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# The Logarithmic Triangle and Zêta function

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Imagine a perfect golden triangle. How would it be possible to make it so that is a modulated triangle which could work in diversed supset&#39;s but that would need to be first included in a infited sized lenght such as the triangle has no boundaries
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The Logarithmic Triangle and Zˆeta function Thomas, Julien Naret  ∗ 11 University Paris 13, Epinay-Villetaneuse, FranceAugust 12, 2015 Abstract Imagine a perfect golden triangle. How would it be possible to makeit so that is a modulated triangle which could work in diversed supset’sbut that would need to be ﬁrst included in a inﬁted sized lenght such asthe triangle has no boundaries and is also capable of being a line? 1 Introduction Let us deﬁne two sets of prime numbers. P  α  and  P  β P  α  is a subset of   P  β  that can only be factorised with factors in  D whereas  P  β  can be both factorised with  D  and  N  factors.Let us deﬁne an array of combinations of all  P  α  and  P  β  called  Q . Q  is a generated theorerically inﬁnite-sized-array that uses all possible com-binaisons of   P  α  and  P  β  and generates the result of the factorial combinations.The number of rows of this array depend on the sum of all P  α  and P  β  consiredpowered to the same number giving  Q ( x m ) and the number of collumns is thesum of numbers in  P  α  and  P  β  combined, called  Q ( y m )Where  N  ( P  α ) and  N  ( P  β ) are the number of numbers considered in eachsubsets. and Where  Q ( x m ) =  P  α  +  P  β  + 1 and  Q ( y m ) =  P  α  ×  P  β . 2 Demonstration →  P  β  ⊃  P  α  ∧  P  β  → ∃  x:P(x)x:P(x)  ∃ →  x:P(x)    Q  →  Q  = ⇒  Q  =  Q ( x m ) Q ( y m ) ∴  Q  ∃  P   :=  Q  ∨  NP   ∧  P   :=  Q  ∀  x  &  y = ⇒  Q ( x m ) Q ( y m )  ≡  Q ( x m ) 2 +  Q ( y m ) 2 =  Z  2 =  ∞ ∞ = ⇒  Z   ∃ ∨  (x,y)= ⇒  Z   ∃ ∞ = ⇒  Z  ( x )  ∃ ∀  (x) & (y) ∗  julien.naret@edu.univ-paris13.fr 1  = ⇒  :=  Z  ( x )  ∃ ! tan  θ = ⇒  tan  θ  =  Q ( x m ) − 0 + Q ( y m ) − 0 +   = 0= ⇒  Z   ∃  R  ∀  Z  ( x ) =  12  +  i t ⇔ ∃  tan  θ  =  ϕ  ∃  !  ∀ ⊃  n  ∃ ⊃  n 2 = ⇒  (x,y)  ∀  x =  ⊃  n/2 & y =  ⊃  n 3 Conclusion Z  ( x ) is equal to  12  +  i t . So accordingly, the Riemann Hypothesis is proved tobe accurate but only with tangente operator of   ϕ  which is constructable froma supset which is equivalent to twice the size of another. This can be possibleusing supsets of any inversed quantiﬁcations and restricting one of them to onlyone ﬁnite set such as supset n and supset  n 2  are commutative in forming the ϕ  angle leading to the 1/2 portion of the  Z  ( x ) equation leaving the  i  and  t  beproven by both  Z  ( x ) been equivalent to both  ∞  and of degree 0 depending onthe vector-space’s format.2
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