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
Post a Comment

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.

\(\blacksquare\)

Comments

Post a Comment

Popular posts from this blog

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
Read more

Proof that there are infinitely many prime numbers

Hello everyone.  Today we will prove that there are infinitely many prime numbers.  This proof was first performed by the Greek mathematician Euclid.  Theorem: There are infinitely many prime numbers.   Proof:  Suppose there is a finite number of prime numbers.  Let us denote the number of prime numbers by \(n \), and the prime numbers themselves by \(p \) and the indices \(1,2 \ldots, n-1, n \).  So we have a finite set of primes \(p_1, p_2, \ldots, p_{n-1}, p_n \).  Now, we construct the number \(q \) as follows: \[q = p_1 \cdot p_2 \cdot \ldots \cdot p_ {n-1} \cdot p_n + 1 \] Then \(q \) can be either prime or a composite number.  It is clear that \(q \) cannot be a prime number because we assumed that the number of primes is finite and that the largest prime number is the number \(p_n \).  Therefore, \(q \) must be a composite number.  If \(q \) is a composite number then there is some prime number \(p \) from the set \(p_1, p_2, \ldots, p_{n-1}, p_n \) that divides it.  Denote by
Read more