Mathemarium

   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 , \(\left(\dfrac{q!}{n!}\right)^2 \le \dfrac{1}{(q+1)^{2(n-q)}}\) , hence \(\display

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. \(\blacksquare\)

Hello everyone.  Today we will prove that any natural number greater than \(1 \) can be expressed as a product of a prime number and number one or as a product of few prime numbers.  We will use the method of mathematical induction as a proof method.  Theorem: Let \(n \) be a natural number greater than \(1 \).  Then \(n \) can be expressed as the product of one prime number and number one or as the product of few prime numbers. Proof: Note that if \(n \) is a prime number, the statement is automatically proved because any number can be written as the product of that number and number one.  1. Base case (n = 2)  Since \(2 \) is a prime number the statement is automatically proved.  2. Induction hypothesis (n = m)  Suppose that \(\forall k \in \mathbb {N}, \quad 2 \le k \le m \), \(k \) can be expressed as the product of a prime number and number one or as a product of few prime numbers.  3. Inductive step (n = m + 1)  Using the assumption from the second step, we will prove that: \

Hello everyone.  Today we will prove the theorem regarding Fibonacci prime numbers.  We will use the method of contradiction as a proof method.  Theorem: Let \(F_n \) be the nth Fibonacci number and let \(F_n \) be a prime number.  Then \(n \) is also a prime number, except in the case of \(F_4 = 3 \).  Proof:   For the case when \(n = 2 \) we have that \(F_2 = 1 \), and as we know the number \(1 \) is neither simple nor composite, so this case does not refute the truth of the theorem.  For the case when \(n = 3 \) we have that \(F_3 = 2 \), so this case is in accordance with the statement.  Now, suppose that for \(n> 4 \), \(F_n \) is a prime number and that \(n = rs \) for some natural numbers \(r, s \) which are greater than \(1 \), that is, that \(n \) is a composite number.  Since \(n> 4 \)  at least one of the numbers \(r \) and \(s \) is greater than \(2 \).  Then, according to the divisibility theorem of Fibonacci numbers which reads: \[\forall m, n \in \mathbb {Z} _

Hello everyone.  Today we will prove the formula for the sum of the first \(n \) Fibonacci numbers.  We will use the method of mathematical induction as a proof method.  Theorem: \(\forall n \in \mathbb {N} _0, \quad \displaystyle \sum_ {j = 0} ^ nF_j = F_ {n + 2} -1 \)  Proof:   1. Base case (n = 0)  \[\displaystyle \sum_ {j = 0} ^ 0F_j = F_ {2} -1 \] \[F_ {0} = F_ {2} -1 \] \[0 = 1-1 \] \[0 = 0 \]  2. Induction hypothesis (n = m)  Suppose that: \[\displaystyle \sum_ {j = 0} ^ mF_j = F_ {m + 2} -1 \]  3. Inductive step (n = m + 1)  Using the assumption from the second step, we will prove that: \[\displaystyle \sum_ {j = 0} ^ {m + 1} F_j = F_ {m + 3} -1 \]  So,  \[\displaystyle \sum_ { j = 0} ^ {m + 1} F_j = \displaystyle \sum_ {j = 0} ^ {m} F_j + F_ {m + 1} = \] \[F_ {m + 2} -1 + F_ {m +1} = \] \[F_ {m + 1} + F_ {m + 2} -1 = \] \[F_ {m + 3} -1 \] \(\blacksquare\)

Hello everyone.  Today we will prove the formula for the area of ​​a circle.  We will use the method of direct proof as a proof method.  Theorem: Denote by \(P \) the area of ​​a circle and by \(r \) the radius of a circle.  Then the following equation holds: \(P = r ^ 2 \pi \)  Proof:   The equation of a circle in Cartesian  coordinate system is \(x ^ 2 + y ^ 2 = r ^ 2 \).  Hence we have that \(y = \pm \sqrt {r ^ 2-x ^ 2} \).  Based on the geometric interpretation of a certain integral, it follows: \[P = \int \limits _ {- r} ^ r \left(\sqrt {r ^ 2-x ^ 2} - \left (- \sqrt {r ^ 2-x ^ 2} \right) \right) \, dx \]  So, \[P = \int \limits _ {- r} ^ r 2 \sqrt {r ^ 2-x ^ 2} \, dx \] \[P = \int \limits _ {- r} ^ r 2 \sqrt {r ^ 2 \left (1- \frac {x ^ 2} {r ^ 2} \right)} \, dx \] \[P = \int \limits _ {- r} ^ r 2r \sqrt {1- \frac {x ^ 2} {r ^ 2}} \, dx \]  Let us now introduce the substitution \(x = r \cos \theta \).  Hence, we have that \(dx = -r \sin \theta d \theta \).  So we can write the