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 square root of a prime number is an irrational number

Hello everyone. Today we will prove that the square root of a prime number is an irrational number. We will use the method of contradiction as a proof method. 

Theorem: If \(p \) is a prime number, then \(\sqrt{p} \) is an irrational number. 

Proof: 

Suppose the opposite, ie. that \(\sqrt{p} \) is a rational number. Then \(\sqrt{p} \) can be written in the form of a fraction \(\frac{a}{b} \), where \(a \) and \(b \) are two coprime integers and \(b \neq 0 \). By squaring the equation \(\sqrt{p} = \frac{a}{b} \) we get the equation \(p = \frac{a^2}{b^2} \), ie. \(a^2 = pb^2 \). 

Let us now write the numbers \(a \) and \(b \) in the form of the product of powers of their prime factors. \[a = p_1^ {n_1} \cdot p_2^{n_2} \cdot p_3^{n_3} \cdot \ldots \cdot p_j^{n_j} \] \[b = q_1^{m_1} \cdot q_2^{ m_2} \cdot q_3^{m_3} \cdot \ldots \cdot q_k^{m_k} \] Squaring these two equations we get: \[ a^2 = p_1^{2n_1} \cdot p_2^{2n_2} \cdot p_3^{2n_3} \cdot \ldots \cdot p_j^{2n_j} \] \[b^2 = q_1^{2m_1} \cdot q_2^{2m_2} \cdot q_3^{2m_3} \cdot \ldots \cdot q_k^{ 2m_k} \] Since \(a^2 = pb^2 \) we conclude that the right side of equality, ie. \(pb^2 \) must consist of products of powers of unique prime factors with even exponents. 

Let us now consider the following two possible cases: 

First case: Let the number \(p \) be in the factorization of the number \(b^2 \). This means that we have \(p \cdot p_i^{2n_i} = p^{2n_i + 1} \) for some \(i \), \(1 \le i \le j \). Since \(2n_i + 1 \) is an odd number, this contradicts the previous conclusion that \(pb^2 \) must consist of the product of powers of unique prime factors with even exponents. 

Second case: Let the number \(p \) not be in the factorization of the number \(b^2 \). Since \(p = p^1 \) and since \(1 \) is an odd number, we again have a contradiction with the conclusion that \(pb^2 \) must consist of the product of powers of unique prime factors with even exponents. 

Since we arrived at a contradiction in both cases, we conclude that the initial assumption that \(\sqrt{p}\) is a rational number is incorrect, hence \(\sqrt{p} \) must be an irrational number. 

\(\blacksquare \)

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 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. \(\blacksquare\)
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