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 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. 

\(\blacksquare \)

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