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 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

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{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}

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 this in mind, we can write the following equation: \[\operatorname

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, \[\displaystyle \sum_{k = 1} ^ {m + 1} k = \displaystyle \sum_{k = 1} ^ {m} k + m + 1 = \] \[\frac {m (m +1)} {2} + m + 1 = \] \[\frac {m ^ 2 + m + 2m + 2} {2} = \] \[\frac {m ^ 2 + 2m +

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 chord are equal  we have that \(\angle CA'B = \al

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 hypotenuse \(c \) the following equation holds:  \

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 isosceles we have that \(\angle RPO = \angle ORP \), so: \[\a

   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  we have proved that the contrapositive of the original statement is co

  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 \(n ^ 2 \) is an odd number. Since  we have proved that the

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^{

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 \(\operatorname{sup}(\mathbb{N}) = S \).  Now consider the difference \(S-1 \).  It is clear th

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

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: \[a ^ 2 + 2ab + b ^ 2 = 2ab + c ^ 2 \]  That is, when we subtrac

  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.e.  \(4k^2 = 2q^2 \), or \(q^2 = 2k^2 \).  From the last equati