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...

Proof that there are infinitely many prime numbers

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 nn , and the prime numbers themselves by pp and the indices 1,2,n1,n1,2 \ldots, n-1, n So we have a finite set of primes p1,p2,,pn1,pnp_1, p_2, \ldots, p_{n-1}, p_n

Now, we construct the number qq as follows: q=p1p2pn1pn+1q = p_1 \cdot p_2 \cdot \ldots \cdot p_ {n-1} \cdot p_n + 1 Then qq can be either prime or a composite number. It is clear that qq cannot be a prime number because we assumed that the number of primes is finite and that the largest prime number is the number pnp_n Therefore, qq must be a composite number. If qq is a composite number then there is some prime number pp from the set p1,p2,,pn1,pnp_1, p_2, \ldots, p_{n-1}, p_n that divides it. Denote by PP product p1p2pn1pnp_1 \cdot p_2 \cdot \ldots \cdot p_{n-1} \cdot p_n Then pp simultaneously divides both PP and P+1=qP + 1 = q which means that pp must also divide the difference (P+1)P(P + 1) -P , i.e.  it follows that pp divides the number 11 But since no prime number divides the number 11 we conclude that pp cannot be in the set p1,p2,,pn1,pnp_1, p_2, \ldots, p_{n-1}, p_n This means that no matter what the number nn is, there is always at least one more prime number outside the finite set, therefore there are infinitely many prime numbers. 

\blacksquare  

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