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 of Archimedes' theorem

Hello everyone. Today we will prove Archimedes' theorem. We will use the method of contradiction as a proof method. 

Theorem: For every two real numbers \(a \) and \(b \) where \(a> 0 \) there exists a natural number \(n \) such that \(a \cdot n> b \). 

Proof: 

Considering that \(a> 0 \)  we can divide the inequality \(a \cdot n> b \)  by the number \(a \) so that we get a new inequality \(n> \frac{b}{a} \) . Let us denote the real number \(\frac{b}{a} \) by \(x \), then we can reformulate the theorem we need to prove as follows:

 For any real number \(x \) there exists a natural number \(n \) such that \(n> x \). 

Suppose the opposite, ie. that there exists a real number \(x \) such that \(x \ge n \) for any natural number \(n \). This would mean that the set of natural numbers \(\mathbb {N} \) is bounded from above. Let us denote this least upper bound by \(\operatorname{sup}(\mathbb{N}) = S \). Now consider the difference \(S-1 \). It is clear that \(S-1 \) is not the upper bound of the set \(\mathbb{N} \), because if it were then \(S \) that is greater than \(S-1 \) would not be the least upper bound. This means that there exists a natural number \(m \) such that \(m> S-1 \), that is, we have \(m + 1> S \). As the addition operation is closed on the set of natural numbers we have that \(m + 1 \in \mathbb{N} \), ie. the number \(m + 1 \) is a natural number greater than \(\operatorname{sup} (\mathbb{N}) \). This contradicts the initial assumption which means that for any real number \(x \) there exists a natural number \(n \) such that \(n> x \). 

\(\blacksquare\)

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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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