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\title{\textbf{Natural numbers}}
\author{Wang-Shiuan Pahngerei}
\maketitle
\date



To begin construction on the system of natural numbers, there're some essential notions required, which are: $1$ (one), natural numbers, $\cdot '$ (the next number of $\cdot$, the successor of $\cdot$), sets, and $=$ (identity). We agree that we're all familiar with properties of sets. Moreover, the notion of identity comes from logic, and satisfies: 
\begin{align}
(R)&\qquad  x=x  &&\forall x \notag\\
(S)&\qquad  x=y\Rightarrow y=x&&\forall x,y\notag\\
(T)&\qquad  \text{if } x=y \text{ and } y=z\text{,  then } x=z&&\forall x,y,z\notag
\end{align}
We do not discuss what $1$, a natural number, $\cdot '$ do really mean in detial. We only care about the properties of them from   cognitive intuition. They are, in fact, called $\mathit{axioms}$, as follow: (Variables always mean natural number in this article.)
\begin{itemize}
\item[(I)] $1$ is a natural number. 
\item[(II)] if $x=y$, then $x' = y'$.
\item[(III)] $\forall x$, $x'\ne 1$. 
\item[(IV)] If $x' =y'$ then $x=y$. 
\item[(V)] Let $\mathfrak{m}$ be a set of natural numbers and satisfies
\begin{itemize}
\item[(a)]  $1\in \mathfrak{m}$
\item[(b)]  $\forall x\in\mathfrak{m}$,  $x'\in\mathfrak{m}$.
\end{itemize}
Then $\mathfrak{m}$ contains all natural numbers ($\ast$).
\end{itemize}

We denote $\textbf{N}$ the set of all natural numbers. Hence, ($\ast$) means $\mathfrak{m} = \textbf{N}$. 

Next, we define:
\begin{align}
&2=1'\qquad 3=2'\qquad 4=3' \qquad 5=4' \qquad 6=5'\notag\\
&7=6'\qquad 8=7'\qquad 9=8' \qquad \mathbf{T}=9'\notag
\end{align}

With them, we can exhibit additional operation on natural number. The following is its standard  definition. 
\begin{itemize}
\item[(i)] $x+ 1 = x'$.
\item[(ii)] $x+y' = (x+y)'$.
\end{itemize}
According to this, it's not hard to reach some $\mathit{great}$ theorems: 
\begin{thm}
(i) $1+1=2$.$\qquad$ (ii) $2+3=5$.
\end{thm}
\begin{proof}
By definition, $1+1 = 1'$ and $2=1'$. By (S) and (T) we deduce $1+1=2$. Hence (i) is done.

By definition, $2+3 = 2+2' = (2+2)'$. Again $2+2=2+1' = (2+1)'$, and still again $2+1=2' =3$. By (II) we get $(2+1)'=3'=4$, which means 2+2=4. Again $(2+2)' = 4' =5$, which means $2+3=5$. 
\end{proof}

Through the process of the proof, we additionally get that 
\begin{align}
2+1=3, \qquad 2+2=4.\notag
\end{align}

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[note] 

We now may create an addition table for those pairs of numbers who are not too "large", for if we hope to evalute $5+7$, by similar argument as the above theorem, we might find that we have not yet possessed a notation for the result. (in fact, $5+7=9'''$, but we can not write $12$ now).

[note] 

The system of decimal expression and addition on decimals of natural numbers will be discuss far later. 



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We now proceed to show the so-called commutative law and associative law of natural numbers. 

\begin{thm}
For all $y$, $1+y = y+1$.
\end{thm}
\begin{proof}
We want to prove it by (V). Let
\begin{align}
\mathfrak{m} = \{ y : \quad 1+y = y+1\}.\notag
\end{align}
By (R), $1+1 = 1+1$, so $1\in\mathfrak{m}$. Assume that $k\in\mathfrak{m}$, and we hope to reach the case that $k'\in\mathfrak{m}$. Since $1+k=k+1$, we evaluate 
\begin{align}
1+k' = (1+k)' = (k+1)' = (k')' = k'+1.\notag
\end{align}
Hence $k'\in\mathfrak{m}$. Therefore $m=\textbf{N}$, which ends the proof.
\end{proof}

\begin{thm}
For all $x$, $y$, $x+y=y+x$. 
\end{thm}
\begin{proof}
Let 
\begin{align}
\mathfrak{m} = \{x : x+y=y+x\quad\forall y \}.\notag
\end{align}
By above theorem, we have that $1+y=y+1\quad\forall y$, so $1\in\mathfrak{m}$. Assume that $k\in\mathfrak{m}$, namely, 
\begin{align}
k+y=y+k\qquad\quad\forall y.\notag
\end{align}
Then, we hope to show that $k'+y=y+k'\quad \forall y$.
Denote
\begin{align}
\mathfrak{n} = \{y: k'+y=y+k'\}.\notag
\end{align}
Then we know that $1\in\mathfrak{n}$. Assume that $t\in\mathfrak{n}$, namely
\begin{align}
k'+t=t+k'.\notag
\end{align}
We hope to show $k'+t'=t'+k'$. We do it as follow:
\begin{align}
k'+t' &= (k'+t)'= (t+k')'=(t+k)'' \notag\\
&=(k+t)'' = (k+t')' = (t'+k)' = t'+k'.\notag
\end{align}
Hence $t'\in\mathfrak{n}$. This means $\mathfrak{n} = \textbf{N}$, i.e.
\begin{align}
\forall y\qquad k'+y=y+k'.\notag
\end{align}
We find immediately that $k'\in\mathfrak{m}$, and therefore $\mathfrak{m}=\textbf{N}$. It is proved. 
\end{proof}

\begin{thm}
$(x+y)+z = x+(y+z)\quad\forall x,y,z$.
\end{thm}
[hint] Let $\mathfrak{m} = \{z: (x+y)+z = x+(y+z)\quad\forall x \forall y \}$. 

\begin{thm}
Given $x$, $y$, exactly one of the following holds:
\begin{align}
x&=y+u \qquad \exists u\notag\\
x&=y\notag\\
y&=x+v\qquad \exists v\notag
\end{align}
\end{thm}
[hint] Let $\mathfrak{m} = \{x:$ The statement holds for all $y$ $\}$. Then for fixed $x$, set $\mathfrak{n} = \{y:$ The statement holds $\}$ . (V) gives a proof. 

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$\{$ To be continued. This article will be updated later!! $\}$

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