# Proof that the set of irrational numbers is dense in reals

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.

• 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 $y = xz$... Commented Sep 18, 2014 at 1:45
• $z = \sqrt{2}$ is not a positive rational number. Commented Jun 17 at 20:02

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 $\mathbb{R}$.

• 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. 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 $(x, y)$.

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

• $zx$ is irrational since it is the product of an irrational number and a nonzero rational number. 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

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

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

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\in \mathbb{Z}$$

Claim: $$k+1

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, $$k+1 and set, $$k+1=m$$

And that implies $$na

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

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

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.

• This implicitly uses the fact that $(\mathbb{N}, \le)$ is well-ordered. 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.

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