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Proof Concerning Central Line \(X_5X_6\) of Triangle

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Hello everyone. Today we will prove theorem about central line \(X_5X_6\) of triangle. Proof of the Central Line in a Golden Rectangle Construction Statement of the Theorem Let \(ABCD\) be a golden rectangle where \( \frac{AB}{BC} = \phi \), and construct the square \( BCQP \) inside it. Reflect \( P \) over \( D \) to obtain \( E \). Then, the line \( EB \) coincides with the central line \( X_5X_6 \) of triangle \( ABP \). Step-by-Step Proof 1. Define the Coordinates We assign coordinates as follows: \( A = (0,0) \), \( B = (\phi x, 0) \), \( C = (\phi x, x) \), \( D = (0, x) \). The square \( BCPQ \) ensures \( P = (\phi x, 2x) \). The reflection of \( P \) across \( D \) is \( E = (-\phi x, 2x) \). 2. Compute the Nine-Point Center \( X_5 \) The nine-point center \( X_5 \) is the circumcenter of the medial triangle, which consists of the midpoints: \[ M_1 = \left(\frac{0 + \phi x}{2}, 0\right) = \left(\frac{\phi x}{2}, 0\right), \] \[ M_2 = \left(\f...
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Proof that a natural number can be expressed as a product of prime numbers

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...
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Proof of the theorem regarding Fibonacci prime numbers

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 ...
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Proof of the formula for the sum of the first n Fibonacci numbers

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,  \[\displaystyle \sum_ { j = 0} ^ {m + 1} F_j = \displaystyle \sum_ {j = 0} ^ {m} F_j + F_ {m + 1} = \] \[F_ {m + 2} -1 + F_ {m +1} = \] \[F_ {m + 1} + F_ {m + 2} -1 = \] \[F_ {m + 3} -1 \] \(\blacksquare\)
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Proof of the formula for the area of a circle

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...
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Proof of the formula for the nth derivative of the natural logarithm

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...
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Proof that the neighboring Fibonacci numbers are coprime

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...
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Proof of the formula for the sum of the first n natural numbers

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...
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