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 the square root of 2 is an irrational number

 Hello everyone. Today we will prove that \(\sqrt{2} \) is an irrational number. We will use the method of contradiction as a proof method. This proof was first given by the Greek philosopher Aristotle. 

Theorem: \(\sqrt{2} \) is an irrational number. 

Proof: 

Suppose that \(\sqrt {2} \) is a rational number. Then \(\sqrt {2} \) can be written in the form \(\frac{p}{q} \) where \(p \) and \(q \) are coprime integers such that \(q \neq 0 \). Note that since \(\frac{p}{q} \) is an irreducible fraction \(p \) and \(q \) cannot be even at the same time, otherwise the fraction would not be irreducible. 

From the equation \(\sqrt{2} = \frac{p}{q} \) it follows that \(2 = \frac{p^2}{q^2} \) or \(p^2 = 2q^2 \). So \(p^2 \) is an even number from which it follows that \(p \) is also an even number so we can write the number \(p \) in the form \(p = 2k \) , where \(k \) is an integer . 

So, we have that \((2k)^2 = 2q^2 \) i.e. \(4k^2 = 2q^2 \), or \(q^2 = 2k^2 \). From the last equation we conclude that \(q^2 \) is an even number, so \(q \) is an even number too. 

We have shown that \(p \) and \(q \) are even numbers, which contradicts the assumption that \(\frac{p}{q} \) is an irreducible fraction. Thus, the initial assumption that \(\sqrt{2} \) is a rational number is incorrect, which means that \(\sqrt{2} \) is an irrational number. 

\(\blacksquare \) 

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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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