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 of Archimedes' theorem

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 aa and bb where a>0a> 0 there exists a natural number nn such that a⋅n>ba \cdot n> b . 

Proof: 

Considering that a>0a> 0   we can divide the inequality a⋅n>ba \cdot n> b   by the number aa so that we get a new inequality n>ban> \frac{b}{a} . Let us denote the real number ba\frac{b}{a} by xx , then we can reformulate the theorem we need to prove as follows:

 For any real number xx there exists a natural number nn such that n>xn> x . 

Suppose the opposite, ie. that there exists a real number xx such that x≥nx \ge n for any natural number nn . This would mean that the set of natural numbers N\mathbb {N} is bounded from above. Let us denote this least upper bound by sup⁡(N)=S\operatorname{sup}(\mathbb{N}) = S . Now consider the difference S−1S-1 . It is clear that S−1S-1 is not the upper bound of the set N\mathbb{N} , because if it were then SS that is greater than S−1S-1 would not be the least upper bound. This means that there exists a natural number mm such that m>S−1m> S-1 , that is, we have m+1>Sm + 1> S . As the addition operation is closed on the set of natural numbers we have that m+1∈Nm + 1 \in \mathbb{N} , ie. the number m+1m + 1 is a natural number greater than sup⁡(N)\operatorname{sup} (\mathbb{N}) . This contradicts the initial assumption which means that for any real number xx there exists a natural number nn such that n>xn> x . 

■\blacksquare

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