Proof of the product formula for \(\dfrac{\pi}{2\sqrt{3}}\)

  Hello everyone.  Today we will prove the product formula for  \(\dfrac{\pi}{2\sqrt{3}}\).  We will use the method of direct proof as a proof method.  Theorem 1:  \[\frac{\pi}{2\sqrt{3}}=\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)}{n}\]\[\text{where} \quad \chi(n)=\begin{cases} 1, & \text{if } n \equiv 1 \pmod{6}\\-1, & \text{if } n \equiv -1 \pmod{6}\\0, & \text{otherwise}\end{cases}\] Theorem 2:  We have\[\frac{\pi}{2\sqrt{3}}=\frac{5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdots}{6 \cdot 6 \cdot 12 \cdot 12 \cdot 18 \cdot 18 \cdot 24 \cdot 30 \cdots}\]expression whose numerators are the sequence of the odd prime numbers greater than \(3\) and whose denominators are even–even numbers one unit more or less than the corresponding numerators. Proof: By Theorem 1 we know that \[\frac{\pi}{2\sqrt{3}}=1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\frac{1}{19}-\cdots\] we will have \[\frac{1}{5} \cdo...
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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. 

\(\blacksquare \)

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