Proof Concerning Central Line X5X6X_5X_6 of Triangle

Image
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...
Post a Comment

Proof that the set of prime numbers is infinite

Hello everyone. Today we will prove that the set of prime numbers is infinite. We will use the method of contradiction as a proof method. We'll also use the theorem of French mathematician Eduardo Lucas. 

Theorem (Lucas): Every prime factor of Fermat number Fn=22n+1F _ n = 2 ^ {2 ^ n} + 1; (n>1n > 1) is of the form k2n+2+1k2 ^{n + 2} + 1. 

Theorem: The set of prime numbers is infinite.

Proof:

Suppose opposite, that there are just finally many prime numbers and we denote the largest prime by pp. Then FpF_p must be a composite number because Fp>pF_p>p. By Lucas theorem we know that there is a prime number qq of the form k2p+2+1k2 ^{p + 2} + 1 that divides FpF_p. But q>pq>p , thus we arrived at  a contradiction. Hence, the set of prime numbers is infinite.

■\blacksquare

Comments

Post a Comment

Popular posts from this blog

Proof of the product formula for π23\dfrac{\pi}{2\sqrt{3}}

  Hello everyone.  Today we will prove the product formula for  π23\dfrac{\pi}{2\sqrt{3}}.  We will use the method of direct proof as a proof method.  Theorem 1:  π23=∑n=1∞χ(n)n\frac{\pi}{2\sqrt{3}}=\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)}{n}whereχ(n)={1,if n≡1(mod6)−1,if n≡−1(mod6)0,otherwise\text{where} \quad \chi(n)=\begin{cases} 1, & \text{if } n \equiv 1 \pmod{6}\\-1, & \text{if } n \equiv -1 \pmod{6}\\0, & \text{otherwise}\end{cases} Theorem 2:  We haveπ23=5⋅7⋅11⋅13⋅17⋅19⋅23⋅29⋯6⋅6⋅12⋅12⋅18⋅18⋅24⋅30⋯\frac{\pi}{2\sqrt{3}}=\frac{5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdots}{6 \cdot 6 \cdot 12 \cdot 12 \cdot 18 \cdot 18 \cdot 24 \cdot 30 \cdots}expression whose numerators are the sequence of the odd prime numbers greater than 33 and whose denominators are even–even numbers one unit more or less than the corresponding numerators. Proof: By Theorem 1 we know that π23=1−15+17−111+113−117+119−⋯\frac{\pi}{2\sqrt{3}}=1-\frac{1}{5}+\frac{1}{7}-\frac{1}{11}+\frac{1}{13}-\frac{1}{17}+\frac{1}{19}-\cdots we will have \[\frac{1}{5} \cdo...
Read more

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 PP the area of ​​a circle and by rr the radius of a circle.  Then the following equation holds: P=r2πP = r ^ 2 \pi   Proof:   The equation of a circle in Cartesian  coordinate system is x2+y2=r2x ^ 2 + y ^ 2 = r ^ 2 .  Hence we have that y=±r2−x2y = \pm \sqrt {r ^ 2-x ^ 2} .  Based on the geometric interpretation of a certain integral, it follows: P=∫−rr(r2−x2−(−r2−x2)) dxP = \int \limits _ {- r} ^ r \left(\sqrt {r ^ 2-x ^ 2} - \left (- \sqrt {r ^ 2-x ^ 2} \right) \right) \, dx   So, P=∫−rr2r2−x2 dxP = \int \limits _ {- r} ^ r 2 \sqrt {r ^ 2-x ^ 2} \, dx P=∫−rr2r2(1−x2r2) dxP = \int \limits _ {- r} ^ r 2 \sqrt {r ^ 2 \left (1- \frac {x ^ 2} {r ^ 2} \right)} \, dx P=∫−rr2r1−x2r2 dxP = \int \limits _ {- r} ^ r 2r \sqrt {1- \frac {x ^ 2} {r ^ 2}} \, dx   Let us now introduce the substitution x=rcos⁡θx = r \cos \theta .  Hence, we have t...
Read more

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 nn Fibonacci numbers.  We will use the method of mathematical induction as a proof method.  Theorem: ∀n∈N0,∑j=0nFj=Fn+2−1\forall n \in \mathbb {N} _0, \quad \displaystyle \sum_ {j = 0} ^ nF_j = F_ {n + 2} -1   Proof:   1. Base case (n = 0)  ∑j=00Fj=F2−1\displaystyle \sum_ {j = 0} ^ 0F_j = F_ {2} -1 F0=F2−1F_ {0} = F_ {2} -1 0=1−10 = 1-1 0=00 = 0   2. Induction hypothesis (n = m)  Suppose that: ∑j=0mFj=Fm+2−1\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: ∑j=0m+1Fj=Fm+3−1\displaystyle \sum_ {j = 0} ^ {m + 1} F_j = F_ {m + 3} -1   So,  ∑j=0m+1Fj=∑j=0mFj+Fm+1=\displaystyle \sum_ { j = 0} ^ {m + 1} F_j = \displaystyle \sum_ {j = 0} ^ {m} F_j + F_ {m + 1} = Fm+2−1+Fm+1=F_ {m + 2} -1 + F_ {m +1} = Fm+1+Fm+2−1=F_ {m + 1} + F_ {m + 2} -1 = Fm+3−1F_ {m + 3} -1 ■\blacksquare
Read more