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:
$$\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 – \sqrt{5}}.$$
Solution.
Let $a = \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 – \sqrt{5}}$. The third power of a number $a$ is equal to:
$$a^3 = 2 + \sqrt{5} + 3 \cdot \sqrt[3]{(2 + \sqrt{5})^2 \cdot(2 + \sqrt{5})} + 3 \cdot \sqrt[3]{(2 + \sqrt{5}) \cdot (2 + \sqrt{5})^2} + 2 – \sqrt{5}$$
$$a^3 = 4 – 3 \cdot \left ( \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 – \sqrt{5}} \right).$$
Since $a = \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 – \sqrt{5}}$, 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, $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 – \sqrt{5}}$ is a rational number.
