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} \cdot \frac{\pi}{2\sqrt{3}}=\frac{1}{5
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Proof that a natural number can be expressed as a product of prime numbers

Hello everyone. Today we will prove that any natural number greater than \(1 \) can be expressed as a product of a prime number and number one or as a product of few prime numbers. We will use the method of mathematical induction as a proof method. 

Theorem: Let \(n \) be a natural number greater than \(1 \). Then \(n \) can be expressed as the product of one prime number and number one or as the product of few prime numbers.

Proof:

Note that if \(n \) is a prime number, the statement is automatically proved because any number can be written as the product of that number and number one. 

1. Base case (n = 2) 

Since \(2 \) is a prime number the statement is automatically proved. 

2. Induction hypothesis (n = m) 

Suppose that \(\forall k \in \mathbb {N}, \quad 2 \le k \le m \), \(k \) can be expressed as the product of a prime number and number one or as a product of few prime numbers. 

3. Inductive step (n = m + 1) 

Using the assumption from the second step, we will prove that: \(\forall k \in \mathbb {N}, \quad 2 \le k \le m + 1 \), \( k \) can be expressed as the product of one prime number and number one or as the product of few prime numbers. 

For all \(k \) less than \(m + 1 \) the truth of the statement automatically follows from the inductive hypothesis. Let us now consider the case when \(k = m + 1 \). If \(m + 1 \) is a prime number, the statement is automatically proved. Otherwise \(m + 1 \) is a composite number which can be expressed in the form \(m + 1 = pq \), where \(p \) and \(q \) are natural numbers such that  \(2 \le p <m+1\) and \(2 \le q <m+1\) , i.e. \(2 \le p \le m\) and \(2 \le q \le m\) . According to the inductive hypothesis, both numbers \(p \) and \(q \) can be expressed as products of few primes or as products of one prime number and number one, therefore the same is true for the number \(m + 1 \). 

\(\blacksquare \)

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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} \cdot \frac{\pi}{2\sqrt{3}}=\frac{1}{5
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