Proof Concerning Central Line X5X6X_5X_6 of Triangle

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Hello everyone. Today we will prove theorem about central line X5X6X_5X_6 of triangle. Proof of the Central Line in a Golden Rectangle Construction Statement of the Theorem Let ABCDABCD be a golden rectangle where ABBC=ϕ \frac{AB}{BC} = \phi , and construct the square BCQP BCQP inside it. Reflect P P over D D to obtain E E . Then, the line EB EB coincides with the central line X5X6 X_5X_6 of triangle ABP ABP . Step-by-Step Proof 1. Define the Coordinates We assign coordinates as follows: A=(0,0) A = (0,0) , B=(ϕx,0) B = (\phi x, 0) , C=(ϕx,x) C = (\phi x, x) , D=(0,x) D = (0, x) . The square BCPQ BCPQ ensures P=(ϕx,2x) P = (\phi x, 2x) . The reflection of P P across D D is E=(−ϕx,2x) E = (-\phi x, 2x) . 2. Compute the Nine-Point Center X5 X_5 The nine-point center X5 X_5 is the circumcenter of the medial triangle, which consists of the midpoints: M1=(0+ϕx2,0)=(ϕx2,0), 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 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 FnF_n represent the nth Fibonacci number. Then: ∀n≥2,NZD⁡(Fn,Fn+1)=1\forall n \ge 2, \quad \operatorname{NZD} \left (F_n, F_ {n + 1} \right) = 1  

Proof: 

1. Base case (n = 2) 

NZD⁡(F2,F3)=NZD⁡(1,2)=1 \operatorname{NZD} \left(F_2, F_{3} \right) = \operatorname{NZD} (1,2) = 1  

2. Induction hypothesis (n = m) 

Suppose that: NZD⁡(Fm,Fm+1)=1\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: NZD⁡(Fm+1,Fm+2)=1\operatorname{NZD}\left(F_{m+1},F_{m+2}\right)=1Since the greatest common divisor of any natural numbers aa and bb is equal to the greatest common divisor of any linear combination of numbers aa and b b we have that NZD⁡(a,b)=NZD⁡(a,ba)\operatorname {NZD} (a, b) = \operatorname {NZD} (a, ba) . With this in mind, we can write the following equation: NZD⁡(Fm+1,Fm+2)=NZD⁡(Fm+1,Fm+2−Fm+1)\operatorname {NZD} \left (F_ {m + 1}, F_ {m + 2} \right) = \operatorname {NZD} \left(F_ {m + 1} , F_ {m + 2} -F_ {m + 1} \right) Next,  by the definition of the Fibonacci sequence Fm+2=Fm+Fm+1F_ {m + 2} = F_ {m} + F_ {m + 1} we have that: NZD⁡(Fm+1,Fm+2)=NZD⁡(Fm+1,Fm)\operatorname {NZD} \left (F_ {m + 1}, F_ {m + 2} \right) = \operatorname {NZD} \left (F_ {m + 1}, F_ {m} \right) NZD⁡(Fm+1,Fm+2)=NZD⁡(Fm,Fm+1)=1\operatorname {NZD} \left (F_ {m + 1}, F_ {m + 2} \right) = \operatorname {NZD} \left (F_ {m}, F_ {m + 1} \right) = 1  

■\blacksquare

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