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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} \cdo...
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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} \cdo...
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Proof that \(I_0(\sqrt{2})\) is an irrational number

   Hello everyone.  Today we will prove that   \(I_0(\sqrt{2})\) is an irrational number, where \( I_0\) denotes a modified Bessel function of the first kind. We will use the method of contradiction as a proof method. Theorem 1:    \(I_0(\sqrt{2})=\displaystyle\sum_{n=0}^{\infty} \frac{1}{(n!)^22^n}\) Theorem 2:    \(I_0(\sqrt{2})\)   is an irrational number. Proof: Suppose that  \(I_0(\sqrt{2})\)  is a rational number. Then  \(I_0(\sqrt{2})\)  can be written in the form \( \dfrac{p}{q} \) where \( p \)  and \( q\) are coprime integers such that \(q \ge 1\). By Theorem 1 we can write the following equality: \(q!(q-1)!p2^q=\displaystyle\sum_{n=0}^{\infty} \frac{(q!)^22^q}{(n!)^22^n}\) . Since left hand side of this equality is an integer the sum \(\displaystyle\sum_{n=q+1}^{\infty} \frac{(q!)^22^q}{(n!)^22^n}\) which is greater than zero \(0\)  also must be an integer. Clearly for \(n \ge q+1\) we have , \(\...
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Proof that the set of prime numbers is infinite

Hello everyone. Today we will prove that the set of prime numbers is infinite.  We will use the method of contradiction as a proof method.  We'll also use the theorem of French mathematician Eduardo Lucas.  Theorem (Lucas): Every prime factor of Fermat number \(F _ n = 2 ^ {2 ^ n} + 1\); (\(n > 1\)) is of the form \(k2 ^{n + 2} + 1\).  Theorem: The set of prime numbers is infinite. Proof: Suppose opposite, that there are just finally many prime numbers and we denote the largest prime by \(p\). Then \(F_p\) must be a composite number because \(F_p>p\). By Lucas theorem we know that there is a prime number \(q\) of the form \(k2 ^{p + 2} + 1\) that divides \(F_p\). But \(q>p\) , thus we arrived at  a contradiction. Hence, the set of prime numbers is infinite. \(\blacksquare\)
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