# Irrational numbers

An irrational number is a non-terminating, non-repeating decimal. A set of all irrational numbers is denoted by $\mathbb{I}$. The examples of rational numbers are $\sqrt{2}, \sqrt{3}, \Pi, e, …$.

By the definition of a rational number, we know that a number is rational if we can write it in a fraction form $\frac{a}{b}$, where $a$ and $b$ are integers and $b \neq 0$. If we prove that a number is not rational, then it is irrational number. Here are a few examples.

Example 1. We will show that $\sqrt{3}$ is an irrational number.

Proof. Suppose that $\sqrt{3}$ is a rational number. That means we can write it in a form of a fraction:

$$\sqrt{3} = \frac{a}{b},$$

where $a, b \in \mathbb{N}$ are relatively prime numbers. That means that $gcd(a, b) = 1$.

After squaring both sides of the above equation, we have:

$$\left( \frac{a}{b} \right)^2 = 3$$

$$a^2 = 3 b^2,$$

from which we conclude that $a$ is divisible by $3$ what we can write as $a = 3k, k \in \mathbb{N}$. We need to prove this claim, that is, we need to prove the following:

If $a^2$ is divisible by $3$, then $a$ is divisible by $3$.

Proof.

We will prove the contrapositive, that is, if $a$ is not divisible by $3$, then $a^2$ is not divisible by $3$. For this purpose, suppose that $a$ is not divisible by $3$. Then we have $a = 3n +1$ or $a = 3n +2$, for some integer $n$. This means that possible remainders after dividing a number $a$ by $3$ are $1$ or $2$. We consider the following two cases separately:

$1^{ \circ }$ If $a = 3n +1$, then we have

$$a^2 = (3n +1)^2 = 9n^2 + 6 n +1 = 3n(3n + 2) +1.$$

We can see that, in this case, $a^2$ is not divisible by $3$, because the number $3n(3n + 2)$ is divisible by $3$ and $1$ is not divisible by $3$, so their sum is not divisible by $3$.

$2^{\circ}$ If $a = 3n +2$, then we have

$$a^2 = (3n + 2) ^2 = 9 n ^2 + 12 n + 4 = 3(3n^2 + 4 n + 1) + 1.$$

For the same reason as in the previous case, the number $a^2 = 3(3n^2 + 4 n + 1) + 1$ is not divisible by $3$.

We have proved that if $a$ is not divisible by $3$, then $a^2$ is not divisible by $3$.

Back to proof that $\sqrt{3}$ is an irrational number. Now we have:

$$(3 k)^2 = 3 b^2$$

$$9 k^2 = 3 b^2$$

$$b^2 = 3k^2,$$

so, $b$ is divisible by $3$:

$$b = 3l, l \in \mathbb{N}.$$

The numbers $a$ and $b$ are both divisible by $3$, which is in contradiction with the assumption that $a$ and $b$ are relatively prime numbers and we conclude that the number $\sqrt{3}$ is not a rational number. Therefore, $\sqrt{3}$ is an irrational number.

Example 2. Is the following number irrational or rational:

$$\sqrt{1 + \sqrt{2}+ \sqrt{3}} ?$$

Solution.

Let $a = \sqrt{1 + \sqrt{2}+ \sqrt{3}}$. Then we have:

$$a = \sqrt{1 + \sqrt{2}+ \sqrt{3}} ^2$$

$$a ^2 = 1 + \sqrt{2 + \sqrt{2}}$$

$$a^2 – 1 = \sqrt{2 + \sqrt{2}} /^2$$

$$(a^2 – 1)^2 = 2 + \sqrt{3}$$

$$\sqrt{3} = (a^2 – 1)^2 – 2.$$

Suppose that $a \in \mathbb{Q}$. According to the properties of the operations $+$ and $\cdot$ on $\mathbb{Q}$ the number $a^2$ is a rational number. Now we have:

$$a^2 \in \mathbb{Q} \Rightarrow a^2 – 1 \in \mathbb{Q} \Rightarrow (a^2 – 1)^2 \in \mathbb{Q} \Rightarrow (a^2 – 1)^2 – 2 \in \mathbb{Q}.$$

Since $\sqrt{3} = (a^2 – 1)^2 – 2$ and $(a^2 – 1)^2 – 2 \in \mathbb{Q}$ it follows that $\sqrt{3} \in \mathbb{Q}$, which is not true. Therefore, $\sqrt{1 + \sqrt{2}+ \sqrt{3}}$ is an irrational number.

Example 3. Prove that the following number is a rational number:

$$\sqrt{6 + 4\sqrt{2}} + \sqrt{6 – 4 \sqrt{2}}?$$

Solution.

Let $a = \sqrt{6 + 4\sqrt{2}} + \sqrt{6 – 4 \sqrt{2}}$. Then we have

$$a = \sqrt{6 + 4\sqrt{2}} + \sqrt{6 – 4 \sqrt{2}}/ ^2$$

$$a^2 = 6 + 4\sqrt{2} + 2 \cdot \sqrt{6^2 – (4\sqrt{2})^2} + 6 – 4 \sqrt{2}$$

$$a^2 = 12 + 2 \cdot \sqrt{36 -32}$$

$$a^2 = 12 + 2 \cdot 2$$

$$a^2 = 16$$

$$a = \pm 4$$

Considering the original equality, we can see that $a >0$, thus $a = 4$, that is, the given number $\sqrt{7 + 4\sqrt{3}} + \sqrt{7 – 4 \sqrt{3}}$ is a rational number.

Example 4. Is $\cos 15^{\circ}$ an irrational number?

Solution.

We will use the half angle formula $\cos \frac{\alpha}{2}= \pm \sqrt{\frac{1 + \cos \alpha}{2}}$. Let $x = \cos \frac{\alpha}{2} = \cos \frac{30^{\circ}}{2} =\cos 15^{\circ}$. Then we have:

$$2 \cos ^2 \alpha = 1 + \cos \alpha$$

$$2 x^2 = 1 + \cos 30 {\circ}$$

$$2 x^2 = 1 + \frac{\sqrt{3}}{2}$$

$$4 x^2 =2 + \sqrt{3}$$

$$4x^2 – 2 = \sqrt{3}$$

Suppose that $x$ is a rational number. Then we have:

$$x \in \mathbb{Q}\Rightarrow x^2 \in \mathbb{Q} \Rightarrow 4x^2 \in \mathbb{Q} \Rightarrow 4x^2 – 2 \in \mathbb{Q},$$

that is $\sqrt{3} \in \mathbb{Q}$, which is not true. Therefore, $\cos 15^{\circ}$ is an irrational number.

Example 5. Prove that $\tan 15^{\circ}$ is an irrational number.

Solution.

We will use the half angle formula  $\tan \frac{\alpha}{2} = \frac{1 – \cos \alpha}{\sin \alpha}$. For $\alpha = 30^{\circ}$ we have

$$\tan 15^{\circ} =\tan \frac{30^{\circ}}{2} = \frac{1 – \cos 30^{\circ}}{\sin 30 ^{\circ}}$$

$$\tan 15^{\circ} = 2 – \sqrt{3},$$

which is an irrational number.

Example 6. Prove that $\log_3 2$ is an irrational number.

Solution.

Suppose the opposite, that is, $\log_3 2 = \frac{\log 2}{\log 3}$ is a rational number. Then we have:

$$\log_ 3 2 = \frac{\log 2}{\log 3} = \frac{a}{b},$$

where $a, b \in \mathbb{N}$ are relatively prime numbers. Now we have

$$b \cdot \log 2 = a \cdot \log 3$$

$$\log 2^b = \log 3^m$$

$$2^b = 3^a$$

Since on the left side of the equal sign is an even number, and on the right side is odd number, we have reached a contradiction, so  $\log_3 2$ is an irrational number.

Example 7. Prove that the following number is rational:

$$\sqrt{2 + \sqrt{5}} + \sqrt{2 – \sqrt{5}}.$$

Solution.

Let $a = \sqrt{2 + \sqrt{5}} + \sqrt{2 – \sqrt{5}}$. The third power of a number $a$ is equal to:

$$a^3 = 2 + \sqrt{5} + 3 \cdot \sqrt{(2 + \sqrt{5})^2 \cdot(2 + \sqrt{5})} + 3 \cdot \sqrt{(2 + \sqrt{5}) \cdot (2 + \sqrt{5})^2} + 2 – \sqrt{5}$$

$$a^3 = 4 – 3 \cdot \left ( \sqrt{2 + \sqrt{5}} + \sqrt{2 – \sqrt{5}} \right).$$

Since $a = \sqrt{2 + \sqrt{5}} + \sqrt{2 – \sqrt{5}}$, we have the following:

$$a^3 = 4 – 3a$$

$$a^3 + 3a – 4 = 0.$$

Now we use the rational root theorem: If a rational number $\frac{p}{q}$ is a root of the polynomial equation $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0, a_n \neq 0$ with the integer coefficients, then $p$ divides  $a_0$ and $q$ divides the leading coefficient $a_n$.

Referring to the rational root theorem, possible candidates for numbers $\frac{p}{q}$, in our equation, are:

$$\pm 1, \pm 2, \pm 4.$$

By directly substituting these $6$ numbers into the equation, we obtain that one of the solutions is number $1$. The remaining two solutions we obtain by dividing the polynomial  $a^3 + 3a – 4$ by $a -1$ and we find zeroes of the obtained quadratic polynomial $a^2 + a +4$. Since $a$ is a positive number, it follows that $a$ is not the solution to the equation $a^2 + a +4 =0$. Therefore, the only possible solution is $a =1$, so $a$ is a rational number, that is, $\sqrt{2 + \sqrt{5}} + \sqrt{2 – \sqrt{5}}$ is a rational number.