The Symmetry Number Structure about Line-1/2

1. The Symmetry of P/2n and Prime Numbers Conjectures

We have

$${\displaystyle \frac{P}{2n}}=\left\{\begin{array}{c}{\displaystyle \frac{1}{2^{N+1}}}n=2^{N}P\\ {\displaystyle \frac{3}{4}}n=2P=3\\ 1n=1P=2\end{array}\right.$$

Figure 1. P/2n number structure with points [0 1/2^N+1 3/4 1].

Figure 1. P/2n number structure with points [0 1/2^N+1 3/4 1].

$N~\left(0,1,2,3,4,\dots \dots \dots .\right)$ All natural numbers

$n~\left(1,2,3,4,\dots \dots \dots .\right)$ All natural numbers excepted 0

$\mathrm{P}~\left(2,3,5,7,\dots \dots \dots .\right)$ All prime numbers

And

$${\displaystyle \frac{P}{2n}}=\left\{\begin{array}{c}{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2n}}P=n-1(n\ge 3)\\ {\displaystyle \frac{1}{2}}P=n\\ {\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2n}}P=n+1\end{array}\right.$$

And We have

p0∈$\mathrm{P}$ ~(0, n] $(n\ge 2)$

And based on Bertrand -Chebyshev Theorem: when $n\ge 2$, there are at least a prime number between n and 2n.

pn∈$\mathrm{P}$ ~[n, 2n) $(n\ge 2)$

So we have:

$$0<{\displaystyle \frac{p0}{2n}}\le {\displaystyle \frac{1}{2}}$$

$${\displaystyle \frac{1}{2}}\le {\displaystyle \frac{pn}{2n}}<1$$

So we have

$${\displaystyle \frac{\mathrm{p}0}{2\mathrm{n}}}={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2\mathrm{n}}}(\mathrm{n}\ge 3)$$

$${\displaystyle \frac{\mathrm{p}\mathrm{n}}{2\mathrm{n}}}={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2\mathrm{n}}}$$

So

$$({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2n}})+({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2n}})={\displaystyle \frac{p0}{2n}}+{\displaystyle \frac{pn}{2n}}$$

$$2n=p0+pn(n\ge 3)$$

This is the proof of Goldbach conjecture.

And

$${\displaystyle \frac{pn}{2n}}-{\displaystyle \frac{p0}{2n}}=\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2n}}\right)-({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2n}})$$

$$pn-p0=2$$

This is the proof of Twin Primes Conjecture

And we also have

$$0<{\displaystyle \frac{p0}{2n}}={\displaystyle \frac{2k_1+1}{2n}}\ll 1/2(n\ge 3)$$

$$0<2k_1+1\ll n$$

$$0<k_1\ll {\displaystyle \frac{n-1}{2}}$$

$$k_{1}isapositiveinteger$$

$$sok1~\mathrm{1,2},3······\left[{\displaystyle \frac{n-1}{2}}\right](n\ge 3)$$

And

$${\displaystyle \frac{1}{2}}\ll {\displaystyle \frac{pn}{2n}}={\displaystyle \frac{2k_2+1}{2n}}<1$$

$$n\ll 2k_2+1<2n$$

$${\displaystyle \frac{n-1}{2}}\ll k_2<{\displaystyle \frac{2n-1}{2}}(n\ge 3)$$

$$k_{2}isapositiveinteger$$

$$so{k}_2~\left[{\displaystyle \frac{n-1}{2}}\right],\left[{\displaystyle \frac{n-1}{2}}\right]+1,······\left[{\displaystyle \frac{2n-1}{2}}\right](n\ge 3)$$

we have

$${\displaystyle \frac{pn}{2n}}-{\displaystyle \frac{p0}{2n}}={\displaystyle \frac{{2k}_2+1}{2n}}-{\displaystyle \frac{2k_1+1}{2n}}$$

$$pn-p0=2(k2-k1)$$

$${(pn-p0)}_{max}=2({k2}_{max}-{k1}_{min})=2\left(\left[{\displaystyle \frac{2n-1}{2}}\right]-1\right)=2(n-2)(n\ge 3)$$

This is the proof of Polignac’s conjecture.

So we get a symmetry structure of P/2n as Figure 2

$${\displaystyle \frac{\mathrm{p}0}{2\mathrm{n}}}={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2\mathrm{n}}}(\mathrm{n}\ge 3)$$

$${\displaystyle \frac{\mathrm{p}\mathrm{n}}{2\mathrm{n}}}={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2\mathrm{n}}}$$

$${\displaystyle \frac{\mathrm{p}\mathrm{n}\pm \mathrm{p}0}{4\mathrm{n}}}={\displaystyle \frac{1}{2}}$$

2. A Concise Proof of The Fermat’ Last Theorem

The Fermat’ Last Theorem:

$${\mathit{x}}^{\mathit{n}}+{\mathit{y}}^{\mathit{n}}={\mathit{z}}^{\mathit{n}}(x,y,z\in n,xyz\ne 0n>2)\text{has no solution}$$

$$\mathit{n}~\left(\mathrm{1,2},\mathrm{3,4},5,\mathit{}6\dots \dots \dots .\right)\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{e}\mathrm{x}\mathrm{c}\mathrm{e}\mathrm{p}\mathrm{t}\mathrm{e}\mathrm{d}0$$

The equivalent proposition of this conjecture is

$$({{{\displaystyle \frac{\mathit{x}}{\mathit{z}}})}^{\mathit{n}}+\left({\displaystyle \frac{\mathit{y}}{\mathit{z}}}\right)}^{\mathit{n}}=1$$

$(x,y,z\in n,xyz\ne 0n>2)$ has no solution.

We have

$$({{{\displaystyle \frac{x}{z}})}^n+\left({\displaystyle \frac{y}{z}}\right)}^n=1=\sum {\displaystyle \frac{1}{2^n}}$$

$$n~\left(\mathrm{1,2},\mathrm{3,4},5,6\dots \dots \dots .\right)\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{e}\mathrm{x}\mathrm{c}\mathrm{e}\mathrm{p}\mathrm{t}\mathrm{e}\mathrm{d}0$$

$$({{{\displaystyle \frac{x}{z}})}^n+\left({\displaystyle \frac{y}{z}}\right)}^n=1$$

$$=({\displaystyle \frac{1}{2^n}}-{\displaystyle \frac{1}{2^n}})+1$$

$$=({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2^n}})+({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2^n}})$$

$$={\displaystyle \frac{1}{2^n}}+(1-{\displaystyle \frac{1}{2^n}})$$

Only When $n=1$ we have

$$\left({\displaystyle \frac{1}{2^n}}-{\displaystyle \frac{1}{2^n}}\right)=\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2^n}}\right)=0$$

$$1=({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2^n}})=({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}})$$

And Only When $n=2$

$$\left({\displaystyle \frac{1}{2^n}}\right)=\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2^n}}\right)={\displaystyle \frac{1}{4}}$$

$$\left(1-{\displaystyle \frac{1}{2^n}}\right)=\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2^n}}\right)={\displaystyle \frac{3}{4}}$$

And We can get the figures as Figure 3.

$$\mathit{n}=1\mathrm{and}\mathit{n}=2$$

In fact we have

$$1={\displaystyle \frac{1}{2^1}}+{\displaystyle \frac{1}{2^1}}={\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{3}{2^2}}={\displaystyle \frac{1}{2^3}}+{\displaystyle \frac{7}{2^3}}\mathit{o}\mathit{r}{\displaystyle \frac{3}{2^3}}+{\displaystyle \frac{5}{2^3}}$$

Figure 4. a symmetry structure of $({{\displaystyle \frac{\mathit{p}}{\mathit{q}}})}^{\mathit{n}}$about line-1/2.

Figure 4. a symmetry structure of $({{\displaystyle \frac{\mathit{p}}{\mathit{q}}})}^{\mathit{n}}$about line-1/2.

$({{\displaystyle \frac{\mathit{p}}{\mathit{q}}})}^{\mathit{n}}$p, q is relatively prime and $n~\left(1,2,3,4,\dots \dots \dots .\right)$

$$1/2\left[\right({{{\displaystyle \frac{\mathit{x}}{\mathit{z}}})}^{\mathit{n}}+\left({\displaystyle \frac{\mathit{y}}{\mathit{z}}}\right)}^{\mathit{n}}]=1/2\leftrightarrow ({{{\displaystyle \frac{\mathit{x}}{\mathit{z}}})}^{\mathit{n}}+\left({\displaystyle \frac{\mathit{y}}{\mathit{z}}}\right)}^{\mathit{n}}=1$$

$$({{\displaystyle \frac{x}{z}})}^n={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2^n}}$$

$${\left({\displaystyle \frac{y}{z}}\right)}^n={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2^n}}$$

3. A Concise Proof of Collatz Conjecture

Collatz Conjecture:

$$f\left(n\right)=\left\{\begin{array}{c}{\displaystyle \frac{n}{2}}ifn\equiv 0\left(mod2\right)\\ 3n+1ifn\equiv 1\left(mod2\right)\end{array}\right.$$

$k\in N\to f^k\left(n\right)=1$

$\mathrm{n}~\left(1,2,3,4,\dots \dots \dots .\right)\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{e}\mathrm{x}\mathrm{c}\mathrm{e}\mathrm{p}\mathrm{t}\mathrm{e}\mathrm{d}0$

$${\displaystyle \frac{\left[\frac{n+1}{2}\right]+\left[\frac{n-1}{2}\right]}{\left[\frac{n}{2}\right]}}={\displaystyle \frac{2n+2}{n+1}}={\displaystyle \frac{\left(n-1\right)+\left(3n+1\right)}{2n}}={\displaystyle \frac{4n}{2n}}={\displaystyle \frac{4}{2}}={\displaystyle \frac{2}{1}}={\displaystyle \frac{1}{\frac{1}{2}}}$$

$$=\sum {\displaystyle \frac{1}{2^N}}$$

Figure 5. a symmetry structure of $4\mathit{n}$ about line-1/2.

Figure 5. a symmetry structure of $4\mathit{n}$ about line-1/2.

$$2\mathit{n}=1/2[\left(\mathit{n}-1\right)+\left(3\mathit{n}+1\right)]$$

$$\underset{\mathit{n}\to \mathit{\infty }}{\mathbf{lim}}\left({\displaystyle \frac{\mathit{n}-1}{4\mathit{n}}}\right)=1/4$$

$$\underset{\mathit{n}\to \mathit{\infty }}{\mathbf{lim}}\left({\displaystyle \frac{3\mathit{n}+1}{4\mathit{n}}}\right)=3/4$$

4. The Proof of Riemann Hypothesis

Riemann Zeta-Function

$$\xi \left(s\right)={\sum }_{n=1}{\displaystyle \frac{1}{n^s}}\begin{array}{ccc}=\prod {\displaystyle \frac{1}{1-p^s}}& & \end{array}(s=a+bi)$$

$$\mathit{S}>1\mathit{}\xi \left(s\right)\to \mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}$$

The trivial zero-points of Riemann Zeta-Function is -2n (n~1,2,3,…….)

Riemann Hypothesis: all the Non-trivial zero-point of Zeta-Function $\mathit{Re}(s)={\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

We can get a symmetrical structure including all numbers about the line-1/2 as Figure 6

$$\mathit{s}={\displaystyle \frac{1}{2}}+\mathit{t}\mathit{i}\mathit{t}\in \mathit{R}$$

$$\mathit{z}\mathit{p}1={\displaystyle \frac{1}{2}}-\mathit{a}+\mathit{b}\mathit{i}\mathit{}\mathit{z}\mathit{p}0={\displaystyle \frac{1}{2}}+\mathit{b}\mathit{i}\mathit{}\mathit{z}\mathit{p}2={\displaystyle \frac{1}{2}}+\mathit{a}+\mathit{b}\mathit{i}$$

$$\mathit{z}\mathit{p}1+\mathit{}\mathit{z}\mathit{p}2=\left({\displaystyle \frac{1}{2}}-\mathit{a}+\mathit{b}\mathit{i}\right)+\left({\displaystyle \frac{1}{2}}+\mathit{a}+\mathit{b}\mathit{i}\right)=1+2\mathbf{b}\mathbf{i}$$

$$\mathit{z}\mathit{p}2-\mathit{}\mathit{z}\mathit{p}1=\left({\displaystyle \frac{1}{2}}+\mathit{a}+\mathit{b}\mathit{i}\right)-\left({\displaystyle \frac{1}{2}}-\mathit{a}+\mathit{b}\mathit{i}\right)=2\mathit{a}$$

$$\mathit{a},\mathit{b}\mathit{}\in \mathit{R}\mathit{}0\le \mathit{a}\le {\displaystyle \frac{1}{2}}$$

As the Figure 7. If we have zero points of $\xi \left(s\right)$ on line-1/2±a as

$$\mathit{z}\mathit{p}1={\displaystyle \frac{1}{2}}-\mathit{a}+\mathit{b}\mathit{i}\mathit{}\mathit{z}\mathit{p}2={\displaystyle \frac{1}{2}}+\mathit{a}+\mathit{b}\mathit{i}$$

And $\mathit{s}={\displaystyle \frac{1}{2}}+\mathit{t}\mathit{i}$ $\mathit{t}\in \mathit{R}$ is the first zero point on line-1/2

We can get a zero point as

$$\mathit{z}\mathit{p}0={\displaystyle \frac{1}{2}}+\mathit{b}\mathit{i}\mathit{b}<\mathit{t}\mathit{}\mathit{b},\mathit{}\mathit{t}\mathit{}\in \mathit{R}$$

It is contrary to that $\mathit{s}={\displaystyle \frac{1}{2}}+\mathit{t}\mathit{i}$ $\mathit{t}\in \mathit{R}$ is the first zero point on line-1/2

As the Figure 8. If we have zero points of $\xi \left(s\right)$ on line 1/2±a as

$$\mathit{z}\mathit{p}1={\displaystyle \frac{1}{2}}-\mathit{a}+\mathit{b}\mathit{i}\mathit{}\mathit{z}\mathit{p}2={\displaystyle \frac{1}{2}}+\mathit{a}+\mathit{b}\mathit{i}$$

And ${\mathit{s}}_{\mathit{n}}={\displaystyle \frac{1}{2}}+{\mathit{t}}_{\mathit{n}}\mathit{i}$ $\mathit{t}\in \mathit{R}$ is the No. n zero point on line-1/2

${\mathit{s}}_{\mathit{n}+1}={\displaystyle \frac{1}{2}}+{\mathit{t}}_{\mathit{n}+1}\mathit{i}$ $\mathit{t}\in \mathit{R}$ is the No. n+1 zero point on line-1/2

We can get a zero point between ${\mathit{s}}_{\mathit{n}}\mathit{a}\mathit{n}\mathit{d}{\mathit{s}}_{\mathit{n}+1}\mathit{o}\mathit{n}\mathit{l}\mathit{i}\mathit{n}\mathit{e}-1/2$ as

$$\mathit{z}\mathit{p}0={\displaystyle \frac{1}{2}}+\mathit{b}\mathit{i}{\mathit{t}}_{\mathit{n}}<\mathit{b}<{\mathit{t}}_{\mathit{n}+1}\mathit{}\mathit{b},\mathit{}\mathit{t}\mathit{}\in \mathit{R}$$

It is contrary to that ${\mathit{s}}_{\mathit{n}}\mathit{a}\mathit{n}\mathit{d}{\mathit{s}}_{\mathit{n}+1}$are the adjacent zero points on line-1/2

So on complex plane, We can have the symmetry structure about the line-1/2 with zp=1/2±a $(0\le \mathit{a}\le {\displaystyle \frac{1}{2}}\mathit{a}\in \mathit{R})$show as on Figure 9.

$$\mathit{S}={\displaystyle \frac{1}{2}}+\mathit{t}\mathit{i}\mathit{}(\mathit{t}\mathit{}\in \mathit{R})$$

$$\mathit{z}\mathit{p}={\displaystyle \frac{1}{2}}\pm \mathit{a}(0\le \mathit{a}\le {\displaystyle \frac{1}{2}}\mathit{}\mathit{a}\mathit{}\in \mathit{R})$$

$$\mathit{z}\mathit{p}1={\displaystyle \frac{1}{2}}-\mathit{a}\mathit{}\mathit{z}\mathit{p}0={\displaystyle \frac{1}{2}}\mathit{}\mathit{z}\mathit{p}2={\displaystyle \frac{1}{2}}+\mathit{a}$$

$$\mathit{z}\mathit{p}1+\mathit{}\mathit{z}\mathit{p}2=\left({\displaystyle \frac{1}{2}}-\mathit{a}\right)+\left({\displaystyle \frac{1}{2}}+\mathit{a}\right)=1$$

$$\mathit{z}\mathit{p}2-\mathit{}\mathit{z}\mathit{p}1=\left({\displaystyle \frac{1}{2}}+\mathit{a}\right)-\left({\displaystyle \frac{1}{2}}-\mathit{a}\right)=2\mathit{a}$$

This is mean that there are no zero points on line-1/2±$\mathit{a}(0\le \mathit{a}\le {\displaystyle \frac{1}{2}}\mathit{a}\in \mathit{R})$.

Hardy and Littlewood give a proof that there are infinite zero points on line-1/2 (Hardy and Littlewood. 1914)

So we give a proof that all the non-trivial Zero points of Riemann zeta-function are on the Line-1/2. This is the proof of Riemann Hypothesis.

5. The Symmetry Number Structure About Line-1/2 Including All Numbers

In fact, we have a symmetrical number structure about line-1/2 as figure.10.

$$\mathit{S}={\displaystyle \frac{1}{2}}+\mathit{t}\mathit{i}\mathit{}(\mathit{t}\mathit{}\in \mathit{R})$$

$$\mathit{z}\mathit{p}={\displaystyle \frac{1}{2}}\pm \mathit{\epsilon }\mathit{}(\mathit{\epsilon }=\mathit{a}+\mathit{b}\mathit{i}\mathit{}\mathit{a},\mathit{b}\mathit{}\in \mathit{R}\mathit{}0\le \mathit{a}\le {\displaystyle \frac{1}{2}})$$

Figure 10. symmetry structure about the line-1/2 with zp=1/2±ε.

Figure 10. symmetry structure about the line-1/2 with zp=1/2±ε.

And we can get a symmetry number structure about line-1/2 as Figure 11. We should call it Reimann dynamic space.

$$1+i^2=0$$

$$1+1/2(i+1)\left(i-1\right)=0$$

$$\mathit{S}={\displaystyle \frac{1}{2}}+\mathit{t}\mathit{i}\mathit{}(\mathit{t}\mathit{}\in \mathit{R})$$

$$\mathit{z}\mathit{p}={\displaystyle \frac{1}{2}}\pm \mathit{\epsilon }\mathit{}(\mathit{\epsilon }=\mathit{a}+\mathit{b}\mathit{i}\mathit{}\mathit{a},\mathit{b}\mathit{}\in \mathit{R}\mathit{}0\le \mathit{a}\le {\displaystyle \frac{1}{2}})$$

$${\displaystyle \frac{P}{2n}}=\left\{\begin{array}{c}{\displaystyle \frac{1}{2^{N+1}}}n=2^{N}P\\ {\displaystyle \frac{3}{4}}n=2P=3\\ 1n=1P=2\end{array}\right.$$

$N~\left(0,1,2,3,4,\dots \dots \dots .\right)$ All natural numbers

$n~\left(1,2,3,4,\dots \dots \dots .\right)$ All natural numbers excepted 0

$\mathrm{P}~\left(2,3,5,7,\dots \dots \dots .\right)$ All prime numbers

Figure 12. Reimann dynamic space with p1 p2 p3 p4.

Figure 12. Reimann dynamic space with p1 p2 p3 p4.

We can have point p1 p2 p3 p4 and

$$\mathit{p}1\in \left({\displaystyle \frac{1}{2}}-\mathit{\epsilon },{\displaystyle \frac{\mathit{p}}{2\mathit{n}}}\right)$$

$$\mathit{p}2\in \left({\displaystyle \frac{\mathit{p}}{2\mathit{n}}},{\displaystyle \frac{1}{2}}+\mathit{\epsilon }\mathit{}\right)$$

$$\mathit{p}3\in \left({\displaystyle \frac{1}{2}}-\mathit{\epsilon },{\displaystyle \frac{1}{2}}\right)$$

$$\mathit{p}4\in \left({\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}+\mathit{\epsilon }\right)$$

And we can get Figure 13.

1. $\mathit{z}\mathit{p}={\displaystyle \frac{1}{2}}+\mathit{b}\mathit{i}$ $0<\mathit{b}<\mathit{t}\mathit{b},\mathit{t}\in \mathit{R}$ (the proof of RH)

$2.({{{\displaystyle \frac{\mathit{x}}{\mathit{z}}})}^{\mathit{n}}+\left({\displaystyle \frac{\mathit{y}}{\mathit{z}}}\right)}^{\mathit{n}}=1$

$$({{\displaystyle \frac{x}{z}})}^n\leftrightarrow {\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2^n}}$$

$${\left({\displaystyle \frac{y}{z}}\right)}^n\leftrightarrow {\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2^n}}\text{(the proof of F.L.T)}$$

3. ${\displaystyle \frac{\mathrm{p}0}{2\mathrm{n}}}\leftrightarrow {\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2\mathrm{n}}}$

$${\displaystyle \frac{\mathrm{p}\mathrm{n}}{2\mathrm{n}}}\leftrightarrow {\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2\mathrm{n}}}\text{(the proof of GC/BC/TPC)}$$

$${\displaystyle \frac{\mathrm{p}}{2\mathrm{n}}}\leftrightarrow {\displaystyle \frac{1}{2}}$$

And we have

$$1/2=1/20=1/2-1/21=1/2+1/2$$

$$1+(\pm {i)}^2=0$$

$$1+1/2(i+1)\left(i-1\right)=0$$

$$\infty =1+1+1+1+\dots$$

We called it L^1/2±ε _{【0 1/2 1】} and analytic continuation to$\left[\begin{array}{cc}+\infty & -\infty \\ -\infty & +\infty \end{array}\right]$ we can get Figure 14.

So we have:

$$1+\left[\begin{array}{c}+\infty \mathrm{i}-\infty \\ 01/21\\ -\infty -\mathrm{i}+\infty \end{array}\right]{\left[\begin{array}{c}11/20\\ {\displaystyle \frac{1}{2}}-\mathsf{\epsilon }1/2{\displaystyle \frac{1}{2}}+\mathsf{\epsilon }\\ 11/20\end{array}\right]}^{-1}=0$$

$$\mathit{z}\mathit{p}={\displaystyle \frac{1}{2}}\pm \mathit{\epsilon }$$

$$\mathit{\epsilon }=\mathit{a}+\mathit{b}\mathit{i}\mathit{}(\mathit{a},\mathit{b}\mathit{}\in \mathit{R}\mathit{}0\le \mathit{a}\le {\displaystyle \frac{1}{2}})$$

$$\mathit{z}\mathit{p}1={\displaystyle \frac{1}{2}}-\mathit{\epsilon }\mathit{}\mathit{z}\mathit{p}2={\displaystyle \frac{1}{2}}+\mathit{\epsilon }$$

We have

$$\mathit{z}\mathit{p}1+\mathit{}\mathit{z}\mathit{p}2=\left({\displaystyle \frac{1}{2}}-\mathit{\epsilon }\right)+\left({\displaystyle \frac{1}{2}}+\mathit{\epsilon }\right)=1$$

$$\mathit{z}\mathit{p}2-\mathit{}\mathit{z}\mathit{p}1=\left({\displaystyle \frac{1}{2}}+\mathit{\epsilon }\right)-\left({\displaystyle \frac{1}{2}}-\mathit{\epsilon }\right)=2\mathit{\epsilon }=2(\mathit{a}+\mathit{b}\mathit{i})$$

And we have

$$n^2={\displaystyle \frac{1}{2}}·n·2n=\sum ^N{\displaystyle \frac{1}{2}}\sum ^N{\displaystyle \frac{1}{2^N}}[\left({\displaystyle \frac{1}{2}}-\mathit{\epsilon }\right)+\left({\displaystyle \frac{1}{2}}+\mathit{\epsilon }\right)]$$

$N~\left(0,1,2,3,4,\dots \dots \dots .\right)$ All natural numbers

$n~\left(1,2,3,4,\dots \dots \dots .\right)$ All natural numbers excepted 0

We can get a matrix ($n\times n$)

$$\left[\begin{array}{c}1/2\dots \dots ..{\displaystyle \frac{1}{2^n}}(1/2+\epsilon )\\ \dots \dots 1/2\dots \dots \dots ..\\ {\displaystyle \frac{1}{2^n}}(1/2-\epsilon )\dots \dots ..1/2\end{array}\right](n\times n)$$

The tr(A)=1/2*n

We have

$$0={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}1={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}2=1+1$$

$$1+i^2=01/2+{\mathit{i}}^{4\mathit{N}+1}=1/2+\mathit{i}$$

$$\infty =1+1+1+1+\dots$$

$$\mathrm{P}0\in \mathrm{P}\le 2\mathrm{n}\mathrm{pn}\in \mathrm{P}\ge 2\mathrm{n}$$

$N~\left(0,1,2,3,4,\dots \dots \dots .\right)$ all the natural numbers.

$n~\left(1,2,3,4,\dots \dots \dots .\right)$ All natural numbers excepted 0

$\mathrm{P}~\left(2,3,5,7,\dots \dots \dots .\right)$ All odd prime number

$$\mathit{S}={\displaystyle \frac{1}{2}}+\mathit{t}(\mathit{t}\in \mathit{R})$$

$$\mathit{z}\mathit{p}={\displaystyle \frac{1}{2}}\pm \mathit{\epsilon }(\mathit{\epsilon }=\mathit{a}+\mathit{b}\mathit{i}\mathit{a},\mathit{b}\in \mathit{R}0\le \mathit{a}\le {\displaystyle \frac{1}{2}})$$

And we find that

1.$1+{\mathit{e}}^{\mathit{\pi }\mathit{i}}=$0 (Euler’s Formula)

$$1+i^2=01+{\displaystyle \frac{1}{2}}\left(i+1\right)\left(i-1\right)=0(1+\mathrm{i})(1-\mathrm{i})=\sum {\displaystyle \frac{1}{2^N}}$$

$$1+{\mathit{e}}^{\mathit{\pi }\mathit{i}}=01+{\displaystyle \frac{1}{2}}({\mathit{e}}^{\mathit{i}\mathit{p}\mathit{\pi }}-{\mathit{e}}^{\mathit{i}2\mathit{N}\mathit{\pi }})=0$$

$N~\left(0,1,2,3,4,\dots \dots \dots .\right)$ all the natural numbers.

$\mathrm{p}~\left(3,5,7,\dots \dots \dots .\right)$ All odd prime number

$$2(\mathit{n}\pm 1)=\mathit{p}\mathit{n}\pm \mathit{p}0$$

2.$\mathit{p}\mathit{n}-2\mathit{n}+\mathit{p}0=2$

And

$$2\mathit{n}-\mathit{p}\mathit{n}+\mathit{p}0=2$$

It is like the Euler’s Polyhedron Formula

We can get Figure 15. This is a symmetry number structure about line-1/2 including all numbers.