between those two rational numbers there is an irrational number. This proves that the irrationals are dense in the reals.

  • $\begingroup$ Why from $|x/\sqrt{2}| < r < |y/\sqrt{2}|$ didn´t you get $|x| < r\sqrt{2} < |y|$ instead? $\endgroup$
    – user561334
    Commented Apr 17, 2020 at 21:33
  • $\begingroup$ @Wybie: ah, that's just a mistake, sorry. But I'm sure you can see how to fix the argument. $\endgroup$
    – Will R
    Commented Apr 17, 2020 at 21:38

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