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User talk:Musictheory2math
Goldbach's conjecture
Main theorem: Let $\mathbb{P}$ is the set prime numbers and $S$ is a set that has been made as below: put a point at the beginning of each member of $\Bbb{P}$ like $0.2$ or $0.19$ then $S=\{0.2,0.3,0.5,0.7,...\}$ is dense in the interval $[0.1,1]$ of real numbers.
$\,$This theorem is a base for finding formula of prime numbers, because for each member of $S$ like $a$ with its special and fixed location into $(0.1,1)$ and a small enough neighborhood like $(a-\epsilon ,a+\epsilon )$, but $a$ is in a special relation with members of $(a-\epsilon ,a+\epsilon )$ but there exists a special order on $S$ into $(0.1,1)$ and of course formula of prime numbers has whole properties related to prime numbers simultaneous. There is a musical note on the natural numbers that can be discovered by the formula of prime numbers. Musictheory2math (talk) 16:29, 25 March 2017 (CET)
- True, $S$ is dense in the interval $(0.1,1)$; this fact follows easily from well-known results on Distribution of prime numbers. But I doubt that this is "This theorem is a base for finding formula of prime numbers". Boris Tsirelson (talk) 22:10, 16 March 2017 (CET)
- Dear Professor Boris Tsirelson , in principle finding formula of prime numbers is very lengthy. and I am not sure be able for it but please give me few time about two month for expression my theories. Musictheory2math (talk) 16:29, 25 March 2017 (CET)
- You mean, how to prove that $S$ is dense in $(0.1,1)$, right? Well, on the page "Distribution of prime numbers", in Section 6 "The difference between prime numbers", we have $ d_n \ll p_n^\delta $, where $p_n$ is the $n$-th prime number, and $ d_n = p_{n+1}-p_n $ is the difference between adjacent prime numbers; this relation holds for all $ \delta > \frac{7}{12} $; in particular, taking $ \delta = 1 $ we get $ d_n \ll p_n $, that is, $ \frac{d_n}{p_n} \to 0 $ (as $ n \to \infty $), or equivalently, $ \frac{p_{n+1}}{p_n} \to 1 $. Now, your set $S$ consists of numbers $ s_n = 10^{-k} p_n $ for all $k$ and $n$ such that $ 10^{k-1} < p_n < 10^k $. Assume that $S$ is not dense in $(0.1,1).$ Take $a$ and $b$ such that $ 0.1 < a < b < 1 $ and $ s_n \notin (a,b) $ for all $n$; that is, no $p_n$ belongs to the set
\[ X = (10a,10b) \cup (100a,100b) \cup (1000a,1000b) \cup \dots \, ; \]
- all $ p_n $ belong to its complement
\[ Y = (0,\infty) \setminus X = (0,10a] \cup [10b,100a] \cup [100b,1000a] \cup \dots \]
- Using the relation $ \frac{p_{n+1}}{p_n} \to 1 $ we take $N$ such that $ \frac{p_{n+1}}{p_n} < \frac b a $ for all $n>N$. Now, all numbers $p_n$ for $n>N$ must belong to a single interval $ [10^{k-1} b, 10^k a] $, since it cannot happen that $ p_n \le 10^k a $ and $ p_{n+1} \ge 10^k b $ (and $n>N$). We get a contradiction: $ p_n \to \infty $ but $ p_n \le 10^k a $.
- And again, please sign your messages (on talk pages) with four tildas: ~~~~. Boris Tsirelson (talk) 20:57, 18 March 2017 (CET)
Theorem $1$: For each natural number like $a=a_1a_2a_3...a_k$ that $a_j$ is $j$_th digit for $j=1,2,3,...,k$, there is a natural number like $b=b_1b_2b_3...b_r$ such that the number $c=a_1a_2a_3...a_kb_1b_2b_3...b_r$ is a prime number. Musictheory2math (talk) 16:29, 25 March 2017 (CET)
- Ah, yes, I see, this follows easily from the fact that $S$ is dense. Sounds good. Though, decimal digits are of little interest in the number theory. (I think so; but I am not an expert in the number theory.) Boris Tsirelson (talk) 11:16, 19 March 2017 (CET)
Now I want state philosophy of This theorem is a base for finding formula of prime numbers: However we loose the induction axiom for finite sets (Induction axiom is unable for discovering formula of prime numbers.) but I thought that if change space from natural numbers with cardinal $\aleph_0$ to a bounded set with cardinal $\aleph_1$ in the real numbers then we can use other features like axioms and important theorems in the real numbers for working on prime numbers and I think it is a better and easier way. Musictheory2math (talk) 16:29, 25 March 2017 (CET)
- I see. Well, we are free to use the whole strength of mathematics (including analysis) in the number theory; and in fact, analysis is widely used, as you may see in the article "Distribution of prime numbers".
- But you still do not put four tildas at the end of each your message; please do. Boris Tsirelson (talk) 11:16, 19 March 2017 (CET)
Importance of density in the Main theorem is similar to definition of irrational numbers from rational numbers.
Goldbach's conjecture is one of the oldest and best-known unsolved problems in number theory and all of mathematics. It states: Every even integer greater than $2$ can be expressed as the sum of two primes.
Assume $S_1=\{a/10^n\, |\, a\in S$ for $n=0,1,2,3,...\}$ & $L=\{(a,b)\,|\,a,b \in S_1$ & $0.01 \le a+b \lt 0.1$ & $\exists m \in \Bbb N,\, a \cdot 10^m,\,b \cdot 10^m$ are prime numbers & $a\cdot 10^m\neq 2\neq b\cdot 10^m\}$
Theorem: $S_1$ is dense in the interval $(0,1)$ and $S_1\times S_1$ is dense in the $(0,1)\times (0,1)$.
If $p,q$ are prime numbers and $n$ is the number of digits in $p+q$ and $m=$max(number of digits in $p$, number of digits in $q$), let $\varphi : L \to \Bbb N,$ $\varphi ((p,q)) = \begin{cases} m+1 & n=m \\m+2 & n=m+1 \end{cases}$
Theorem: For each $p,q$ belong to prime numbers and $\alpha \in \Bbb R$ that $0 \le \alpha,$ now if $\alpha = q/p$ then $L \cap \{(x,y)\,|\,y=\alpha x \}=\{10^{-\varphi ((p,q))}(p,q)\}$ and if $\alpha \neq q/p$ then $L \cap \{(x,y)\,|\,y=\alpha x \}=\emptyset $ and if $\alpha = 1$ then $L \cap \{(x,x)\}$ is dense in the $\{(x,x)\,|\,0.005 \le x \lt 0.05 \}$.
Definition: Assume $L_1=\{(a,b)\,|\,(a,b) \in L$ & $b \lt a \}$ of course members in $L$ & $L_1$ are corresponding to prime numbers as multiplication and sum and minus and let $E=(0.007,0.005)$ (and also $5$ points to form of $(0.007+\epsilon _1,0.005-\epsilon _2)$ that $\epsilon _2 \approx 2\epsilon _1$) is a base for homotopy groups! and let $A:=\{(x,y)\,|\, 0 \lt y\lt x,$ $0.01 \le x+y \lt 0.1\}$ & $V:=\{(a+b)\cdot 10^m \,|$ $(a,b) \in ((S_1 \times S_1) \cap A) \setminus L,$ there is a least member like $m$ in $\Bbb N$ such that $(a+b)\cdot 10^m \in \Bbb N \}$ & $r:\Bbb N\to (0,1)$ is a function given by $r(n)$ is obtained as put a point at the beginning of $n$ like $r(34880)=0.34880$ and similarly consider $\forall k\in\Bbb N\cup\{0\}$ $r_k: \Bbb N \to (0,1)$ by $r_k(n)=10^{-k}\cdot r(n)$.
Conjecture $1$: For each even natural number like $t=t_1t_2t_3...t_k$, then $\exists (a,b),(b,a)\in L \cap$ $\{(x,y)\,|$ $x+y=0.0t_1t_2t_3...t_k\}$ such that $0.0t_1t_2t_3...t_k=a+b$ & $10^{k+1} \cdot a,10^{k+1} \cdot b$ are prime numbers.
- Conjecture $1$ is an equivalent to Goldbach's conjecture, this conjecture has two solutions $1)$ Homotopy groups $\pi _n(X)$ (by using cognition $L_1$ from homotopy groups this conjecture is solved of course we must attend to two spheres because $S^2$ minus the tallest point in north pole as topological and algebraic is an equivalent with plane $\Bbb R^2$ (except $\infty$) and also every mapping is made between these two spheres easily if these spheres aren't concentric.) and $2)$ Algebraic methods.
- Assuming conjecture $1$, it guides us to finding formula of prime numbers at $(0,1) \times (0,1),$ in natural numbers based on each natural number is equal to half of an even number so in natural numbers main role is with even numbers but when we change space from $\Bbb N$ to $r(\Bbb N)$ then main role will be with $r( \{2k-1\, | \, k \in \Bbb N \} )$, because $r(\{2k-1\,|\,k\in \Bbb N\})\subset r(\{2k\,|\,k\in \Bbb N\})$ or in principle $r(\Bbb N)=r(\{2k\,|\,k\in \Bbb N\})$ for example $0.400=0.40=0.4$ or $0.500=0.50=0.5$ but however a smaller proper subset of $r( \{2k-1\, | \, k \in \Bbb N \} \cup \{2\} )$ namely $S$ is helpful, but for finding formula of prime numbers we need to all power of Main theorem not only what such that is stated in above conjecture namely for example we must attend to the set $V$ too!
- Conjecture $2$: For each even natural number like $t=t_1t_2t_3...t_k,$ $\exists x\in \{\alpha (a^2+b^2)^{0.5}\,|$ $(a,b) \in L,$ $\alpha \in (1,\sqrt 2] \} \cap r_1(\{2k\,|\, k\in \Bbb N \} )$ such that $t=10^{k+1} x$.
- Conjecture $2$ is an equivalent to conjecture $1$, because $\forall t=t_1t_2t_3...t_k \in \Bbb N$ that $t$ is even, $\forall (a,b)\in \{(x,y)\,|\, x+y=0.0t_1t_2t_3...t_k,\, 0\lt y\le x \}$ we have: $(a^2+b^2)^{0.5}\lt 0.0t_1t_2t_3...t_k\le \sqrt 2\cdot (a^2+b^2)^{0.5}$ so by intermediate value theorem we have $0.0t_1t_2t_3...t_k=\alpha (a^2+b^2)^{0.5}$ that $1\lt \alpha \le \sqrt 2$. But now if $a=10^{-k-1} p,b=10^{-k-1} q$ for $p,q$ belong to prime numbers we have:$$\alpha = \frac{t}{\sqrt {p^2+q^2}}$$
Theorem: $\forall p,q,r,s$ belong to prime numbers & $q \lt p$ then $(p,q)$ is located at the direct line contain the points $(0,0),10^{-\varphi ((p,q))}(p,q)$ and if $(r,s)$ is belong to this line then $p=r$ & $q=s$.
Let $S^2_2$ be a sphere with center $(0,0,r_2)$ and radius $r_2$ and $S^2_1$ be a sphere with center $(0.007,0.005,c)$ and radius $r_1$ such that $S^2_1$ is into the $S^2_2$ now suppose $f_1,f_2$ are two mapping from $A$ to $S^2_2$ such that $1)$ if $x \in A,$ $f_1 (x)$ is a curve on $S^2_2$ that is obtained as below: from $x$ draw a direct line that be tangent on $S^2_1$ and stretch it till cut $S^2_2$ in curve $f_1 (x)$ and $2)$ if $x \in A,$ $f_2 (x)$ is a curve on $S^2_2$ that is obtained as below: from $x$ draw a direct line that be tangent on $S^2_1$ and then in this junction point draw a direct line perpendicular at $S^2_2$ till cut $S^2_2$ in curve $f_2 (x)$.
Let $f_3: L_1 \to S^1$ is a mapping with $f_3 ((a,b)) = (a^2+b^2)^{-0.5}(a,b)$ and $f_4:L_1 \to S^2$ is a mapping with $f_4 ((a,b)) = (a^4+a^2b^2+b^2)^{-0.5}(a^2,ab,b) $
Guess $1$: $f_3 (L_1) $ is dense in the $S^1 \cap \{ (x,y)\,|\, 0 \le y$ & $2^{-0.5} \le x \}$.
Let $U: S^2_2 \setminus \{(0,0,2r_2)\} \to \{(x,y,0)\,|\, x,y \in \Bbb R \}$ is a mapping such that draw a direct line by $(0,0,2r_2)$ & $(x,y,z)$ till cut the plane $\{(x,y,0)\,|\, x,y \in \Bbb R \}$ in the point $(x_1,y_1,0)$. Now must a group be defined on the all the points of $S^2_2 \setminus \{(0,0,2r_2)\}$.
Let $G=S^2_2 \setminus \{(0,0,2r_2)\}$ be a group by operation $g_1 + g_2 = U^{-1} (U(g_1)+U(g_2))$ that second addition is vector addition in the vector space $(\Bbb R^2,\Bbb Q,+,.)$ and now we must attend to subgroups of $G$ particularly $y=\pm x,\,y=0,\,x=0$
Theorem: Let $K_3 =\{p+q+r\,|\, p,q,r \in \Bbb P \}$ then $r(K_3)$ is dense in the interval $(0.1,1)$ of real numbers. Proof from Goldbach's weak conjecture
Guess $2$: Let $K_2 =\{p+q\,|\,p,q \in \Bbb P \}$ then $r(K_2)$ is dense in the interval $(0.1,1)$ of real numbers.
Let $F= \Bbb Q$ so what are Galois group of polynomials $x^4+b^2x^2+b^2$ and $(1+a^2)x^2 +a^4$.
Theorem $2$: If $(a,b),(c,d)\in \{(u,v)\,|\, u,v\in S_1$ & $0.01\le u+v\lt 0.1$ & $0\lt v\lt u \}$ and $(a,b),(c,d),(0,0)$ are located at a direct line then $(a,b)=(c,d)$.
- Proof: Suppose $A_1=\{(x,y)\,|\, y \lt x\lt 0.01,\, x+y\ge 0.01\}$ & $A_2=\{(x,y)\,|\, y\lt x\lt 0.1,$ $x+y\ge 0.1\}$ so $\forall (x,y) \in A_2 :\,\, 0.1(x,y)\in A_1$ & $\forall (x,y) \in A_1 :\,\, 10(x,y)\in A_2$ so theorem can be proved in $A_3=\{(x,y)\,|\, 0\lt y\lt x\lt 0.1,\, x\ge 0.01\}$ instead $A$, but in $A_3$ we have: $\forall (x_1,y_1),(x_2,y_2)\in A_3\cap (S_1\times S_1)$ so $x_1=10^{-r_1}p_1,\, y_1=10^{-s_1}q_1,$ $x_2=10^{-r_2}p_2,$ $y_2=10^{-s_2}q_2$ and if ${{y_1}\over {x_1}}$=${{y_2}\over {x_2}}$ then ${{10^{-s_1}q_1}\over {10^{-r_1}p_1}}$=${{10^{-s_2}q_2}\over {10^{-r_2}p_2}}$ so $p_1=p_2,\, q_1=q_2$ so $x_1=x_2$ so $y_1=y_2$ therefore $(x_1,y_1)=(x_2,y_2)$.
Let $Y=\{(a,b)\,|\, (a,b)\in (S_1\times S_1)\setminus L,\, 0.01\le a+b\lt 0.1\}$ & $\forall i\in \Bbb N,$ $E_i=\{(a,b)\,|$ $(a,b)\in S_1\times S_1,$ $a+b=r_1(2i)\}$ & $O_i=\{(a,b)\,|\, (a,b)\in S_1\times S_1,\, a+b=r_1(2i-1)\}$.
In theorem $2$ I obtained a cognition to $(S_1\times S_1)\cap A$ from $(0,0)$ but now I want do it from $\infty$; in the trapezoid shape with vertices $\{(0.1,0),(0.01,0),(0.05,0.05),(0.005,0.005)\}$ intersection of two direct lines contain points $\{(0.1,0),(0.01,0)\}$ & $\{(0.05,0.05),(0.005,0.005)\}$ is $(0,0)$ so we can describe $(S_1\times S_1)\cap A$ from $(0,0)$ but when we look at two parallel lines contain points $\{(0.1,0),(0.05,0.05)\}$ & $\{(0.01,0),(0.005,0.005)\}$ there isn't any point as a criterion for description of $Y$ or $L$ only inaccessible $\infty$ remains to description or the same these parallel lines contain points $\{(0.1,0),(0.05,0.05)\}$ & $\{(0.01,0),(0.005,0.005)\}$.
Theorem $3$: $1)\, \forall x\in [0.01,0.1)\setminus r_1 (\Bbb N),\,\, \{(u,v)\,|\, u+v=x\}\cap (S_1\times S_1)=\emptyset ,\,\,$ $2)\, \forall i\in \Bbb N,$ $E_i\subsetneq L,\, O_i\cap L\neq \emptyset \neq O_i\cap Y\neq O_i,\qquad Y=(\bigcup _{i\in \Bbb N} O_i )\setminus L,$ $L=(\bigcup _{i\in \Bbb N} E_i)\cup (\bigcup _{i\in \Bbb N} (O_i\cap L)),\,\,$ $3)\, \forall i\in \Bbb N,$ cardinal$(O_i \setminus L)\in \Bbb N$.
- Proof: $1,2)\, \forall (u,v)\in \{(x,y)\,|\, 0\lt y,x,\, 0.01\le x+y\lt 0.1\}$ be aware to summation $u+v$ at the lines $x+y=c$ for $0.01\le c\lt 0.1 \,\,\,3)\, \forall i\in\Bbb N,\, 2i-1$ can be written as utmost $2i-1$ summation to form of $a\cdot 10^m+b\cdot 10^n$ that $m\neq n,\, a,b\in S_1,\, a\cdot 10^m,b\cdot 10^n$ are prime numbers and or to form of $2+b\cdot 10^n$ that $b\in S_1,\, b\cdot 10^n$ is a prime number.
Guess $3$: $\forall i\in \Bbb N,$ cardinal$(E_i)=\aleph_0 =$cardinal$(O_i \cap L)$
Guess $4$: $L_1$ is dense in $\{(x,y)\,|\, 0\le y\le x,\, 0.01\le x+y\le 0.1\}$.
Guess $5$: $\forall i\in \Bbb N,\, E_i\cap \{(x,y)\,|\, y\le x\}$ is dense in the $\{(x,y)\,|\, x+y=r_1(2i),\, 0\le y\le x\}$ and $O_i\cap L\cap \{(x,y)\,|\, y\le x\}$ is dense in the $\{(x,y)\,|\, x+y=r_1(2i-1),$ $0\le y\le x\}$.
A rectangle: Suppose $B$ is a rectangle with vertices $(0.105,-0.005),(0.05,0.05),(0.005,0.005),$ $(0.06,-0.05)$ that one of usages of this rectangle is for writing each even natural number as minus of two prime numbers but during calculations we must use $\{(x,0)\,|\, 0.01\le x\lt 0.1\}$, but the question is however this rectangle as topological isn't equivalent to the plane $\Bbb R^2$ and each point in this rectangle is corresponding to a infinite set with cardinal $\aleph_0$ in $\Bbb R^2$ but which concept on the plane $\Bbb R^2$ is corresponding to the density concept on this rectangle.
Now I want find a relation between $L_1$ & $W:=((S_1\times S_1) \cap A) \setminus L$.
Theorem: Let $K =\{2k \,|\, k \in \Bbb N \}$ so $r(K)$ is dense in the interval $(0.1,1)$ of real numbers. Proof from the Main theorem and this $r(p)=r(p\cdot 10)$ that $p$ is a prime number then $p\cdot 10$ is an even number and $\{ p\cdot 10 \,|\, p \in \Bbb P \} \subset K$.
Main theorem as a result of prime number theorem is a fundamental concept in number theory also multiplication operation is a base in normal definition of prime numbers so logarithm function as an inverse of $f(a)=a^n$ has some or whole prime numbers properties that has been used in prime number theorem and consequently in the Main theorem. But I want offer a new theory with researching on logarithmic functions that it can be a useful discussion in number theory.
Now a new definition of prime numbers based on mapping $r$ is necessary, presently I have an idea consider $\forall k\in \Bbb N,$ the sequence $b_k:\Bbb N \to \{1,2,3,4,5,6,7,8,9\},\,b_k(n)$ is the last digit in $k^n,$ so if $k=k_1k_2k_3...k_r$ then $b_k(1)=k_1$ and if $k^n=t_1t_2t_3...t_s$ so $b_k(n)=t_1,$ but for primes $k,$ it is a special different pattern than composite numbers and of course I want find some properties on $r$ for example $r(m \cdot n)$ when last digit is $1,2$ or $3$ or $4,5,6,7,8,9$, of course for $3$ penultimate digit (and probably two to last digit) is important and in addition is there any way for assessment location $r(m\cdot n)$ from $r(m)$ & $r(n)$.
Our weakness is from basic concepts, I want obtain a cognition of $(S_1\times S_1)\cap A$ and $L_1$ from point $(0.02,0.03)$ like theorem $2$ from $(0,0)$, but this time it is an equivalent to a new definition of $S_1$, because intersection of two direct lines contain points $\{(0.1,0),(0.01,0)\}$ & $\{(0.05,0.05),(0.005,0.005)\}$ is the point $(0,0)$ but however two direct lines contain points $\{(0.1,0),(0.05,0.05)\}$ & $\{(0.01,0),(0.005,0.005)\}$ are parallel so imposition of point $(0.02,0.03)$ as a criterion only can be equilibrated by concept of the set $S_1$!
Hypothesis $1$: $\forall m,t\in\Bbb N$ if $t=t_1t_2t_3...t_k\cdot 10^{m-1}$ for $t_k=2,4,6,8$ or $t=t_1t_2t_3...t_k\cdot 10^m$ for $t_k=1,3,5,7,9$ then if $\forall p,q\in \Bbb P$ that $p,q\lt t$ implies $t\neq p+q$ then $\exists M\subseteq \Bbb N$ that cardinal$(M)=\aleph_0$ so $\forall i\in M$ if $\forall r,s\in\Bbb P$ that $r,s\lt t\cdot 10^{i-m}$ then $t\cdot 10^{i-m}\neq r+s$.
- I believe the Goldbach's conjecture is truth so I think with proof by contradiction we can prove the Goldbach's conjecture.
- I think with assuming this hypothesis a contradiction will be obtained and consequently the Goldbach's conjecture will be proved.
From Dirichlet's theorem on arithmetic progressions that says: for any two positive coprime integers $a$ and $d$, there are infinitely many primes of the form $a+nd$, where $n$ is a non-negative integer, and from existence an one-to-one correspondence between the sets $A_1$ & $A_2$ in theorem $2$, I think there is a fixed in the following sets, $\forall p\in\Bbb P$ if $0.01\le r_1(p)\le 0.05$ the set $\{(a,b)\in L_1\,|\, a=r_1(p)\},$ and if $0.05\lt r_1(p)\lt 0.1$ the set $\{(a,b)\in L_1\,|\, a=r_1(p)\}\cup\{(a,b)\in L_1\,|\, a=r_2(p)\}$.
Guess $6$: $\forall t,r,s\in\Bbb N$ that $t=t_1t_2t_3...t_k$ is even and $r,s$ are odd and $s\le r$ and $t=r+s$ then $(10^{-k-1}r,10^{-k-1}s)\in A\cup \{(x,x)\,|$ $0.005\le x\lt0.05\}$.
If $\{\mathrm I_{\theta}\}_{\theta\in\mathcal I}$ is a partition for $U:=\{(p,q)\,|\, p,q\in\Bbb P,$ $q\le p\}$ then $\{J_{\theta}\}_{\theta\in\mathcal I}$ is a partition for $L_2:=L_1\cup \{(a,a)\in L\}$ such that $\forall\theta\in\mathcal I,$ $J_{\theta}=\{(r_s(p),r_t(q))\in L_2\,|\, r_t(q)\le r_s(p),$ $(p,q)\in\mathrm I_{\theta},$ $s,t\in\Bbb N\}\cup\{(r_t(q),r_s(p))\in L_2\,|\, r_s(p)\le r_t(q),$ $(p,q)\in\mathrm I_{\theta},$ $s,t\in\Bbb N\}$.
In the following I need to some partitions for $L_2$ or the same cognitions to $L_2,$ however we should be aware to the details of trapezoid shape with vertices $\{(0.1,0),(0.01,0),(0.05,0.05),(0.005,0.005)\}$, the problem is which partition is a good cognition to $L_2$
Guess $7$: $\forall a,d\in\Bbb N$ that gcd$(a,d)=1$ then cardinal$(\{(r_s(n),r_t(a+nd))\in L_1\,|\, n=0,1,2,3,...,$ $r,s\in\Bbb N\}\cup\{(r_t(a+nd),r_s(n))\in L_1\,|\, n=0,1,2,3,...,\, r,s\in\Bbb N\})=\aleph_0$
Theorem $4$: $\forall t\in\Bbb N\,\forall p\in\Bbb P\,1)\,$if $0.01\le r_1(p)\le 0.05$ then cardinal$(\{(r_1(p),u)\in L_2\})\in\Bbb N,\,2)$ if $0.05\lt r_1(p)\lt 0.1$ then cardinal$(\{(r_1(p),u)\in L_2\}\cup\{(r_2(p),u)\in L_2\})\in\Bbb N,\, 3)$ if $r_1(p)\lt 0.05$ then cardinal$(\{(u,r_t(p))\in L_2\})\in\Bbb N$ & cardinal$(\{(u,v)\in L_2\,|\, v\in\bigcup _{m\in\Bbb N} \{r_m(p)\}\})=\aleph_0$
- Proof: Be aware to number of digits in prime numbers corresponding to coordinates of each member in $L_2$ and Dirichlet theorem on arithmetic progressions.
Guess $8$: $\forall c\in (\bigcup _{k=2}^{\infty} r_k(\Bbb N))\cup\{0.01\},\,$cardinal$(\{(u,v)\in L_1\,|\, u-v=c\})=\aleph_0$
Hypothesis $2$: $\forall k,m\in\Bbb N,\,$that gcd$(2,m)=1$ and $\forall p,q\in\Bbb P$ with $p,q\lt 2^km$ that $2^km\neq p+q$ then $\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ cardinal$(\{t\in\Bbb N\,|\, 2^tm\neq p+q$ for $\forall p,q\in\Bbb P$ that $p,q\lt 2^tm\})=\aleph_0$
Now by using Chen's theorem that says: Every sufficiently large even number can be written as the sum of either two primes, or a prime and a semiprime(the product of two primes) and its extension by Tomohiro Yamada that says: Every even number greater than ${\displaystyle e^{e^{36}}\approx 1.7\cdot 10^{1872344071119348}}$ is the sum of a prime and a product of at most two primes, and Chen's theorem $II$ that is a result on the twin prime conjecture, It states that if $h$ is a positive even integer, there are infinitely many primes $p$ such that $p+h$ is either prime or the product of two primes and Ying Chun Cai's theorem that says: there exists a natural number $N$ such that every even integer $n$ larger than $N$ is a sum of a prime less than or equal to $n^{0.95}$ and a number with at most two prime factors, I want make some partitions for $L_2$ (in principle I want write Gaussian integers with prime coordinates in several equations.):
Theorem: $r(\{2p\,|\, p\in\Bbb P\})$ and $r(\{5p\,|\, p\in\Bbb P\})$ are dense in the interval $[0.1,1]$.
- Proof: Under Euclidean topology, mapping $f_1:\{(x,x)\,|\, x\in [0.005,0.05)\}\to\{(x,0)\,|\, x\in [0.01,0.1)\}$ by $f_1((a,a))=(2a,0)$ & $f_2:\{(x,0)\,|\, x\in [0.01,0.1)\}\to \{(x,x)\,|\, x\in [0.005,0.05)\}$ by $f_2((a,0))=(0.5a,0.5a)$ are homeomorphism and also $\forall r_1(p)\in [0.01,0.02]$ we have $0.5r_1(p)=r_2(5p)$ & $\forall r_1(p)\in (0.02,0.1)$ we have $0.5r_1(p)=r_1(5p)$ of course by another topology we should transfer density to $[0.1,1]$.
- Question $1$: According to gcd$(2,5)=1$ & $2\cdot 5=10$ & $(5-2)-2=1$ so $\forall q\in\Bbb P,$ is $r(\{pq\,|\, p\in\Bbb P\})$ dense in the $[0.1,1]$
Assume $\forall m,n\in\Bbb N$
$\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\end{cases}$
and $p_n\star _1p_m=p_{n\star m}$ that $p_n$ is $n$_th prime & $\forall (p_n,p_m),(p_s,p_t)\in\Bbb P\times\Bbb P,$ $(p_n,p_m)\star _2(p_s,p_t)=(p_n\star _1p_s,p_m\star _1p_t)$.
It is clear $(\Bbb N,\star)$ & $(\Bbb P,\star _1)$ & $(\Bbb P\times\Bbb P,\star _2)$ are groups and $<2>=(\Bbb N,\star)\cong (\Bbb Z,+)\cong (\Bbb P,\star _1)=<3>$ & $<(3,2),(2,3)>=(\Bbb P\times\Bbb P,\star _2)\cong (\Bbb Z\times\Bbb Z,+)$ & $\pi _n(\Bbb P\times\Bbb P)\cong\pi _n(\Bbb Z)\times\pi _n(\Bbb Z)$.
and let $Q_1=\{{m\over n}\,|\, m,n\in\Bbb N\}$, it is clear $(Q_1,\star _{Q_1})$ is a group as below:
$\begin{cases} \forall m,n,u,v\in\Bbb N,\, {m\over n}\star _{Q_1} 1={m\over n}\\ ({m\over n})^{-1}={m^{-1}\over n^{-1}}\\ {m\over n}\star _{Q_1} {u\over v}={m_1\over n_1}\star _{Q_1} {u_1\over v_1}={{m_1\star u_1}\over {n_1\star v_1}}\quad\text{if}\,\,\begin{cases} {m\over n}={m_1\over n_1},\,\, {u\over v}={u_1\over v_1},\,\, {mu\over nv}={m_1u_1\over n_1v_1}\\ \text{gcd}(m_1,n_1)=1=\text{gcd}(m_1,v_1)=\text{gcd}(u_1,n_1)=\text{gcd}(u_1,v_1)\end{cases}\end{cases}$
Of course each sequence $\{a_n\}$ that $\forall n,m\in\Bbb N,\,n\neq m$ then $a_n\neq a_m$ is a cyclic group as: $a_n\star _aa_m=a_{n\star m},\, e=a_1,\, G=<a_2>$
Manner of making of new groups on $\Bbb N$: I want explain it with an example write integers like a sequence as: $$0,1,2,-1,-2,3,4,-3,-4,5,6,-5,-6,7,8,-7,-8,9,10,-9,-10,11,12,-11,-12,...$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,...$$ and let $e=1$ and $(\Bbb N,\star _{\Bbb N})=<2>$ then begin for discovering of rule of $\star _{\Bbb N}$ as below:
$\begin{cases} 0+8=8 & 1\star _{\Bbb N} 15=15\\ 1+8=9 & 2\star _{\Bbb N} 15=18\\ 2+8=10 & 3\star _{\Bbb N} 15=19\\ (-1)+8=7 & 4\star _{\Bbb N} 15=14\\ (-2)+8=6 & 5\star _{\Bbb N} 15=11\\ 3+8=11 & 6\star _{\Bbb N} 15=22\\ 4+8=12 & 7\star _{\Bbb N} 15=23\\ (-3)+8=5 & 8\star _{\Bbb N} 15=10\\ (-4)+8=4 & 9\star _{\Bbb N} 15=7\\ 5+8=13 & 10\star _{\Bbb N} 15=26\\ 6+8=14 & 11\star _{\Bbb N} 15=27\\ (-5)+8=3 & 12\star _{\Bbb N} 15=6\\ (-6)+8=2 & 13\star _{\Bbb N} 15=3\\ 7+8=15 & 14\star _{\Bbb N} 15=30\\ 8+8=16 & 15\star _{\Bbb N} 15=31\\ (-7)+8=1 & 16\star _{\Bbb N} 15=2\\ (-8)+8=0 & 17\star _{\Bbb N} 15=1\end{cases}$
that its group is:
$\begin{cases} m\star _{\Bbb N} 1=m\\ (4m)\star _{\Bbb N} (4m-2)=1=(4m+1)\star _{\Bbb N} (4m-1)\\ (4m)\star _{\Bbb N} (4m+2)=3=(4m+1)\star _{\Bbb N} (4m+3)\\ (4m-2)\star _{\Bbb N} (4n-2)=4m+4n-5\\ (4m-2)\star _{\Bbb N} (4n-1)=4m+4n-2\\ (4m-2)\star _{\Bbb N} (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\end{cases}\\ (4m-2)\star _{\Bbb N} (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star _{\Bbb N} (4n-1)=4m+4n-1\\ (4m-1)\star _{\Bbb N} (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star _{\Bbb N} (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\end{cases}\\ (4m)\star _{\Bbb N} (4n)=4m+4n-3\\ (4m)\star _{\Bbb N} (4n+1)=4m+4n\\ (4m+1)\star _{\Bbb N} (4n+1)=4m+4n+1\end{cases}$
that this group $(\Bbb N,\star _{\Bbb N})$ is helpful for twin prime conjecture.
and the Klein four-group $(\Bbb Z _2\times\Bbb Z _2,+)$ is a fundamental concept in the group theory that its usage is for propositions rejection so I made below group somehow similar to that group in terms of members production for proof of the Goldbach's conjecture with proof by contradiction that is: $$0,1,2,-2,-1,3,-3,4,5,-5,-4,6,-6,7,8,-8,-7,9,-9,10,11,-11,-10,12,-12,...$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,...$$ and $e=1$ and $(\Bbb N,\star _{K_4})=<2>$ so we have:
$\begin{cases} 0+(-7)=-7 & 1\star _{K_4}17=17\\ 1+(-7)=-6 & 2\star _{K_4}17=13\\ 2+(-7)=-5 & 3\star _{K_4}17=10\\ (-2)+(-7)=-9 & 4\star _{K_4}17=19\\ (-1)+(-7)=-8 & 5\star _{K_4}17=16\\ 3+(-7)=-4 & 6\star _{K_4}17=11\\ (-3)+(-7)=-10 & 7\star _{K_4}17=23\\ 4+(-7)=-3 & 8\star _{K_4}17=7\\ 5+(-7)=-2 & 9\star _{K_4}17=4\\ (-5)+(-7)=-12 & 10\star _{K_4}17=25\\ (-4)+(-7)=-11 & 11\star _{K_4}17=22\\ 6+(-7)=-1 & 12\star _{K_4}17=5\\ (-6)+(-7)=-13 & 13\star _{K_4}17=29\\ 7+(-7)=0 & 14\star _{K_4}17=1\\ 8+(-7)=1 & 15\star _{K_4}17=2\\ (-8)+(-7)=-15 & 16\star _{K_4}17=31\\ (-7)+(-7)=-14 & 17\star _{K_4}17=28\end{cases}$
that its group is:
$\begin{cases} m\star _{K_4}1=m\\ (6m-4) \star _{K_4}(6m-1)=1=(6m-3) \star _{K_4}(6m-2)=(6m) \star _{K_4}(6m+1)\\ (6m-4) \star _{K_4}(6n-4)=6m+6n-9\\ (6m-4) \star _{K_4}(6n-3)=6m+6n-6\\ (6m-4) \star _{K_4}(6n-2)=\begin{cases} 6m-6n-3 & 6m-4\gt 6n-2\\ 6n-6m+5 & 6n-2\gt 6m-4\end{cases}\\ (6m-4) \star _{K_4}(6n-1)=\begin{cases} 6m-6n & 6m-4\gt 6n-1\\ 6n-6m+1 & 6n-1\gt 6m-4\end{cases}\\ (6m-4) \star _{K_4}(6n)=6m+6n-4\\ (6m-4) \star _{K_4}(6n+1)=\begin{cases} 6m-6n-4 & 6m-4\gt 6n+1\\ 6n-6m+4 & 6n+1\gt 6m-4\end{cases}\\ (6m-3) \star _{K_4}(6n-3)=6m+6n-4\\ (6m-3) \star _{K_4}(6n-2)=\begin{cases} 6m-6n & 6m-3\gt 6n-2\\ 6n-6m+1 & 6n-2\gt 6m-3\end{cases}\\ (6m-3) \star _{K_4}(6n-1)=\begin{cases} 6m-6n+2 & 6m-3\gt 6n-1\\ 6n-6m-2 & 6n-1\gt 6m-3\\ 2 & m=n\end{cases}\\ (6m-3) \star _{K_4}(6n)=6m+6n-3\\ (6m-3) \star _{K_4}(6n+1)=\begin{cases} 6m-6n-3 & 6m-3\gt 6n+1\\ 6n-6m+5 & 6n+1\gt 6m-3\end{cases}\\ (6m-2) \star _{K_4}(6n-2)=6m+6n-1\\ (6m-2) \star _{K_4}(6n-1)=6m+6n-5\\ (6m-2) \star _{K_4}(6n)=\begin{cases} 6m-6n-2 & 6m-2\gt 6n\\ 6n-6m+2 & 6n\gt 6m-2\end{cases}\\ (6m-2) \star _{K_4}(6n+1)=6m+6n-2\\ (6m-1) \star _{K_4}(6n-1)=6m+6n-8\\ (6m-1) \star _{K_4}(6n)=\begin{cases} 6m-6n-1 & 6m-1\gt 6n\\ 6n-6m+3 & 6n\gt 6m-1\end{cases}\\ (6m-1) \star _{K_4}(6n+1)=6m+6n-1\\ (6m) \star _{K_4}(6n)=6m+6n\\ (6m) \star _{K_4}(6n+1)=\begin{cases} 6m-6n & 6m\gt 6n+1\\ 6n-6m+1 & 6n+1\gt 6m\end{cases}\\ (6m+1) \star _{K_4}(6n+1)=6m+6n+1\end{cases}$
By using theorem $1$ of Polignac's conjecture we can define function $f:\{(c,d)\,|\, (c,d)\subseteq [0.01,0.1)\}\to\Bbb N$ that $f((c,d))$ is the least $n\in\Bbb N$ that $\exists t\in(c,d),\,\exists k\in\Bbb N$ that $p_n=t\cdot 10^{k+1}$ that $p_n$ is $n$_th prime and $\forall m\ge f((c,d))\,\,\exists u\in (c,d)$ that $u\cdot 10^{m+1}\in\Bbb P$
and $g:(0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\to\Bbb N,$ is a function by $\forall\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))$ $g(\epsilon)=max(\{f((c,d))\,|\, d-c=\epsilon,$ $(c,d)\subseteq [0.01,0.1)\})$.
Guess $9$: $g$ isn't an injective function.
Question $2$: Assuming guess $9$ let $[a,a]:=\{a\}$ and $\forall n\in\Bbb N,\, h_n$ is the least subinterval of $[0.01,0.1)$ like $[a,b]$ in terms of size of $b-a$ such that $\{\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\,|\, g(\epsilon)=n\}\subsetneq h_n$ and it is clear $g(a)=n=g(b)$ now the question is $\forall n,m\in\Bbb N$ that $m\neq n$ is $h_n\cap h_m=\emptyset$?
- Guidance given by @reuns from stackexchange.com:
- For $n \in \mathbb{N}$ then $r(n) = 10^{-\lceil \log_{10}(n) \rceil} n$, ie. $r(19) = 0.19$. We look at the image by $r$ of the primes $\mathbb{P}$.
- Let $F((c,d)) = \min \{ p \in \mathbb{P}, r(p) \in (c,d)\}$ and $f((c,d)) = \pi(F(c,d))= \min \{ n, r(p_n) \in (c,d)\}$ ($\pi$ is the prime counting function)
- If you set $g(\epsilon) = \max_a \{ f((a,a+\epsilon))\}$ then try seing how $g(\epsilon)$ is constant on some intervals defined in term of the prime gap $g(p) = -p+\min \{ q \in \mathbb{P}, q > p\}$ and things like $ \max \{ g(p), p > 10^i, p+g(p) < 10^{i+1}\}$
Now I want define a group on $L$ like $(L,\star _L)$.
Let $P_1=\{v_n\,|\,\forall n\in\Bbb N,\, v_n$ is $(n+1)$_th prime$\}$ and $\forall n,m\in\Bbb N,\, v_n\star _3 v_m=v_{n\star m}$ and $\forall (v_n,v_m),(v_s,v_t)\in P_1\times P_1,$ $(v_n,v_m)\star _4(v_s,v_t)=(v_n\star _3 v_s,v_m\star _3 v_t)$.
it is clear $(P_1,\star _3)$ & $(P_1\times P_1,\star _4)$ are groups and $<5>=(P_1,\star _3)\cong (\Bbb Z,+)$ & $<(5,3),(3,5)>=(P_1\times P_1,\star _4)\cong (\Bbb Z\times\Bbb Z,+)$.
and let $h(v_n)$ is the number of digits in $v_n$ and $g((v_n,v_m))=max(h(v_n),h(v_m))+1$.
Question $3$: To define an Abelian group structure on $L$ I need to know: If $g((v_n,v_m))=k_1$ and $g((v_s,v_t))=k_2$ then $g((v_n,v_m)\star_4(v_s,v_t))=?$
- Although I guess $(L,\star _L)\cong (\Bbb Q,+)$.
Conjecture $3$: $\forall n\in\Bbb N,\, n\ge 3\,\,\exists s_1,s_2\in P_1$ such that $v_{2n-1}=v_{s_1}\star _3v_{s_2}$.
- According to equivalency of these two $(\Bbb N,⋆)$ & $(P_1,⋆_3)$ from aspect of being group, this conjecture is an equivalent to Goldbach's conjecture.
- We can pay attention to subgroups as $<v_{s_1}⋆_3v_{s_2}>$ as a solution for Goldbach also we can use quotient groups $P_1/<v_s>$.
- There exists an one-to-one correspondence between equations in $(\Bbb Z,+)$ & $(\Bbb N,⋆)$ & $(P_1,⋆_3)$ & $(\Bbb N,⋆_{K_4})$ and prime numbers properties exist in those groups although other sequences may have the same structures but prime numbers structures are specific because the own prime numbers are specific.
- There doesn't exist another way to define another group structure on $\Bbb P$ or $P_1$ for using strength of finite groups or infinite groups unless we knew the formula of prime numbers and on the other hand we can't know the formula before than knowing Goldbach and Polignac conjecture.
Question $4$: Is there any subgroup of $P_1$ like $H$ such that $\forall v_s\in H,\, s\notin P_1$?
- I want connect subgroups of $P_1$ with Goldbach's conjecture.
Question $5$: To define an Abelian group structure on $\Bbb N$ that is not a finitely generated Abelian group and is isomorph to $(\Bbb Q,+)$, I need to know what is rule of this sequence in $\Bbb N\times\Bbb N$: $(1,1),(1,2),(2,1),(1,3),(2,2),(3,1),(1,4),(2,3),(3,2),(4,1),...,(1,2k-2),(2,2k-3),...$ $,(k-1,k),(k,k-1),...,(2k-3,2),(2k-2,1),(1,2k-1),(2,2k-2),...,(k,k),...$ $,(2k-2,2),(2k-1,1),...$
- Answer given by Professor Daniel Lazard:
- $\begin{cases} (a_1,b_1)=(1,1)\\ (a_{n+1},b_{n+1})=\begin{cases} (1,a_n+1) & b_n=1\\ (a_n+1,b_n-1) & \text{else}\end{cases}\end{cases}$ Alireza Badali 22:21, 8 May 2017 (CEST) =='"`UNIQ--h-1--QINU`"' Polignac's conjecture == In number theory, Polignac's conjecture was made by Alphonse de Polignac in 1849 and states: For any positive even number $n$, there are infinitely many prime gaps of size $n$. In other words: There are infinitely many cases of two consecutive prime numbers with difference $n$. (Tattersall, J.J. (2005), Elementary number theory in nine chapters, Cambridge University Press, ISBN: 978-0-521-85014-8, p. 112) Although the conjecture has not yet been proven or disproven for any given value of n, in 2013 an important breakthrough was made by Zhang Yitang who proved that there are infinitely many prime gaps of size n for some value of n < 70,000,000.(Zhang, Yitang (2014). "Bounded gaps between primes". Annals of Mathematics. 179 (3): 1121–1174. MR 3171761. Zbl 1290.11128. doi:10.4007/annals.2014.179.3.7. _ Klarreich, Erica (19 May 2013). "Unheralded Mathematician Bridges the Prime Gap". Simons Science News. Retrieved 21 May 2013.) Later that year, James Maynard announced a related breakthrough which proved that there are infinitely many prime gaps of some size less than or equal to 600.(Augereau, Benjamin (15 January 2013). "An old mathematical puzzle soon to be unraveled? Phys.org. Retrieved 10 February 2013.) Assuming Polignac's conjecture there isn't any rhythm for prime numbers and so there isn't any formula for prime numbers! Let $B=\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1\}$ & $C=(S_1\times S_1)\cap \{(x,x)\,|\, 0.01\le x\lt 0.1\}$ & $T=\{(a,b)\,|\, a,b \in S_1,\, 0.01 \lt b \lt a\lt 0.1,\, \exists m \in \Bbb N,\, a \cdot 10^m, b \cdot 10^m$ or $a\cdot 10^{m-1}, b\cdot 10^m$ are consecutive prime numbers$\}$ & $\forall n \in \Bbb N,\, J_n :=\{(a,b) \,|\, (a,b) \in T,\, \exists k \in \Bbb N, a-b=r_k (2n)\}$. :It is clear $\bigcup _{n\in \Bbb N} J_n=T$ and Polignac's conjecture is equivalent to $\forall n \in \Bbb N,$ cardinal$(J_n)=\aleph_0$. '''Guess''' $1$: $\forall (a,a) \in C$ there are some sequences in $T$ like $\{a_n\}$ that $a_n\to (a,a)$ where $n\to \infty$ and there are some sequences in $T$ like $b_n$ that $b_n\to (0.1,0.01)$ where $n\to \infty$. '''Guess''' $2$: $\exists N_1 \subseteq \Bbb N,\, \forall n\in N_1,$ cardinal$(J_n)=\aleph_0=$cardinal$(N_1)$ It is clear $\exists \epsilon,\epsilon_1,\epsilon_2 \in \Bbb R,\epsilon\gt 0, \epsilon_2\gt \epsilon_1\gt 0$ that $\forall (a,b) \in T \cap \{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,$ $x-y\lt \epsilon\}$ that $\exists m\in \Bbb N$ that $a\cdot 10^m$,$b\cdot 10^m$ are consecutive prime numbers, but $a\cdot 10^m$,$b\cdot 10^m$ are large natural numbers, and $\forall (a,b)\in T\cap$ $\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,$ $0.11-\epsilon_1 \lt x+y\lt 0.11+\epsilon_1,$ $x-y\gt 0.09-\epsilon_2 \}$ that $\exists m\in \Bbb N$ that $a\cdot 10^{m-1},b\cdot 10^m$ are consecutive prime numbers but $a\cdot 10^{m-1}$,$b\cdot 10^m$ are large natural numbers and $\forall (a,b)\in T\setminus (\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,$ $x-y\lt \epsilon\}$ $\cup$ $\{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,$ $0.11-\epsilon_1 \lt x+y\lt 0.11+\epsilon_1,\, x-y\gt 0.09-\epsilon_2 \})$ that $\exists m\in \Bbb N$ that $a\cdot 10^{m-1}$ or $a\cdot 10^m$ & $b\cdot 10^m$ are consecutive prime numbers but $a\cdot 10^m$ or $a\cdot 10^{m-1}$ & $b\cdot 10^m$ aren't large natural numbers. Theorem: $\forall c\in r_1(\Bbb P)$ cardinal$(T\cap \{(x,c)\,|\, x\in \Bbb R \})=1=$ cardinal$(T\cap \{(c,y)\,|\, y\in \Bbb R\})$ '''Guess''' $3$: $\forall k\in \Bbb N,$ $\forall c\in r_k (\Bbb N),\, c\lt 0.09$ then cardinal$(T\cap \{(x,y)\,|\, x-y=c\}) \in \Bbb N \cup \{0\}$. '''Theorem''' $1$: For each subinterval of $[0.01,0.1)$ like $(a,b),\,\exists m\in \Bbb N$ that $\forall k\in \Bbb N$ with $k\ge m$ then $\exists t\in (a,b)$ that $t\cdot 10^{k+1}\in \Bbb P$. :Proof given by [https://math.stackexchange.com/users/149178/adayah @Adayah] from stackexchange.com: Without loss of generality (by passing to a smaller subinterval) we can assume that $(a, b) = \left( \frac{s}{10^r}, \frac{t}{10^r} \right)$, where $s, t, r$ are positive integers and $s < t$. Let $\alpha = \frac{t}{s}$. :The statement is now equivalent to saying that there is $m \in \mathbb{N}$ such that for every $k \geqslant m$ there is a prime $p$ with $10^{k-r} \cdot s < p < 10^{k-r} \cdot t$. :We will prove a stronger statement: there is $m \in \mathbb{N}$ such that for every $n \geqslant m$ there is a prime $p$ such that $n < p < \alpha \cdot n$. By taking a little smaller $\alpha$ we can relax the restriction to $n < p \leqslant \alpha \cdot n$. :Now comes the prime number theorem: '"`UNIQ-MathJax7-QINU`"' :where $\pi(n) = \# \{ p \leqslant n : p$ is prime$\}.$ By the above we have '"`UNIQ-MathJax8-QINU`"' :hence $\displaystyle \lim_{n \to \infty} \frac{\pi(\alpha n)}{\pi(n)} = \alpha$. So there is $m \in \mathbb{N}$ such that $\pi(\alpha n) > \pi(n)$ whenever $n \geqslant m$, which means there is a prime $p$ such that $n < p \leqslant \alpha \cdot n$, and that is what we wanted. ::Clearly the [[User_talk:Musictheory2math#Goldbach.27s_conjecture|Main theorem]] of Goldbach's conjecture is a result of theorem $1$. Theorem: $\forall \epsilon_1,\epsilon_2$ that $0\lt \epsilon_1\lt \epsilon_2\lt 0.09$ then cardinal$(T\setminus \{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,\, x-y\gt \epsilon_1\})$ $=$ cardinal$(T\setminus \{(x,y)\,|\, 0.01\lt y\lt x\lt 0.1,\, x-y\lt \epsilon_2\})=\aleph_0$. :Proof: Be aware to number of digits in prime numbers corresponding to coordinates of each member in $T$. :'''Guess''' $4$: cardinal$(\{(a,b)\,|\, (a,b)\in T,\, 0.01\lt b\lt a\lt 0.1,$ $\epsilon_1\lt a-b\lt \epsilon_2\})\in \Bbb N\cup \{0\}$ Alireza Badali 13:17, 21 August 2017 (CEST) =='"`UNIQ--h-2--QINU`"' Landau's forth problem == Landau's forth problem: Are there infinitely many primes $p$ such that $p−1$ is a perfect square? In other words: Are there infinitely many primes of the form $n^2 + 1$? In analytic number theory the Friedlander–Iwaniec theorem states that there are infinitely many prime numbers of the form ${\displaystyle a^{2}+b^{4}}$. Suppose $H=\{(a,b)\,|\, a\in\bigcup_{k\in\Bbb N} r_k(\{n^2\,|\, n\in\Bbb N\}),\, b\in\bigcup_{k\in\Bbb N} r_k(\{n^4\,|\, n\in\Bbb N\})$ & $\exists t\in\Bbb N$ that $a\cdot 10^t\in\{n^2\,|\, n\in\Bbb N\},$ $b\cdot 10^t\in\{n^4\,|\, n\in\Bbb N\},\, (a+b)\cdot 10^t\in\Bbb P\}$ and $H_1=\{(a,b)\in H\,|\, b\in\bigcup_{k\in\Bbb N} r_k(\{1\})\}$ :Friedlander-Iwaniec theorem is the same cardinal$(H)=\aleph_0$ '''A question''': cardinal$(H_1)\in\Bbb N$ or cardinal$(H_1)=\aleph_0$? :This question is the same Landau's forth problem. Alireza Badali 20:47, 21 September 2017 (CEST) =='"`UNIQ--h-3--QINU`"' Grimm's conjecture == In mathematics, and in particular number theory, Grimm's conjecture (named after Karl Albert Grimm) states that to each element of a set of consecutive composite numbers one can assign a distinct prime that divides it. It was first published in American Mathematical Monthly, 76(1969) 1126-1128. Formal statement: Suppose $n + 1, n + 2, …, n + k$ are all composite numbers, then there are $k$ distinct primes $p_i$ such that $p_i$ divides $n+i$ for $1 ≤ i ≤ k$. Let $C=\{(x,y)\,|\, 0.01\le x\lt 0.1,\, 0.01\le y\lt 0.1\}$ '''A conjecture''': Suppose $n+1,n+2,n+3,...,n+k$ are all composite numbers then $\forall i=1,2,3,...,k$ $\exists (r_1(p_i),r_1(t_i))\in C$ that $p_i\in\Bbb P,$ $t_i\in\Bbb N$ & $\forall j=1,2,3,...,k$ that $i\neq j$ implies $p_i\neq p_j$ we have $r_1(p_i)\cdot r_1(t_i)=r_2(n+i)$ or $r_3(n+i)$. :This conjecture is an equivalent to Grimm's conjecture. Alireza Badali 23:30, 20 September 2017 (CEST) =='"`UNIQ--h-4--QINU`"' Lemoine's conjecture == In number theory, Lemoine's conjecture, named after Émile Lemoine, also known as Levy's conjecture, after Hyman Levy, states that all odd integers greater than $5$ can be represented as the sum of an odd prime number and an even semiprime. Formal definition: To put it algebraically, $2n + 1 = p + 2q$ always has a solution in primes $p$ and $q$ (not necessarily distinct) for $n > 2$. The Lemoine conjecture is similar to but stronger than Goldbach's weak conjecture. Theorem: $\forall n\in\Bbb N$ that $m=2n+5,\, \exists (a,b)\in\{(x,y)\,|\, 0\lt y\le x,\, x+y\lt 0.1,\, x,y\in S_1,\, \exists t\in\Bbb N,$ $x\cdot 10^t,y\cdot 10^t\in\Bbb P\},\,\exists k\in\Bbb N,\, \exists c\in r_k(\Bbb P)$ such that $a+b=r_1(m)-c$ :In principle $(r_1(m),-c)\in \{(x,y)\,|\, -x\lt y\lt 0,\, 0.01\le x\lt 0.1\}$ :This theorem is an equivalent to Goldbach's weak conjecture. '''A conjecture''': $\forall n\in\Bbb N$ that $m=2n+5,\, \exists k_1,k_2\in\Bbb N,\, \exists b\in r_{k_1}(\Bbb P)$ that $b\lt 0.05,\, \exists a\in r_{k_2}(\Bbb P)$ such that $2b=r_1(m)-a$ :In principle $(b,b)\in\{(x,x)\,|\, 0\lt x\lt 0.05\}$ & $(r_1(m),-a)\in\{(x,y)\,|\, -x\lt y\lt 0,\, 0.01\le x\lt 0.1\}$ :This conjecture is an equivalent to Lemoine's conjecture. Alireza Badali 00:30, 27 September 2017 (CEST) =='"`UNIQ--h-5--QINU`"' Some new functions related to zeta function == In mathematics, analytic number theory is a branch of number theory that uses methods from mathematical analysis to solve problems about the integers. It is often said to have begun with Peter Gustav Lejeune Dirichlet's 1837 introduction of Dirichlet L-functions to give the first proof of Dirichlet's theorem on arithmetic progressions. It is well known for its results on prime numbers (involving the Prime Number Theorem and Riemann zeta function) and additive number theory (such as the Goldbach conjecture and Waring's problem). In analytic number theory in Riemann zeta function there is an important technique as <big>$\frac{1}{n}$</big>, inverse of natural numbers but I want add another technique to this collection as putting a point at the beginning of natural numbers like $6484070\to 0.6484070$ so we will have a stronger theory, let $S=\{0.2,0.3,0.5,0.7,0.11,...\}$ and $s_1=0.2,\, s_2=0.3,\, s_3=0.5,...$ that $s_k$ is $k$_th member in $S$. Suppose $A_n=\{s_is_j\,|\, s_i,s_j\in \{s_1,s_2,s_3,...,s_n\},$ $s_i\neq s_j\}$ & $\mu _1:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax9-QINU`"' such that $f:\Bbb C\to\Bbb C$ is an injective function that $\forall n\in\Bbb N,\,\forall z\in\Bbb C,\, n^{f(z)}$ isn't a Gaussian integer, for example '"`UNIQ-MathJax10-QINU`"' for large sufficiently $N\in\Bbb N$ of course $f$ maps $\Bbb C$ injectively into a disk around $\sqrt 2$ of radius $N^{-1}$ for large sufficiently $N,$ this disk contains no solution $z$ of $n^z\in\Bbb Z [i]$, according to definition of the function $\mu _1$ I think it is in a near relation with Riemann zeta function. Let $\mu _2:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax11-QINU`"' Let $\mu _3:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax12-QINU`"' Let $\forall j\in\Bbb N,\,\mu _4:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax13-QINU`"' Let $\forall j\in\Bbb N,\,\mu _5:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax14-QINU`"' and $\mu _6:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax15-QINU`"' Let $\mu _7:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax16-QINU`"' let $\mu _8:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax17-QINU`"' Let $p_j$ is $j$_th prime number & $\mu _9:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax18-QINU`"' Let $a_n:\Bbb N\to\Bbb N$ is a sequence & $\mu _{10}:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax19-QINU`"' Let $a_n:\Bbb N\to\Bbb N$ is a sequence & $\mu _{11}:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax20-QINU`"' Let $\omega =0.p_1p_2p_3...=0.23571113171923293137...$ that in principle in $\omega$ prime numbers has been arranged respectively, now assume $a_n:\Bbb N\to (0,1)$ is a sequence that $\sum _{n\in\Bbb N} a_n=\omega$ & $\mu _{12}:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax21-QINU`"' Let $f_1:\Bbb C\to\Bbb C$ & $\mu _{13}:\Bbb C\to\Bbb C$ are functions as: '"`UNIQ-MathJax22-QINU`"' Let $f_2:\Bbb C\to\Bbb C$ & $\mu _{14}:\Bbb C\to\Bbb C$ are functions as: '"`UNIQ-MathJax23-QINU`"' Find $f_1$ & $f_2$ as much as possible simple such that $\mu _{13}(i[\Bbb Q])$ is dense in the $\mu _{13}(\Bbb C)$ and $\mu _{14}(i[\Bbb Q])$ is dense in the $\mu _{14}(\Bbb C)$. Now some theorems on these functions about density should be presented. Let $\mu_{15}:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax24-QINU`"' Let $\mu_{16}:\Bbb C\to\Bbb C$ is a function as: '"`UNIQ-MathJax25-QINU`"' Formula of prime numbers can't be as polynimial and it should be a logarithmic function: Let '"`UNIQ-MathJax26-QINU`"' (i.e. the decimal part of $\omega _1$ is obtained by concatenating the prime numbers) and '"`UNIQ-MathJax27-QINU`"' (i.e. the decimal part of $\omega _2$ is obtained by concatenating the prime numbers, each of them followed by a number of copies of $0$ equal to the number of its digits in base $10$). Questions: $1.$ Is $\omega _1$ or $\omega _2$ or another some similar number transcendental, and if yes is this a contradiction to the existence of a formula for prime numbers? $2.$ For each sequence $a_n: \mathbb N \to \mathbb N$ is there any sequence like $b_n: \mathbb N \to \mathbb N$ such that the number $\theta :=0.a_10 \dots 0a_20 \dots 0a_30 \dots 0 \dots$ obtained by concatenating the numbers $a_n$, each of them followed by a number of copies of $0$ equal to $b_n$, is a transcendental number? Alireza Badali 20:44, 12 October 2017 (CEST) =='"`UNIQ--h-6--QINU`"' Some notes == I see, you like to densely embed natural numbers into a continuum. You may also try to embed them into the unit circle on the complex plane by $n\mapsto i^n=\cos(\log n)+i\sin(\log n)$. Then $(mn)^i=m^i n^i$. Passer By (talk) 13:14, 3 December 2017 (CET)
- Very nice, thank you so much. Alireza Badali 21:33, 3 December 2017 (CET)
To your Question 2 above: the affirmative answer is given by Liouville's theorem on approximation of algebraic numbers. Passer By (talk) 19:05, 4 December 2017 (CET)
- Thank you. Alireza Badali 16:00, 6 December 2017 (CET)
You often mention "Formula of prime numbers". What do you mean? This is not a well-defined mathematical object, but a vague idea, with a lot of non-equivalent interpretations. See for instance [1], [2], [3], [4], [5], [6] etc. Passer By (talk) 19:24, 4 December 2017 (CET)
- Thank you and each one of these wordage about the formula of prime numbers is a theory lonely and probably non-equivalent or even incompatible so please don't consider their relations together. Alireza Badali 16:00, 6 December 2017 (CET)
Musictheory2math. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Musictheory2math&oldid=42439


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I actually grabbed a few dollar store LA COLORS palettes and goofed with them and I was kind of surprised how good they were. But I know less to nothing so... Anyone else ever use the LA COLORS eye shadows?
I should probably learn how to use neutrals first to learn technique before I jump into all these colors. But I do what I want.

subset: commoncrawl


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Event Navigation
EMERGE College Ministry
May 27 @ 7:00 pm - 9:00 pm
Setup is at 4:30 pm.
Helping to meet the unique needs of college-aged young adults (18-25). In an atmosphere where they can be EXPOSED to, EXPERIENCE and EXPRESS Christ, His Kingdom and the biblical purposes of His church.


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Classic Tails for Cat Lovers
Have you reached the bottom of your summer reading stack? Or perhaps your picks fell short this summer, and you need new ideas. To help you rekindle reading momentum this fall, we asked veterinarians, cat experts and pet stores for their cat-related book recommendations. Here's what they suggested:
Chicken Soup for the Cat Lover's Soul, Mark Victor Hansen, et al.
In this tribute to the extraordinary feline spirit, you'll learn about the piano-playing Devon Rex. Then, there's the loyal three-legged kitty that consoles a lonely third grader. Still not inspired? The book then takes you to another entry about a precocious tabby that saved its human family from a gas leak. A heartwarming collection of true stories, it's "the No. 1 best-seller in the store, by far," says Wanda Kelsey-Mendez, owner of Gatos Cat Boutique online.
Dewey: The Small-Town Library Cat Who Touched the World, by Vicki Myron
After losing her family farm and leaving her abusive husband, small-town single mom Myron attempts to rebuild her life. But it's not until the coldest morning of the year that life really turns around — for her and for her small town — when she discovers Dewey, a tiny kitten, freezing in the metal drop box. His loveable feline antics quickly enthrall the sleepy town, awakening even the senses of a disabled child. "It's a fun read," says Dr. Laura Sullivan, a veterinarian at Cascade Hospital for Animals in Grand Rapids, Mich. "All cat owners could relate to the highs and lows of kitty ownership."
Catnip for the Soul, By Jane O'Boyle
What do cats really think of humans? If cats could speak, O'Boyle surmises this would be their answer: "No fur, no paws, no tail. They run away from mice. They never get enough sleep. How can you help but love such an absurd animal?" With humor and plenty of imagination, O'Boyle attempts to tackle the inner workings of the feline mind. Dr. Tracy L. Adams of Plaza Veterinary Hospital in Farmington, Mich., calls it "an enjoyable, funny narration from a cat's perspective."
The Cat Who… series, By Lilian Jackson Braun
Do-gooder duos in TV shows like "Scooby Doo" and "Inspector Rex" have done wonders to showcase dogs' instinct and loyalty. But cats too have the skill to be great mystery-solving partners. The Cat Who… is a book series of 35 mysteries that prominently feature two cats. With the aid of keen feline instinct, their owner, a prizewinning reporter, successfully follows leads and solves crimes. "They are very entertaining [books] and a great read for cat lovers," says Kelsey-Mendez. Start with the first book, The Cat Who Could Read Backwards, and you'll be hooked at least until next fall.
Why Does My Cat…?, By Sarah Heath
Still baffled by kitty's erratic behavior, like its haphazard elimination on the carpet? Author and animal behavioral consultant Sarah Heath provides practical insights on bothersome cat behaviors and answers fun such fun questions as "Why does my cat walk on the paper I am reading?" along the way. It's an informative read and a pleasant journey into the mind of your favorite feline.
The Cat Behavior Answer Book, By Arden Moore
A prolific writer, pet show host and professional speaker, Arden Moore has inspired millions to love, understand and live harmoniously with their pets. Here, she demystifies cat behavior with wit and warmth. "This book does a really good job describing cat behavior and offering concrete suggestions for dealing with some of the most troublesome cat actions," says Kelsey-Mendez. With thorough discussions presented in an easy-to-comprehend question and answer format, this book is a handy exploration of the feline brain and physiology.
The Domestic Cat, edited by Dennis C. Turner and Patrick Bateson
Ready for something more scientific? Try this collection of articles on animal behavior and veterinary science, penned mainly by academics. The book explores feline biology, covering everything from the development of young cats and the mother-kitten relationship, to social life and predatory behavior. "Unfortunately, many of the popular books still contain myths, misconceptions and made-up or unsupported explanations for cat behaviors," confesses Suzanne Hetts, Ph.D., of Animal Behavior Associates in Littleton, Colo. If you are seeking to really understand kitty from the inside out, Hetts recommends this read, which "goes beyond the popular 'easy reads' and delves into what science can tell us about cat behavior."
Whether you select an easier read or a scientific tome on the inner workings of cats, a cat-related book is a perfect source of informative tidbits and entertainment that will forever enhance your bond with kitty.


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A Seattle car owner lost a window but gained a dozen tortillas.
Police said a vandal shattered the window of a car parked near the University of Washington last week but did not steal anything. The vandal instead left behind a dozen tortillas strewn across the floor of the car, The Associated Press reported.
Police responded to the incident with a flash of humor befitting the bizarre nature of the crime.
U.S. & World
"The Seattle Police Department would like to take this opportunity to remind car prowlers that you shouldn't break in to something that's nachos," they said in a news release.


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Educational BackgroundPh.D., Marketing Education and Vocational Education, University of Pittsburgh
Area of Expertise/Research2011: National Association of Women MBA Pittsburgh Chapter (NAWMBA) advisor and mentor for the NAWMBA Pittsburgh Chapter launch on March 26, 2011. This organization encompasses MBA women from Robert Morris University, Duquesne, University of Pittsburgh and Carnegie Mellon. 2010: Member of the Women's Ad-Hoc committee developed through the RMU Board of Trustees. Currently assign the needs of the undergraduate and graduate women at RMU. 2011: Study abroad course in MARK4050 International Marketing, May 9 to 19, 2011 - Germany, France, and Switzerland with 13 students.2011: The Speaker in Business Session by Pittsburgh Symphony Orchestra that exposed students to the business operation of an orchestra. The program?s follow-up included over 100 undergraduate students attending the PSO performance on February 26, 2011.
PublicationsManna, D. R., Marco, G. J., Khalil, B & Meier, S (2011) Sustainable Markets: Case Study of Toyota Motor Sales, U. S. A., Inc. Journal of Business Case Studies.Schimmel, K., Motley, D., Rancic, S., Marco, G., Eschenfelder, M., (2010) The importance of university web page in selecting a higher education institution. Research in Higher Education Journal.Marco, G. J., Clinton, S. R. , & Chu, Y. (2009). Two Years Into the Journey: AACSB Assessment of Learning in a Principles of Marketing Course. American Journal of Business Education.Smith, A. D. & Marco, G. J. (2004). Strategic Perspectives Associated with the Golf Industry. The Journal of American Academy of Business, 5 (1/2), 367-373.Clinton, S. R., Marco, G. J. , & Manna, D. (2008). A Case Series of Today's Vertical Integration. Journal of Business Case Studies.Marco, G. J., Smith, A., & Fratto, V. (2004). Strategic Importance of Business Ethics: Case Study of Abuse. Journal of Global Business, 15 (29), 37-46.Marco, G. J., Schimmel, K. E. , & Eschenfelder, M. J. (2008). Difference in Selection Criteria Among Traditional Students, Adult Continuing Education Students and Graduate Students. American Journal of Business Education.


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Kevin Hart of 'About Last Night' a man on the move
Kevin Hart is yawning.
It's 1 in the afternoon at this junket in the Beverly Hilton, and he's looking extremely put together in a sharp, dark blue, subtly patterned suit. He is affable and engaged, raising his energy to answer each question.
But he's yawning. That's what he gets for vying for the crown of Hardest Working Man in Show Business.
"I'm taking advantage of what I have now. I have heat. It's starting to happen," says the star of current back-to-back-to-back movies "Grudge Match" (with Robert De Niro and Sylvester Stallone), "Ride Along" (nearing $100 million at the time of this writing) and the new "About Last Night."
'Right direction'
"I've got a movie I'm about to start filming with Will Ferrell in March ('Get Hard'). That's huge. I'm excited about this. Between that, between 'Think Like a Man Too,' between 'Wedding Ringer,' 'About Last Night,' 'Ride Along' - we're about to start filming 'Ride Along 2' - everything's moving in the right direction. I'm about to start touring again. I'm thinking about my stand-up, when I'm going to do my next concert film."
With his cocaine-addicted father in and out of jail, the fast-talking, diminutive (5-foot-4) comedian was raised in Philadelphia by his mother. Hart is only 33 years old, but he has amassed more than 50 film and TV credits. His YouTube videos have millions of hits, and his stand-up performances have become enormously successful.
Hart's 2011 "Laugh at My Pain" tour grossed in excess of $15 million, good for the No. 1 spot on Ticketmaster's listing of comedians. His 2013 concert film, "Let Me Explain," made $32 million.
"If you work hard to get results and t hen you start to see those results, you don't relax and go, 'Oh my God, I made it, let me now go on vacation and chill,' " he says, succumbing to another yawn. "That's when you work harder."
He's fighting off sleep today to promote "About Last Night," a remake of the Rob Lowe-Demi Moore romcom that claimed (unconvincingly) to be based on David Mamet's dark, angry and then-shocking play, "Sexual Perversity in Chicago." That movie took the play's nugget of two male friends and two female friends navigating the dangerous waters of modern courtship - and shipwrecking themselves in ugly and angry fashion - and steered it into the now-familiar comic waters of the "rules" of dating.
Closer to Mamet
The new film is also a romcom deriving from the 1986 movie, not the 1974 play, recast with an African American ensemble (Michael Ealy, Regina Hall and Joy Bryant joining Hart), but it feels somehow closer in spirit to Mamet's play than the first film was.
"As raw as ('Sexual Perversity') was, it was dope the way it was done," Hart says. "It was shocking to me that the studio said, 'We're going to go there with this. We're going rated R. We're going to show that couple that's all about f-, then going back to loving one another. Then we want to show the couple that's all about the romantic aspects, then hide when they have problems.' They're a little more reserved on the outside. The other couple (Hart and Hall), their business is out in the street and everybody sees it."
As in his stand-up, the comic relates his comedy to his own life. His routine in "Let Me Explain" even begins with him trumpeting his joy at his divorce being finalized.
"I related to ('About Last Night') so much because that used to be me. Me and my ex-wife. Our business was on display for everybody. We fought, we fought. Then we'd have sex, put a Band-Aid over everything, act like nothing ever happened. You show up around the same people you were just fighting around, and they're like, 'Uh ... is this all right?' Y'all are oblivious to how crazy you are."
Crackling chemistry
The remake keeps those outsize spats funny, largely because of the crackling on-screen chemistry of Hart and Hall.
"We're good friends, first of all," he says of his co-star in the "Think Like a Man" and "Scary Movie" films, and elsewhere. "Second, Regina is pound-for-pound one of the funniest females to do it, man. I'm so glad she got a chance to display how funny she actually is.
"I'm an improv warrior, man. It's what I do. I love it. She didn't miss a beat in any way shape or form. We were ping-ponging off each other the whole time."
The new film also presents Hart with his richest role to date. The actor cites "the moment where you get to see that I really love her. I'm at the bar with Michael and I explain what it is that she does to me and why I feel that it's OK and why I don't care.
"It's a great moment for my character, and it's a great one for me in my career because I'm showing those levels, that I can act. I think, walking away from it, people are going to say, 'He's got real chops. He's not playing around.' "
If that kind of career awareness is surprising, it's not new for Hart.
"My business acumen has definitely grown, for the simple fact that I'm not just a talent, I'm a brand. And when you realize that you're a brand, you focus on ways to build that brand."
Still, with his work schedule and divorce, Hart says that his anchor remains his two kids.
"This could all fall apart. The one thing that I need to make sure that I manage and understand: my kids," he says.
A beautiful thing
"Men embracing fatherhood and understanding how important it is to be with their kids and spending time with their kids and putting their stamp on who they're raising and why ... that's a beautiful thing, man.
"That's what I live for. That's what motivates me to get up each morning and do what I do to the best of my abilities, because I'm building something for them." {sbox}
About Last Night (R) opens Wednesday at Bay Area theatres.
To see a trailer, go to www.aboutlastnightmovie.com.
Michael Ordoña is a freelance writer. E-mail: firstname.lastname@example.com


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In 1898, the American advertising and sales pioneer, E. St. Elmo Lewis developed a practical sales tool using the latest Scientific Management insights. He created his AIDA funnel model on customer studies in the US life insurance market to explain the mechanisms of personal selling. Lewis held that the most successful salespeople followed a hierarchical, four layer process using the four cognitive phases that buyers follow when accepting a new idea or purchasing a new product.
The AIDA model describes the basic process by which people become motivated to act on a purchase and is based on external stimuli from sales representatives. This motivation to make a purchase depends on:
Lewis held that the fourth stage or mental state, ACTION, was a natural result of moving through the first three stages -- that desire leads to action, i.e.
In 1911, Sheldon extended the model with a fifth phase, 'permanent Satisfaction' to stress the importance of repeat sales. In the following decades, the AIDA model served to study how advertising affects consumers and was the basis for numerous motivational driven consumer behaviour research models such as ACAIP, ACCA, and CAB.
The AIDA funnel model was to be used as the backbone for structuring an organisation's sales. The sales funnel helps sales personnel to better target a client by basing their actions on the client's position in the funnel. Salespeople should seek different sales objectives for their suspects, their prospects and their customers. The aggregated information from the sales funnel allows a firm to construct an overview across sales representatives and departments and to deliver better structured sales forecasts.
The AIDA model allowed salespersons and managers to monitor their personal sales activity progress. Structuring the organisation's sales funnel according to the AIDA model's four hierarchical layers permits the insertion of quantitative conversion rates.
The model demonstrated to sales people that timing is important in sales processes since the mental state of prospective buyers changes over time. The buyer requires different information from the sales person at each state to be able to move to the next phase.
The model formed the basis for measuring the effect of advertising. It is the foundation for what became known in consumer behaviour research as the 'hierarchy of effect' models.
The model takes the view that the sales person rather than the buyer has the most control in the interaction. Later research showed that buying and selling are the same process whose outcome is determined by the total sales process rather than the actions and traits of the individual actors.
The model can be applied too strictly: its goal is to stimulate the buying action from potential buyers, not to lead them through five stages at all costs. AIDA as a blue print for a sales presentation requires the salesperson to talk disproportionally to keep feedback from the potential buyer from disrupting the flow of the sales process.
The key for successful AIDA implementation is to understand the state of the prospect. This requires skill and experience on the side of the sales person.
Later research showed that buyers change their mental state in a more dynamic, complex way than sequentially.


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Quick! Somebody get Gov. Brown a pillow and a glass of warm milk.
Because people -- especially wealthy civil rights lawyer Molly Munger and her Prop 38 campaign -- are being mean, terribly mean about Brown's treasured temporary tax ballot initiative, Prop 30.
That's the message from a letter sent by supporters of the Brown measure -- California's two U.S. senators and Democratic legislative leaders -- to Munger's campaign.
U.S. & World
The letter accuses Prop 38 of going negative against Prop 30, and suggests -- the following is not a joke -- that Munger agree to a "Positive Campaign Compact." That is, Prop 38 won't say anything bad about Prop 30, and Prop 30 won't say anything bad about Prop 38.
Let the voters decide.
Prop 38 quickly rejected this overture -- and with good reason, since it's hard to take seriously a call for positive campaigning that's received in a negative letter accusing you of negative attacks. But Brown's backers aren't giving up their push for being nice.
To add to the absurdity of politicians with long histories of negative campaigning demanding that people be nice, the Prop 30 campaign sent out an email to supporters reiterating the points of the letter and demanding that Prop 38 play nice.
Naturally, the email was signed by the brilliant veteran political operative Ace Smith, who has forgotten more about negative campaigning than almost any human being alive will ever know.
What's next -- asking Attorney General Kamala Harris to file elder abuse charges against Munger for how she's treated the 74-year-old governor?
It's enough to make you wish shamelessness and irony could be taxed. California could balance its budget then.
The Prop 30 plea for peace would be easier to take seriously if Brown hadn't spent his term in office dismissing, often in very negative terms, the ideas of virtually anyone in California with an approach to the budget difference than his. His whole campaign is a bum's rush -- you have to embrace my idea, all the other ideas are either bad or unrealistic politically.
Now he wants those whose ideas he's dismissed to play nice?
Here's a cynical prediction: Brown and his allies are putting this out because Prop 38 is gaining some strength in polls (though it remains behind), and because they need to justify what likely will be their own sustained attacks on the measure.
And finally, a word of advice to California's two senators, Democrats Barbara Boxer and Dianne Feinstein, who signed the letter asking Munger to back off.
You might be wise to stay out of this ballot fight, just to avoid embarassment.
Someone might ask why, after 20 years in office, you're backing Brown's meager measure even though it merely maintains the current austerity levels of spending and may make the budget worse long-term by locking into the constitution certain revenues for local government.
After two decades, don't the two of you have anything better to contribute to the discussion of how to fix California than simply asking Molly Munger to play nice?
Lead Prop Zero blogger Joe Mathews is California editor at Zocalo Public Square, a fellow at Arizona State University's Center for Social Cohesion, and co-author of California Crackup: How Reform Broke the Golden State and How We Can Fix It (University of California, 2010).
Send us your thoughts via Twitter @PropZero or add your comment to our Facebook page.


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Mocha Nougatine Cake
Enough about the weather report, onto the meat now.
There is a bakery in a city of Bandung, Indonesia called Bawean or Sweetheart. This bakery is so famous and I've been hearing lots of things about this particular bakery. I for one, never been there and when I heard so many good things about it, especially about what they call "Mocha Nougat Cake", I just had to try making it at home. From what I heard, this is one of their most popular cakes, it has three layers of vanilla sponge, filled with very strong rum custard, covered in mocha buttercream and crushed nougatine. When I say strong custard, I really meant STRONG. I never prove it of course, but I have quite a few friends coming from the city and the first thing they say when describing this cake is that this cake is strong strong strong in alcohol. I mean literally, it's so strong that it's bitter and the alcohol goes straight into your nose on your first bite. Many said that they feel tipsy after a slice or two, but strangely, people kept coming back and makes you wanting for more.
I was really skeptical putting a lot of rum in the filling as I personally don't like cake that has too much alcohol in it. I love putting alcohol in my cake as a taste-booster, but when it starts to overpower the taste, that's when I draw the line. I say if you like your alcohol, that's fine, but why not go drink it straight from the bottle rather than spending hours or days making a cake that taste completely like alcohol? It's just a one-person opinion though :D. I did however put a good dose of rum in it, just not too boozy yet.
I won't post the recipe just yet. The recipe still needs tweaking :)
19 comments:
- Brisbane Baker said...
I love weekends where you have nothing to do :)
Mocha Nougatine, sounds so delicious! And looks it too! Great work!
- April 26, 2010 at 3:26 AM
Can't wait for the recipe. It sounds amazing :D
- April 26, 2010 at 1:40 PM
- Steph said...
Wow, such a pretty cake!
- April 27, 2010 at 7:35 AM
just wanted to say that I discovered your blog through tastespotting. Beautiful cakes! I see many confections I would love to try (one day...) and wish you could post the recipes, but i understand it is time-consuming.
Did you have professional training? I think you should be a professional baker!
- April 28, 2010 at 11:29 PM
- Bertha said...
Thanks for the compliments everyone!
Anonymous: No, I don't have a professional training nor do I have the skill to be one ;). I do try to post every recipes lately, don't you notice? :P
- April 28, 2010 at 11:41 PM
- sana said...
you do the great jobb!!
- April 29, 2010 at 8:47 AM
- Cristina @ TeenieCakes said...
Oh! Such a beautiful cake and it sounds delicious. Looking forward to the recipe - I'd like to give it a try. Beautiful presentation and images! =)
- May 3, 2010 at 7:35 AM
- DG said...
This cake looks beautiful. I always impressed with all your cakes, it looks professional.
Btw, just to let you know that I just baked HK style cake from your recipe. The cake is nice. Thanks for sharing it. :)
- May 11, 2010 at 9:52 PM
- Bertha said...
DG: I just stopped by your blog and I couldn't be happier. Your cakes turned out more beautiful than mine! So glad that you like it :)
- May 17, 2010 at 10:17 PM
- COLARGOL said...
this cake looks absolutely fabulous and very greedy, I love also the individual version.
Your blog is very nice, I did a little tour, the photos are wonderful and the desserts treasures
- May 18, 2010 at 1:03 PM
- Breadpitt said...
nuts , cruchny , caramel= drool...;-)
- May 27, 2010 at 1:37 PM
- Kitchen Corner said...
Your cake looks awesome! Very beautiful. Simple and nice and elegant! I love it!
- June 11, 2010 at 2:08 AM
- Marianna said...
beautiful- adore this! my eyes were especially lured by the cupcake sized ones!
- June 28, 2010 at 8:04 AM
- Unknown said...
Please post your recipe! I'm salivating.
- August 17, 2010 at 5:02 PM
hi, i would like to know if you have perfected your recipe! thanks!
- November 30, 2010 at 5:14 PM
- Bertha said...
Hi Stephanie,
I haven't had a chance to make a second one yet, but I'll post the recipe when I do :)
- November 30, 2010 at 5:31 PM
you never post recipes..:(
- July 27, 2012 at 3:51 PM
Bertha, when would you be able to post the recipe...long time now...3yrs since u first baked this !!!
- March 19, 2013 at 2:05 AM
- Angelica said...
Wow, this cake looks amazing! Did you ever post the recipe for this cake? I'd love to bake it myself one day.
- April 25, 2013 at 4:38 AM


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Little Girl, Ready for Bed
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Hugh Hanley's Circle of Songs CD/Songbook sets were developed to promote adult/child interactive singing either in pairs or small groups. Children also enjoy singing these songs by themselves. Each set contains songs and activities that are appropriate and great fun for children ages two to seven years and for adults who would like to sing with them. The songs and activities are presented slowly and simply on the recording to help participants learn the songs and to invite participation.
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A power outage plunged Damascus and southern Syria into darkness late Saturday, Syria's state news agency said, while anti-regime activists reported a string of tit-for-tat, sectarian kidnappings in the country's north.
The news agency, SANA, quoted Electricity Minister Imad Khamis as saying that the failure of a high voltage line had left the country's south without power.
The blackout affected Syria's capital, Damascus, and the southern provinces of Daraa and Sweida, which abut the Jordanian border.
An Associated Press reporter in Damascus reported dark streets across the capital. A fuel shortage makes it hard for residents to run backup generators.
A similar blackout struck Damascus and southern Syria on Jan. 20, leaving many residents with no way to heat their homes on a cold winter night. The government blamed that outage on a rebel attack, and power was restored to most areas the following day.
The Syrian capital's 2.5 million residents have grown used to frequent power cuts as the country's conflict has damaged infrastructure and sapped the government's finances.
Meanwhile, anti-regime activists reported a string of kidnappings in recent days that have enflamed tensions between Sunni and Shiite Muslim villages that back opposite sides in the country's civil war.
The activists differed on the number kidnapped from both sides, with reports ranging from a few dozen to more than 300.
The kidnappings point to the dark sectarian overtones of Syria's civil war, which pits a predominantly Sunni Muslim rebellion against a regime dominated by President Bashar Assad's minority Alawite sect, an offshoot of Shiite Islam. The country is also home to Christian, Kurdish, Armenian and Shiite communities, all of whom have been swept up in the conflict.
The kidnappings took place between two Shiite villages in the northern Idlib province and a number of Sunni villages that surround them.
The Britain-based Syrian Observatory for Human Rights said 42 Shiites, including mainly women and children, were snatched Thursday from a bus that was traveling from the Shiite villages of Foua and Kfarya to the capital Damascus. Observatory director Rami Abdul-Rahman, said it was not clear who took them, adding that Shiites have refused to give the names of those kidnapped or details about the make or color of the bus.
Since then, however, Shiite gunmen from the two villages have kidnapped more than 300 people from nearby Sunni villages, Abdul-Rahman said.
The kidnappings highlighted how much the civil war has heightened sectarian tensions. Kidnapping for ransom has grown common across Syria since the crisis began in March 2011, but sectarian and political abductions have been rare.
Anti-regime activists in Idlib reached via Skype confirmed the kidnappings, but gave much lower numbers for the number of people involved.
Activist Fadi al-Yassin Al-Yassin said Foua and Kfarya are being used by the regime to bombard nearby villages and towns, saying the regime has turned them into "castles of shabiha," referring to pro-government gunmen.
In retaliation for the bus kidnappings, members of the pro-government Popular Committees set up a checkpoint around the two Shiite villages and on Thursday and Friday were taking people from cars they stopped, the Observatory said. It added that most of the people abducted were from the Sunni villages of Saraqeb, Binnish, Sarmin, Qimnas, Maaret al-Numan and Maaret Musreen.
Al-Yassin confirmed the kidnappings on both sides but added that the 300 figure is high. He said few dozens of people have been abducted in the area.
Abdul-Rahman and al-Yassin said such acts could incite sectarian clashes between Shiites, who have largely sided with the regime, and majority Sunnis in Idlib, where the sects have coexisted for decades.
The high number of women and children allegedly taken prompted the U.N. Special Representative on Sexual Violence in Conflict, Zainab Hawa Bangura, to issue a statement calling for their release.
"Allegations of abduction and rape of women and girls by armed groups have been received," she said. "I am deeply concerned about the well-being of these women and children and would like to remind the armed group responsible for this abduction that acts of sexual violence will not be tolerated."
In the nearby province of Aleppo, rebels and troops fought fierce battles around the military air base of Kweiras, which opposition forces have been trying to capture for weeks, the Observatory said.
In the city of Aleppo, Syria's largest urban center and commercial capital, rebels and troops battled around the international airport and the nearby air base of Nairab, said the Observatory and the Aleppo Media Center.
Fighting also raged in parts of Damascus, where troops bombarded rebel-held southern neighborhoods, according to the Observatory. It said at least one person was killed in the Palestinian refugee camp of Yarmouk by sniper fire.
In the southern region of Quneitra rebels briefly controlled a military checkpoint in the town of Khan Arnaba and captured five officers before they withdrew, the Observatory said. It added that the rebels captured a tank then blew it up.
Quneitra is on the cease-fire line between Syria and Israel, which controls most of the Golan Heights after capturing the strategic territory from Syria in the 1967 war.


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Huawei P Smart Blue Smartphone
- 6.21-inch
- 1080x2340 pixels
- Android 9 Pie
- 13-megapixel + 2-megapixel
- 3400mAh
- Delivered in 24 hours
- Out of stock
WIDER VIEW, COMFORTABLE GRIP The HUAWEI FullView Display features a striking 5.65-inch 2.5D glass screen that reflects vivid, true-to-life colours. But a wider screen does not mean a larger body, the thin, lightweight HUAWEI P smart fits easily in the palm of your hand. HUAWEI P smart front display showing an ice cream in true-to-life colours SMART SPLIT SCREEN The smart Split Screen lets you run two apps side by side at the touch of a button*. So when you get an important email or message while streaming TV or a movie on your HUAWEI P smart, you dont need to pause the fun. *Some applications do not support this feature. DUAL CAMERA, ONE AMAZING SHOT High-spec 13MP + 2MP Dual Camera means top quality images at the touch of a button. The large aperture lenses let in more light for rich, detailed, true-to-life photographs, while the Dual Camera enables perfect shallow depth-of-field bokeh shots. *The photograph displayed has not been taken with this product. HUAWEI P smart dual camera showing the bokeh effect SMART BEAUTY Using the Face Beauty algorithm, the 8MP front camera detects unique facial characteristics and then adds just the right amount of enhancement, helping you look your best without even trying. *The photograph displayed has not been taken with this product. Couple photographed using the smart beauty feature - HUAWEI P smart smart beauty effect €" HUAWEI P smart HANDS-FREE SELFIE It's never been easier to capture the moment. The HUAWEI P smart's gesture-shot function lets you take selfies and groufies with a simple wave of your hand. SPEED SHARING The HUAWEI Share function transfers files in a flash between HUAWEI phones. No internet or manual pairing required. Super fast, super simple. DUAL BLUETOOTH CONNECTION HUAWEI P smart supports two Bluetooth audio devices simultaneously. Connect to your car's audio system and receive calls on your Bluetooth headset at the same time. SMOOTH UX The new, exquisite EMUI 8.0 is engineered to enable a highly responsive, intuitive user experience. With 32GB memory (expandable with microSD up to 256GB) plus an upgraded storage management system, there will always be room for your favorite photos, music and videos. *The actual space available to users may be less than the stated memory capacity due to pre-installed applications and OS, and may change due to updates, user operations, and other related factors.
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Despite the cat already being out of the bag on the new 2010 Ford Taurus, Ford's furiously trying to shove it back in, covering up the new Ford Taurus in all sorts of anti-spy camouflage. And though the sedan is covered and padded to the extreme there's still a lot to learn from these photos. The front bumper appears to have more detail than the previously snapped Taurus, suggesting that wasn't a final product. There's also a tester running around with dual exhausts, which could mean EcoBoost, or Heavens to Betsy, a 2010 SHO Taurus! Like Mulder, we want to believe. Full spy report below the jump.
Here are some shots of Ford's invigorated 2010 Taurus, caught testing in Detroit. This new Taurus is what the car should have been all along, according to Ford's CEO Alan Mulally. Although the car is very heavily camouflaged, the prototype definitely looks related to the camera phone shots of a purported Taurus clay model, which leaked earlier this year. We think we can see some differences on the front bumper—around the fog lights—suggesting that the finer points of the 2010 Taurus' design were still in flux at the time the photo was snapped inside Ford's Design Center. We have seen photos two different prototypes—with single exhaust and dual exhaust. The dual exhaust set-up likely points to the Ecoboosted 3.5L V-6, which will produce 340-hp in the Lincoln MKS.


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Discover Your Family Tree: Genealogy for the Absolute Beginner
I have do not have experience with any of the Family Tree University courses, but this may be a good overview for those new to genealogy. It is an online course where you download the lessons and complete them over two weeks. At the end of each lesson there is an exercise to complete, and then you receive feedback via email from the instructor. The first session of this course will begin on Monday, October 11, 2010.
Here is the course description from the Family Tree University Web site:
This course will start you on the fun and rewarding journey of discovering your roots. Learn how to begin, where to look for information to extend your family tree, what to do with what you find and how to put it all together.
- How to fill out a family tree chart
- How to record and organize information
- Where to find information to extend your family tree
- How to search online and in repositories
- Anyone new to genealogy and family history research
- Genealogy "dabblers" who need additional direction in order to advance their research
- Researchers who want a refresher course on core family history research principles
- Library and archive staff who want to master basic genealogy principles in order to better assist patrons


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It's turned out to be hard not to love a scheming, unfaithful, bisexual, sexually manipulative Roman noblewoman – at least, the way Lucy Lawless plays "Spartacus'" Lucretia.
As the Starz series begins its second season – now titled "Spartacus: Vengeance" – Lawless joins PopcornBiz for a lengthy chat about sinking her acting claws into conflicted schemer, enjoying her third series that boasts a legion of deeply devoted fans, and yes, those boundary-busting sex scenes.
You just keep finding these shows that attract such passionate fan bases.
They find me. I mean, well, two of them were given to me by my husband. But I don't know – They find me.
How much is strategy and how much is luck?
Not strategy. I don't know how much luck there is either in having your husband, you know. Well, he didn't write the role for me. He awarded me with it, but they could easily have nixed it. So I've never had a strategy. I've always just floated on. I would say that I'm like a cow grazing on the side of the road. 'Oh, this looks good. I'll do that.' And strangely enough, it works out just as well as being somebody who goes gunning for things, spends a lot of energy chasing things up. I'm just not that kind of person – and I have often chided myself for not being, but it seems to work out okay.
As you "floated" into this role, and it keeps going on, what is really captivating you about being Lucretia as the character evolves?
It's the chance to bring humanity to what could easily be a villainess role. And I want to make it understandable. I want to at least make you understand her. Maybe love her, but relate to her. And that's such a pleasure for now. Just to inhabit the role: this is who I would be if I had these words to say and if I was in this situation. You have to make it as if that is you. Usually I sort of cogitate on what would make me say these things, but I generally only do research if something specific is required that I've never been exposed to – like, in this case, madness at the beginning of season two. But I'm not sure that that helped me, to be honest with you. Because sometimes real madness doesn't help with the process of shooting madness.
Tell me what you can tease about the upcoming season.
I will say that through many trials and tribulations, Lucretia gets everything she ever wanted. Or no, let me say that again: Through trials and tribulations, all Lucretia's dreams come true.
Is that ultimately always a good thing for her?
Well, that's very subjective isn't it? I don't know. But as an actress it's a very good thing, because the cause of true love or good luck in our show never runs smooth. So it's been a very textured ride, meaning that I've gone to some really dark places sort of beyond my own personal experience. So I thank them for that, a role that has been that demanding.
How about just plain fun? When you first saw something on the pageant thought "Oh, I've never gotten to do that before. That's going to be incredible. That's going to be a great day of filmmaking?"
Well, I have a very perverse sense of humor so usually the most terrible stuff is what makes me laugh the most. So there were quite a few laughs this season, I must say. And sometimes it's a very rueful laughter because, like I said, I went to the heart of darkness in season two. But in the end, for Lucretia, it all turns out worth it.
How much does it dominate your mind that you never know when you're going to have to drop the robes for one of the show's signature sex scenes?
Not too much. See, when you're the cow on the side of the road you just go along and you deal with problems as they crop up. So there's not too much anticipation. Though I have always known Lucretia's arcs because in this show you're future arc is very important to the way that you're playing thing. And I guarantee that by the time you see the last episode you're going to say, "Oh, my God." And you're going to go back and watch the whole thing again with new eyes.
Tell me about any of the behind the scenes challenges of completing this season?
There weren't any. This was very kopasetic, I must say. At least in my role where I got to work very closely with Nick Tarabay, who plays Ashur. He's absolutely genius. So it was wonderful to work with an actor who will feed off you. And will take one another to new heights. And he's one of those. Also, Lucretia enters into a sexual relationship with someone. And that's going to be a tricky one to ride out.
You also have been involved in several shows that have unusually passionate fan bases. What's that experience been like to go, not just from one show like 'Xena' where everybody saw you as that character and adored you as a character, but to go through a few now?
You know, it's amazing. I was just at the Consumer Electronics Show. I went to this meeting of Washington and technology sector –so they're all Washington technology lobbyists – and the leaders of Best Buy and Amazon and you name it were all in this room. And when people figured out who I was – because I don't look like my characters necessarily – I had a lot of people come up and ask for photographs and I would say, 'So what do you know me from? Where does this love come from?' And they will say whether it is 'Xena', 'Battlestar', or 'Spartacus,' so it's interesting that those three have very different appeal. And some people like both but very often they fit into one or the other. And they're really, really zealous. And I love that. It's like I'm not just cannibalizing the same fan base but actually growing it. So I know with women Xena was heavily female skewed. 'Battlestar', heavily male skewed, and young men. And 'Spartacus' is a much broader thing and also male skewed [Sounds like: a ton]. So I've got this wonderful panopoly of fans now from all different walks of life.
Among the famous people you've encountered who has been the most surprising fan of your work? Somebody who you admire and they came up and said, 'I love "Xena"' or 'I love "Galactica"'?
Rupert Everett was one, because I was so madly in love with him when I saw 'Another Country' way back when I was a teenager, and he just blew my mind. Who else? Oh, Madeleine Albright. That was a surprise. I've never met her, but apparently she was called 'Xena' around the office.
What does the average 'Spartacus' fan want to know when they get a chance to ask you about the show? And what do you tell them?
They just talk at me, which is really nice, or they'll say, 'So what is it like to do the sex scenes?' I suppose. And the answer is, 'Horrible. Don't do it. Don't do it. Unless you're really jacked up on opium.'
"Spartacus : Vengeance" premieres tonight, January 27th, on STARZ. But you can watch the first episode online now.


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This Lanka Kade Ostrich Is Painted On Both Sides And Is Handcrafted From Sustainable Rubber Wood. Just One Of Many To Collect From Our Natural Painted World Animals Which Are A Chunky 25mm Thick And Are Great For Children To Create Their Own Wild Life Play Scene. Size: 110 X 82 X 25mm. Suitable For Children 10 Months Plus. Wipe Clean Only. All Products Successfully Tested To En71 Standards. Hand Crafted And Painted By Skilled Artisans Using Non Toxic Paints Made Under Strict Fair Trade Conditions In Sri Lanka. Ref: Np73


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|System: X360, PS3
|Review Rating Legend
|Dev: Dimps
|1.0 - 1.9 = Avoid
|4.0 - 4.4 = Great
|Pub: Atari
|2.0 - 2.4 = Poor
|4.5 - 4.9 = Must Buy
|Release: June 10, 2008
|2.5 - 2.9 = Average
|5.0 = The Best
|Players: 1-2
|3.0 - 3.4 = Fair
|ESRB Rating: Teen
|3.5 - 3.9 = Good
by Jwan Jordan
January 29, 2008 - Well, despite some legal issues, Dragon Ball Z looks like it's preparing to make the leap into the next generation of gaming, and I have to say it is looking very good. Atari apparently had some licensing right issues with FUNimation about creating more Dragon Ball games. Regardless, Atari has opted to not only go ahead with creating the game, but launched a trailer to tease fans as to what's coming this year. Other than the teaser trailer, Atari has not gone into any specifics about the game other than it will have "radical 3D fighting effects" and online play will be possible.
I'll humor you as if you don't know what Dragon Ball Z is and enlighten as to why it has such a strong fan base. Dragon Ball Z is a popular Manga and hit anime series created by Akira Toriyama. It initially begins as a series merely called Dragon Ball about a monkey tailed kid, named Goku, who is sent to earth to destroy it. A series of events eventually causes Goku to adopt the customs of earth and inevitably become the planet's strongest and most beloved hero. The series plays out in a semi-realistic chronological order, meaning Dragon Ball starts off as a very young, naïve Goku, and eventually the series transitions into Dragon Ball Z where he is older, more powerful, has a wife and kid, and well, isn't really that much smarter, yet still possesses his heart of gold. Some argue whether or not the later series, Dragon Ball GT, should be considered a legitimate continuation of the popular series due to its horrible ratings amongst fans. Dragon Ball GT is the final continuation of Dragon Ball, with a story that turns Goku back into his youthful Dragon Ball iteration and leads him on a journey to once again find the legendary dragon balls to save the universe. During Dragon Ball's time frame, it attracted millions of both Asian and American fans who are still to this day very loyal to the series. Dragon Ball has also attached a large assortment of video games spanning many consoles over the years, and they keep coming for better or worse.
So far all we have to go by is the Japanese teaser trailer just released to the public. It showcases a few characters that I think it's safe to assume will be playable in the game. Some of the random clips show, Goku, of course, and others such as Cell, Vegeta, Frieza, and the Ginyu Force. Nothing was shown past the Cell Saga, and the rumor is that it won't go past that, though it does seem like a bad choice not to include the entire Dragon Ball series in its entirety by this point. Regardless, the trailer did not show a trace of the Buu saga or GT, so the rumors may be true. I honestly believe that stopping at the Cell saga is a bad choice, merely because the fan base is so huge and there are so many characters that fans want to see and are accustomed to playing as on the other Dragon Ball titles, especially if it's going to have online possibilities.
Despite my concerns, it is very comforting to see the new generation graphics at work on a Dragon Ball title. My first impression was how similar the graphics look to the Dragon Ball Z Budokai series, which possessed a 3D polygon sort of look. What is unique about the look is the next gen ability to use enhancing lighting and shadow effects on the characters; which gives them a lot more of a polished look and natural feel to them. In a clip, we see Goku charging his ki energy, and thanks to next-gen processing, his hair moves naturally as the ki energy rises above him. As usual in the series, rocks, pebbles, and dust rise from the ground demonstrating his rising power level. Also, his clothes actually gave off more of a natural cloth movement as his energy runs rampant, with his clothes and sash flapping naturally with the wind.
Gameplay is, of course, one on one fighting and looks very similar to the Budokai series, which means it's more of a one dimensional plane and involves moving from the left to right side of the screen. I honestly wonder if this may be a step backwards, but I hope there is a method to their perceived madness. Nonetheless the fighting looks fast paced and hectic, which is a good thing. It appears they may have truly captured the blinding speed of the Dragon Ball series. From the trailer, you can clearly see constant teleporting back and forth between enemies, and true the series, a well launched punch literally goes through your foes in a very agonizing way.
One thing that concerns me is the use of Goku's and others' special abilities and how will it play into the game. The original Budokai title made it impossible to perform special attacks making even fans of Street Fighter feel flustered. I hope they will not make that mistake with Burst Limit and make special attacks too difficult to perform. Everyone knows that Kamehamaha waves, instance transmissions, and special beam cannons are what make Dragon Ball Z so fantastic in the first place. One of the announcements made at a public release was in regards to the special effects, and yes, they look beautiful. Thanks to the new generation, the Kamaehamaha wave has never looked better. Even more so when energy collides; the collision of power looks and feels unbelievable.
So far the text and voices are only in Japanese, but it's safe to assume that the cast will redeem their eternal roles as Dragon Ball Z voice over counterparts. The sound effects seem to all be in their respectable places also. From the instant transmissions, to heavy blows to the rib, and dashes to and fro, the soundtrack doesn't skip a beat and really has the authentic l of a very well done Dragon Ball Z game.
The game looks like it is doing really well. I'm honestly really pumped about it. Though it is unfortunate it only goes to the Cell saga, I'm really accustomed to the Super Sayian 3 Goku by now. It has been proven that DBZ games can be great to watch and just as fun to play. Here's hoping for another enhanced Dragon Ball game!
Jwan Jordan
CCC Freelance Writer
Game Features:


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DIY PAINT - Layered Chocolate
Shipping calculated at checkout.
'Layered Chocolate' DIY Paint
Available in 3 sizes!
- 8 OZ JAR. Perfect for a small project like a chair or small side table.
- 16 OZ PINT. Usually more than enough to complete a good size piece of furniture.
32 OZ QUART.- Covers approximately 150 square feet will cover several pieces of furniture.This size is available by custom order, so please call or email me!
ONLY 9 INGREDIENTS: Love, water, clay, porcelain clay, chalk, alcohol, ester (as a binder), cellulose, pigments, preservative.
Check out our Facebook page for painting ideas, painting techniques and other fun diy projects.


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Embraer ERJ 190
To Italy and back
Embraer ERJ 190
Scheduled service Frankfurt–Florence–Frankfurt: A last-minute gate change slightly delays takeoff so that now there is some time to catch up. Only 35 minutes remain for the turnaround in Florence – that's the time till the airplane takes off again for Frankfurt. In other words, 35 minutes for deboarding and boarding, refueling, loading, the walkaround inspection, cabin cleaning …
The following video contains no captions. You will find a description on the website.