Mathemarium

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 assum...

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 ...

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 t...

Hello everyone.  Today we will prove the formula for the nth derivative of the natural logarithm.  We will use the method of mathematical induction as a proof method.  Theorem: The nth derivative of the function $\ln (x)$ for $n \ge 1$ is given by the formula: $$\frac {\mathrm {d} ^ n} { \mathrm {d} x ^ n} \ln (x) = \frac {(n-1)! (-1) ^ {n-1}} {x ^ n}$$  Proof:  1. Base case (n = 1) $$\frac {\mathrm {d}} {\mathrm {d} x} \ln (x) = \frac {(1-1)! (-1) ^ {1-1}} {x ^ 1}$$ $$\frac {1} {x} = \frac {(0)! (-1) ^ {0}} {x}$$ $$\frac {1} {x} = \frac {1} {x}$$  2. Induction hypothesis (n = m)  Suppose that:  $$\frac{\mathrm{d}^m}{\mathrm{d}x^m}\ln(x)=\frac{(m-1)!(-1)^{m-1}}{x^m}$$ 3. Inductive step (n = m + 1)  Using the assumption from the second step, we will prove that: $$\frac {\mathrm {d} ^ {m + 1}} {\mathrm {d} x ^ {m + 1}} \ln (x) = \frac {m! (-1) ^ {m}} {x ^ {m + 1}}$$ So, \[\frac{\mathrm{d}^{m+1}}{\mathrm{d}x^{m+1}}\ln(x)=\frac{\m...

Hello everyone.  Today we will prove that the neighboring Fibonacci numbers are coprime.  We will use the method of mathematical induction as a proof method.  Theorem: Let $F_n$ represent the nth Fibonacci number.  Then: $$\forall n \ge 2, \quad \operatorname{NZD} \left (F_n, F_ {n + 1} \right) = 1$$  Proof:   1. Base case (n = 2)  $$\operatorname{NZD} \left(F_2, F_{3} \right) = \operatorname{NZD} (1,2) = 1$$  2. Induction hypothesis (n = m)  Suppose that: $$\operatorname { NZD} \left (F_m, F_ {m + 1} \right) = 1$$  3. Inductive step (n = m + 1)  Using the assumption from the second step, we will prove that:  $$\operatorname{NZD}\left(F_{m+1},F_{m+2}\right)=1$$ Since the greatest common divisor of any natural numbers $a$ and $b$ is equal to the greatest common divisor of any linear combination of numbers $a$ and $b$ we have that $\operatorname {NZD} (a, b) = \operatorname {NZD} (a, ba)$.  With thi...

Hello everyone.  Today we will prove the formula for the sum of the first $n$ natural numbers.  We will use the method of mathematical induction as a proof method.  It is believed that the Pythagoreans also knew this formula.  Theorem: For any natural number $n$ the following formula holds: $$\displaystyle \sum_{k = 1} ^ nk = \frac {n (n + 1)} {2}$$  Proof:   1. Base case (n = 1) $$\displaystyle \sum_{k = 1} ^ {1} k = \frac {1 \cdot (1 + 1)} {2}$$ $$1 = \frac {1 \cdot (2)} {2}$$ $$1 = \frac {2} {2}$$ $$1 = 1$$  2. Induction hypothesis (n = m)  Suppose that: $$\displaystyle \sum_{ k = 1} ^ mk = \frac {m (m + 1)} {2}$$  3. Inductive step (n = m + 1)  Using the assumption from the second step, we will prove that: $$\displaystyle \sum_{k = 1} ^ {m + 1} k = \frac {(m + 1) (m + 2)} {2}$$  So, $$\displaystyle \sum_{k = 1} ^ {m + 1} k = \displaystyle \sum_{k = 1} ^ {m} k + m + 1 =$$ $$\frac {m (m +1)} {2} + m + 1 =$$ \[\fra...

Hello everyone.  Today we will prove the sine theorem.  We will use the method of direct proof as a proof method.  This proof was first constructed by the Persian mathematician Tusi.  Theorem: Let $a, b, c$ be the sides of any triangle, and $R$ the radius of the circumscribed circle around that triangle and let the angles $\alpha, \beta, \gamma$ be the angles opposite the sides $a, b, c$ , respectively.  Then the following equations hold: $$\frac {a} {\sin \alpha} = \frac {b} {\sin \beta} = \frac {c} {\sin \gamma} = 2R$$  Proof:   Now we will prove that the following equation holds: $\frac {a} {\sin \alpha} = 2R$.  The equations $\frac {b} {\sin \beta} = 2R$ and $\frac {c} {\sin \gamma} = 2R$ can be proved in an analogous way.  Consider the following diagram showing the triangle $\triangle ABC$ with the circumscribed  circle of radius $R$. Since angles inscribed in a circle and subtended by the same chor...

Hello everyone.  Today we will prove the cosine theorem.  We will use the method of direct proof as a proof method.  This theorem was first formulated by the Persian mathematician Kashani.  Theorem: Let $a, b, c$ be the sides of any triangle and let the angles $\alpha, \beta, \gamma$ be the angles opposite the sides $a, b, c$, respectively.  Then the following equations hold: $$c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma$$ $$b ^ 2 = a ^ 2 + c ^ 2-2ac \cos \beta$$ $$a ^ 2 = b ^ 2 + c ^ 2-2bc \cos \alpha$$  Proof:  We will now prove that the first equation holds i.e. : $$c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma$$ Other two equations can be proved in an analogous way.  There are three possible cases, and these are: $\gamma$ is a right angle, $\gamma$ is an acute angle, and $\gamma$ is an obtuse angle.  First case: $\gamma = 90 ^ {\circ}$  According to the Pythagoras' theorem, we know that for a right triangle with hypot...

Hello everyone.  Today we will prove that the central angle of a circle is equal to twice the corresponding inscribed angle of the circle.  We will use the method of direct proof as a proof method.  Theorem: The central angle of a circle is equal to twice the corresponding inscribed angle of the circle.   Proof:   We can reformulate the statement as follows:  Let $P, Q, R$ be three arbitrary points on the circumference of a circle $k (O, r)$.  Then, $\angle QOP = 2 \angle QRP$.  We will construct the proof by proving three separate possible cases:  First case: The center of the circle is on a side of the inscribed angle.   We can write the following equations : $$\angle POR + \angle ORP + \angle RPO = 180 ^ {\circ}$$ $$\angle QOP = 180 ^{\circ} - \angle POR$$ $$\angle ORP = \angle QRP$$ Combining the first and second equation we get: $$\angle QOP = \angle ORP + \angle RPO$$ Since triangle $\triangle POR$ is isoscel...

   Hello everyone.  Today we will prove that an integer is odd if its square is odd. We will use proof by contraposition as proof method.   Theorem: Let $n$ be an integer.  If $n ^ 2$ is an odd number, then $n$ is also an odd number.  Proof:   The contrapositive of this statement is: Let $n$ be an integer.  If $n$ is an even number, then $n ^ 2$ is also an even number.  Let us now prove the contrapositive. Since  $n$ is an even number, we can write it in the form $n = 2k$ where $k$ is an integer.  By squaring this equation we get: $$n ^ 2 = (2k) ^ 2$$ $$n ^ 2 = 4k ^ 2$$ $$n ^ 2 = 2 \left(2k ^ 2 \right)$$ Since $k$ is an  integer then $2k ^ 2$ must be an integer due to the closedness of multiplication and exponentiation operations on the set of integers. Denote  $2k ^ 2$ by $r$, then we have $n ^ 2 = 2r$,  from which we conclude that $n ^ 2$ is an even number. Since ...

  Hello everyone.  Today we will prove that an integer is even if its square is even.  We will use proof by contraposition as proof method. Theorem: Let $n$ be an integer.  If $n ^ 2$ is an even number, then $n$ is also an even number.  Proof:   The contrapositive of this statement is: Let $n$ be an integer.  If $n$ is an odd number, then $n ^ 2$ is also an odd number.  Let us now prove the contrapositive. Since  $n$ is an odd number, we can write it in the form $n = 2k + 1$ where $k$ is an integer.  By squaring this equation we get: $$n ^ 2 = (2k + 1) ^ 2$$ $$n ^ 2 = 4k ^ 2 + 4k + 1$$ $$n ^ 2 = 2 \left(2k ^ 2 + 2k \right) +1$$ Since $k$ is an integer then $2k ^ 2 + 2k$ must be an integer due to the closedness of addition, multiplication and addition operations on the set of integers. Denote  $2k ^ 2 + 2k$ by $l$,  then we have that $n ^ 2 = 2l + 1$, from which we conclude that ...

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} \c...

Hello everyone. Today we will prove Archimedes' theorem.  We will use the method of contradiction as a proof method.  Theorem: For every two real numbers $a$ and $b$ where $a> 0$ there exists a natural number $n$ such that $a \cdot n> b$.  Proof:   Considering that $a> 0$  we can divide  the inequality $a \cdot n> b$   by the number $a$ so that we get a new inequality $n> \frac{b}{a}$ .  Let us denote the real number $\frac{b}{a}$ by $x$, then we can reformulate the theorem we need to prove as follows:  For any real number $x$ there exists a natural number $n$ such that $n> x$.  Suppose the opposite, ie.  that there exists a real number $x$ such that $x \ge n$ for any natural number $n$.  This would mean that the set of natural numbers $\mathbb {N}$ is bounded from above.  Let us denote this least upper bound by \(\operatorname{sup}(\mathbb{N...

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_...

Hello everyone.  Today we will prove Pythagoras' theorem using elementary algebra. This proof was first appeared in China more than 2000 years ago.  Theorem: Let $a$ and $b$ be the legs, and $c$ the hypotenuse of a right triangle.  Then the equation $c ^ 2 = a ^ 2 + b ^ 2$ holds.  Proof:  Consider the following diagram consisting of a large square with side length $a + b$ and a small square with side length $c$ and notice four right triangles with legs $a, b$ and hypotenuse $c$. Let us denote the area of ​​the large square by $P$. Then $$P = (a + b)(a + b)$$ Let $P_1$ be the area of ​​one of the four right triangles, and $P_2$ the area of ​​a small square with side length $c$.  Then we have that: $$P_1 = \frac {ab} {2}$$ $$P_2 = c ^ 2$$ Since $P = 4P_1 + P_2$ we can write the following equation: $$(a + b) (a + b) = 4 \cdot \frac {ab} {2} + c ^ 2$$  From here we have that: \[a ^ 2 + 2ab + b ^ 2 = 2ab +...

  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...