Proof that the square root of a prime number is an irrational number

Hello everyone. Today we will prove that the square root of a prime number is an irrational number. We will use the method of contradiction as a proof method. 

Theorem: If $p$ is a prime number, then $\sqrt{p}$ is an irrational number. 

Proof: 

Suppose the opposite, ie. that $\sqrt{p}$ is a rational number. Then $\sqrt{p}$ can be written in the form of a fraction $\frac{a}{b}$, where $a$ and $b$ are two coprime integers and $b \neq 0$. By squaring the equation $\sqrt{p} = \frac{a}{b}$ we get the equation $p = \frac{a^2}{b^2}$, ie. $a^2 = pb^2$. 

Let us now write the numbers $a$ and $b$ in the form of the product of powers of their prime factors. $$a = p_1^ {n_1} \cdot p_2^{n_2} \cdot p_3^{n_3} \cdot \ldots \cdot p_j^{n_j}$$ $$b = q_1^{m_1} \cdot q_2^{ m_2} \cdot q_3^{m_3} \cdot \ldots \cdot q_k^{m_k}$$ Squaring these two equations we get: $$a^2 = p_1^{2n_1} \cdot p_2^{2n_2} \cdot p_3^{2n_3} \cdot \ldots \cdot p_j^{2n_j}$$ $$b^2 = q_1^{2m_1} \cdot q_2^{2m_2} \cdot q_3^{2m_3} \cdot \ldots \cdot q_k^{ 2m_k}$$ Since $a^2 = pb^2$ we conclude that the right side of equality, ie. $pb^2$ must consist of products of powers of unique prime factors with even exponents. 

Let us now consider the following two possible cases: 

First case: Let the number $p$ be in the factorization of the number $b^2$. This means that we have $p \cdot p_i^{2n_i} = p^{2n_i + 1}$ for some $i$, $1 \le i \le j$. Since $2n_i + 1$ is an odd number, this contradicts the previous conclusion that $pb^2$ must consist of the product of powers of unique prime factors with even exponents. 

Second case: Let the number $p$ not be in the factorization of the number $b^2$. Since $p = p^1$ and since $1$ is an odd number, we again have a contradiction with the conclusion that $pb^2$ must consist of the product of powers of unique prime factors with even exponents. 

Since we arrived at a contradiction in both cases, we conclude that the initial assumption that $\sqrt{p}$ is a rational number is incorrect, hence $\sqrt{p}$ must be an irrational number. 

$\blacksquare$