# Analysis--Natural Numbers

Posted by GwanSiu on December 27, 2017

At the begining of this year, I plan to learn mathematics analysis and it will be helpful for my future research. Thus, I decide to open a new session in my blog and write some article about that. This’s my first article about analysis and the reference book is Analysis I & II written by Terence Tao. In this article, I talk about Peano axiom, addiction and multiplication.

In fact, English and Chinese are used in this session of analysis due to my limited English ability and I need to make some concept more understandable in Chinese.

## 1. What’s Natural Numbers?

I belive most of us have learn natural number from primary school. At that time, we just know how to calculate natural number or apply some operation, such as addiction, substraction and etc. However, we didn’t know what natural number is and why addiction should be like that. During the period of secondary school, we may receive an informal definition about natural numbers.

Definition 1.1(informal): A natural number is any element of the set

which is the set of all the numbers created by starting with 0 and then counting forward indefinitely. We call $\mathbb{N}$ the set of natural numbers. This definition indeed solve the problem of what natural numbers are. However, it still make us get confused in some sense. For instance, it don’t give us a definition how to keep counting indefinitely without cycling back to 0. Also, how to perform operator such as addiction, multiplication, and exponentiation on natural numbers?

Actually, we can define complicated operations in terms of simple operations. For instance, exponentiation is nothing but repeated multiplication; multiplication is nothing but repeated addiction. What’s the addiction? Addiction is nothing but counting forward, or increment.

Thus, to define natural numbers, we will use two fundamental concept: the zero number 0, and the increment operation. Let’s start Peano axiom!

## 2. Peano’s axiom

Axiom 2.1. 0 is a natural number. Axiom 2.2. If n is a natual number, then n++ is also a natural number.

Axiom 2.1 and Aximo 2.2 规定了自然数的起始点是0，并规定了自然数增量依旧是自然数。因此，我们可以定义1，2，3都是自然数，但是这并不足够描述我们所理解的自然数。如果我们考虑这样一个数字系统仅有{0，1，2，3}(现在请允许使用集合符号，方便解释)，3的增量等于0，相当于这个数字系统形成一个环域，这样的系统也依旧满足Axiom 2.1 and Aximo 2.2,但并不是我们通俗理解的自然数系统，因此，我们需要更多公理去避免环域的情况发生。

Axiom 2.3. 0 is not the successor of any natural number, i.e., we have $n++ \neq 0$ for every natural number n.

Axiom 2.3 仅仅规定了0不能是任何自然数的后继，因此，从Axiom 2.3,我们可以回答4不等于0等一系列的问题。注意我在这里的用词仅仅，这里只是保证自然数在** ++** 后不回到0，假设4++=4这种自循环情况，依旧满足 Axiom 2.1~2.3的情况，因此我们还需要公理2.4。

Axiom 2.4. Different natual numbers must have different successors, i.e., if n,m are natual numbers and $n \neq m$ , then $n++ \neq m++$. Equivalently, if $n++ = m++$, then we must have $n = m$.

Axiom 2.4规定了不同自然数的后继必定是不同的，这保证了自然数在++过程中不会有4++=4，以及6=2的情况。(非正式)证明，假设6=2，那么5++=1++，则5=1,往后一直推导，则有4=0，这违反了 Axiom 2.3，因此，题设不成立。6不等于2。Axiom 2.1~2.4可以足够保证我们可以将不同的自然数分开(即1是1，不会等于2)，但还有若该自然数系统存在一个奇奇怪怪的数字，如pi,0.5,0.67,1.5等。举个(informal)例子半个++增量，$\mathbb{N}={0,0.5,1,1.5,2,2.5,3,3.5,…}$, 这个例子可以看到，增量是原来的半个。我们现有的 Axiom2.1~2.4 仅仅只是说自然数可以由0和增量++得到。但并未就得到这个概念予以明确的定义，因此，我们不能保证我们所定义的自然数系统中不会有一些奇奇怪怪的数字或者符号。

Axiom 2.5.(Principle of mathematical induction): Let $P(n)$ be any property pertaining to a natual number on $n$. Suppose that $P(0)$ is true, and suppose that whenever $P(n)$ is true, $P(n++)$ is also true. Then $P(n)$ is true for every natural number n.

Axiom 2.5就是我们通俗意义上说的数学归纳法，它保证了所定义的自然数系统是不会存在一些奇奇怪怪的数字或者符号。相对于Axiom,其实Axiom 2.5更多是一种Axiom scheme(公式模板)。数学归纳法通常分为两步: (1).base case成立; (2).假设case n是成立的，证明case n++是成立的。

Definition 3.1(Addition of natural numbers). Let m be a natural nummber. To add zero to m, we define 0+m := m. Now suppose inductively that we have defined how to add $n$ to $m$. Then we can add $n++$ to $m$ by defining $(n++)+m := (n+m)++$.

Lemma 3.2. For any natural number n, n+0=n.(数学归纳法) Lemma 3.3. For any natural numbers n and m, n+(m++)=(n+m)++.（数学归纳法）

Lemma 3.2-3.3中可以推出交换律，结合律以及分配率。

Proposition 3.4(Addiction is commutative). For any natural numbers n and m, n+m=m+n.

Proposition 3.5(Addiction is associative). For any natural numbers a,b and c, we have (a+b)+c=a+(b+c).

Proposition 3.6(Cancellation is law). Let a,b,c be natural numbers such that a+b=a+c. Then we have b=c.

Definition 3.7(Ordering of the natural numbers). Let $n$ and $m$ be natural numbers. We say that $n$ is greater than or equal to m, and write $n\geq m$ or $m \leq n$, iff we have $n=m+\alpha$ for some natural number $\alpha$. We say that $n$ is strictly greater than m, and write $n>m$ or $m<n$, iff $n\geq m$ and $n\neq m$.

Proposition 3.8(Basic properties of order for natural numbers). Let a,b,c be natural numbers. Then

• (Order is reflexive) $\alpha \geq \alpha$.
• (Order is transitive) If $\alpha \geq b$ and $b \geq c$, then $\alpha \geq c$.
• (Order is anti-symmetric) If $\alpha \geq b$ and $b \geq \alpha$, then a=b.
• (Addiction preserves order) $\alpha \geq b$ if and only if $\alpha +c \geq b++c$.（加法是具有保序性的）
• $a<b$ if and only if $a++\leq b$.
• $\alpha < b$ if and only if $b=a+d$ for some positive number $d$.

## 4. Multipication

Definition 4.1(Multiplication of natual number). Let $m$ be a natural number. To multiply zero to m, we define $0\times m$ := 0. Now suppose inductively that we have definded how to multiply $n$ to $m$. Then we can multiply $n++$ to $m$ by defining $(n++)\times m := (n\times m)+m$.

Lemma 4.2(Multiplication is commutative) Let n,m be natural numbers. Then $n\times m = m\times n$.

Proposition 4.3(Distributive law) For any natual numbers a,b,c, we have a(b+c)=ab+bc and (b+c)a=ab+bc.

Porposition 4.4(Multiplication is associative). For any natural numbers a,b,c, we have $(a\times b)\times c=a\times (b\times c).$

Proposition 4.5(Multiplication preserves order). If a,b are natural numbers such that $a<b$ and c is positive, then $ac<bc$.

Corollary 4.6(Cancellation law). Let a,b,c be natural numbers such that ac=bc and c is non-zero, Then $a=b$.

Definition 4.7(Euclidean algorithm). Let n be a natural number, and let q be a positive number. Then there exist natual numbers m,r sucha that $0\leq r\leq q$ and $n=mq+r$.

## 5. Exponentiation

Definition 5.1(Exponentiation for natural natural numbers). Let $m$ be a natural number. To raise $m$ to the power 0, we define $m^{0}=1$; in particular we define $0^{0}=1$. Now suppose that $m^{n}$ has been defined for some natural number $n$, then we define $m^{n++}:=m^{n}\timnes m$.

## 6. Positive Numbers

Definition 6.1(Positive natural numbers) A natural number $n$ is said to be pisitive iff it is not equal to 0.

Proposition 6.2 If $a$ is positive and $b$ is a natural number, then $a+b$ is positive.

Corollary 6.3. If $a$ and $b$ are postiive numbers such that $a+b=0$, then $a=0$ and $b=0$.

Lemma 6.4. Let $a$ be a positive number. Then there exists exactly one natural number $b$ such that $b++=a$.(Hint:存在性与唯一性都要进行证明，如何证明存在？如何证明唯一性？)

Proposition 6.5(Positive natural nummbers have no zero divisors). Let n,m be natural numbers. Then $n\times m=0$ if and only if at least one of $n,m$ is equal to zero. in particular, if $n$ and $m$ are both positive, then $nm$ is also positive.

## 7. isomorphic, axiomatic and constructive, Recursive definitions.

### 7.3 Recursive definition

Proposition(Recursive definition). Suppose for each natural number $n$, we have some function $f_{n} : \mathbb{N}\rightarrow \mathbb{N}$ from the natural numbers to the natural numbers. Let $c$ be a natural number. Then we can assign a unique natural number $a_{n}$ to each nutural num ber n, such that $a_{0}=c$ and $a_{n++}=f_{n}(a_{n})$ for each natural number n.