Before we go into anything about irrational numbers, we need to show that irrational number exist. By construction of the definition of rational numbers we know that they exist. To determine whether there are any irrational numbers we need only to show that there are numbers that are not rational. To do this we must show that is at least $1$. Let us consider $\sqrt{2}$.

Theorem 1: $\sqrt{2}$ is an irrational number. In other words, $\sqrt{2}$ cannot be expressed as a ratio of two integers.

Proof: We know that the the even integers are closed under multiplication, and likewise the odd integers. In particular, the square of an even integer is even and the square of an odd integer is odd.

Now suppose that $\sqrt{2}$ is a rational number, then for some $a,b \in \mathbb{Z}$, \[ \sqrt{2}=\frac{a}{b} \] Now, let's presume that $\frac{a}{b}$ is in it's lowest terms. In particular, $a$ and $b$ are not both even. So, squaring the above equation and simplifying, we get \[ 2=\frac{a^2}{b^2} \\ a^2=2b^2. \]

The term $2b^2$ represents an even integer, so $a^2$ is an even integer, and hence $a$ is an even integer, say $a=2c$, where $c$ is also an integer. Replacing $a$ by $2c$ in the equation $a^2=2b^2$, we obtain \[ (2c)^2=2b^2\\ 4c^2=2b^2\\ 2c^2=b^2 \]

The term $2c^2$ represents an even integer, so $b^2$ is an even integer, and hence $b$ is an even integer. But now we have concluded that both $a$ and $b$ are even integers, whereas $\displaystyle\frac{a}{b}$ was presumed to be in lowest terms. This contradiction leads us to conclude that it is not possible to express $\sqrt{2}$ in the rational form $\displaystyle\frac{a}{b}$, and therefore $\sqrt{2}$ is irrational.

So, we have one irrational number and although we don't know how large the set of irrational numbers is, we can say that the set of irrational numbers exists. Under further investigation we shall find out that the set of all irrational numbers can be broken down further into algebraic and non-algebraic (transcendental). For further understanding of these sets indiviually, click on the side panel links.

As I searched through the different books on irrational numbers I always ran into the three famous construction problems.

1. "Duplicating (doubling) the cube" means to construct a cube having twice the volume of a given cube. If we take an edge of the given cube as the unit of length, the problem is to construct a line of length $\sqrt[3]{2}$, because this would be the length of the edge of a cube that has twice the volume of the given cube.
   
2. "Trisecting an angle" means to devise a method, using only the prescribed tools, by which any angle can be trisected. There are certain special angles, $45^{\circ}$ and $90^{\circ}$ for example, which can be trisected with an unmarked straightedge and compass; but the so-called "general" angle cannot be divided into three equal angles with the prescribed tools.
3. "Squaring the circle" means to construct a square equal in area to a given circle or, equivalently, to construct a circle equal in area to a given square.