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.

Solution 1:

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 $\mathbb{R}$.

Solution 2:

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 $(x, y)$.

Solution 3:

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

Solution 4:

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.