The natural numbers are pretty familiar. 自然数很常见，大家都很熟悉。
But if I asked you to tell me what natural numbers are 但如果我问你自然数到底是什么？
and how they work without using the notion of number 还有，如果不用数的概念自然数要怎么工作呢？
in your answer, could you do it? 用你自己的话，你能答得上来吗？
In the late 19th century, mathematicians 在19世纪晚期
were on a quest to put all of mathematics 数学家们开始了一场旨在
on firm logical footing. 为所有数学找到坚实逻辑基础的运动
Kelsey talked about this in a previous video, Kelsey在上个视频中讨论过这场运动
where she mentioned, among other things, 在其它她所提到的事情中
that the number system can be constructed 有数字系统可被分层次构建这一条。
hierarchically starting from natural numbers 即从自然数开始，
and working your way up. 如图一路向高级别构建。
But what about the natural numbers themselves-- zero, 那么自然数本身如何呢？
one, two, three, et cetera? 0,1,2,3等等这些数又怎样呢？
Are they fundamental, or can they also 他们是基础数学元素吗？
be constructed from more basic ingredients? 又或者说它们也能被更为基础的元素构建出来?
Before we can tackle that question, 在我们探究这一问题之前
we need a firmer handle on what exactly 我们得要先扎实的掌握
it means to be a natural number. 一个自然数到底是什么（或者说什么才是一个自然数）
More specifically, we need a way to characterize 更具体的说，就是我们要找出定义
the essential features of natural numbers 自然数的必要特征
without referencing directly or indirectly things 并且不能直接或者间接应用
that we already know about the natural numbers, 我们已知的自然数的数学特征。
because that would be circular. 否则的就是在无意义的绕圈子。
Now it turns out that you can capture 现在看来可以通过
the essence of numberhood in a small set of axioms, 将数字的本质浓缩到一小串数学公理来实现
analogous to Euclid's axioms in geometry, 类似几何学中的欧几里得公理一样
that will allow us to build a set N that behaves just 于是我们就可以假定一个数列N
like the natural numbers without ever explicitly mentioning 其表征就和自然数一样但不明确提及任何
numbers, or counting, or arithmetic as we do so. 数字或者计数或者算术
These axioms were first published in 1889 more or less 这套公理最初出版于1889年
in their modern form by Giuseppe Peano 以当时的“现代模式”发表，作者是Giuseppe Peano（皮亚诺）
building on and integrating earlier work 通过整合与发展 Peirce 和 Dedekind
by Peirce and Dedekind. 早年的研究成果得出的。
I'm going to show you how this is done. 我将要向你展示的这套公理的是怎么运作的
But to appreciate what a massive achievement. 但为了你们自己能获得探索的成就感。
It is, I want you to try doing it yourself first. 我建议你们先自己去试试看。
Just casually try describing numbers 就只是随意的描述自然数
without mentioning numbers or even 不用数字
inherently number-ish concepts, like one, or plus, or next. 或者数字相关的固有概念，像“一个”、“加”或者“下一个”这样的概念
You'll find it's pretty hard to do. 你会发现其实这很难做到
First, some preliminaries. 首先，一些预备知识
To make this somewhat self-contained, 为了确保接下来介绍的某种闭环
I need to assume your familiarity 我需要假设你很熟悉
with some basic logical infrastructure that 一些基本的逻辑基础结构
is not specific to numbers. 并不特定于数字
First of all, you should be familiar with some ideas 首先你要熟悉一些关于数集的一些概念
surrounding sets, namely that the members of a set 也就是：数集的的成员叫作元素（单元）
are called elements, that two sets with an identical roster 两个拥有同样的
of elements are really the same set, 元素组成的数集其实就是同一个数集
and with the idea of a subset of a set-- 以及子集的概念
in particular, the fact that every set 尤其是每个数集都是
is a subset of itself. 它自己的子集
Second, we need a concept of equals. 第二，我们要明确“相等”这一概念
We'll be very bare bones about this, 我们对这一个概念都很直截了当
requiring only that whatever equality means 只需要了解相当的定义
the relation be reflexive, symmetric, and transitive. 即自反性、对称性和传递性
And we'll also demand that the set 而后我们也会定义
N we're trying to build be closed under equality. 数集N能够在相等这一概念下构筑起来
In other words, if x is a thing in N, and x equals y, 换言之，如果X属于N，且X等于Y
then y is also a thing in N because, of course, 则Y也属于N，当然
x and y are the same entity. 因为X与Y实质是相同的。
Now that last part may seem self evident, 最后这一部分看起来似乎是自证明了
but when you're laying out mathematical logic, 但当你展示数学逻辑的时候
it's best not to leave any assumptions unstated since that 最后不要落下任何未声明的假设
makes it harder to track which parts of an argument 因为会让之后论点的论据
actually depend on which other parts. 不清楚
And finally, we'll also rely on the concept of a function, 最后我们也将依赖于函数或者说映射的概念
or a map-- i.e. 即
a rule that associates an output with one or more inputs. 一个输出值对应一个或多个输入值的运算法则
Armed with these basic tools, let's 拥有了这些基本工具后
introduce Peano's axioms incrementally 让我们逐步解析皮亚诺公理
so that you can see exactly why each one needs to be there. 这样你就可以看懂每一条的存在意义了
Now warning-- some parts of the argument are subtle, 友情提醒：这套理论中部分内容比较微妙
and you may need to watch this more than once to keep up 你可能得看不止一遍
with what's going on. 才能看的懂
Let's begin. 让我们开始吧。
Axiom one-- Zelda is in N. This is basically a declaration that 公理1--Z在N当中。
set N isn't empty. 这条只是一个基本声明，表示N不是空集
Zelda is in there. Z在它当中
Who or what is Zelda? 谁或者说什么是Z？
Don't worry about it. 别担心
It'll turn out to be the zero that you know and love. 它会变成你所熟知和热爱的0的~~
But I'm trying to use nomenclature here 这里我尝试使用了命名法
that's not too suggestive so that you 这样不会有很有暗示性
won't be tempted to bring in preexisting concepts. 就不会让你想到先存的概念
Remember, the goal is to arrive at numbers 请记住我们的目标是
without referencing numbers. 得出数字但证明过程中不使用数字
Now, to populate the rest of N, we 继续,为了充实N
can just keep including the next element. 我们可以不停的加入下一个元素
But what is next? 但什么是“下一个”呢？
The whole notion of next is loaded with conceptual baggage “下一个”的概念充斥着概念包袱
that can be hard to disentangle from your preexisting notions 使其难以与你的先存概念脱离
of counting and numbers. 包括计数和数字
So that's what the remaining axioms will do. 所以这就是接下来几条公理要做的事情
They will define next using only more 它们会定义“下一个”
basic and non-numerical ideas. 但使用的是更基本且非数字的概念
Let's see how that works starting with axiom two. 让我们来看看它们是如何运作的，从公理2开始
There exists a function S whose inputs and outputs 已知函数S，其输入值和输出值
are elements of N. In other words, 是N的元素。
if x is an N, then S of x is also in N. Now, on it's own, 换言之，如果x属于N则S（x）也属于N
this doesn't bias anything because we haven't specified 这并不能左右什么
the mechanical rule underlying S. For instance, 因为我们并没有具体界定S的算法
suppose that Zelda is in the set, 举例来说，Z是在这个数集中的
and it turns out that S of Zelda equals Zelda. 而得出S（z）=z
This would still satisfy axioms one and two, 这依然满足公理2
but it looks nothing like the natural numbers. 但却完全不像自然数
So let's fix that. 所以让我们来修正一下它
Axiom three-- Zelda is never the output of S. More 公理3--z永远不是S的输出值
precisely, for any x in N, the statement S of x 准确的来说，对于N中任何一个x而言
equals Zelda is always false. S（x）=z永远是不成立的（因为所有x都属于N）
Now, this is progress. 这就是进步
Axiom two says that S of Zelda is in N, 公理2表示S（z）属于N
and axiom three says that S of Zelda 而公理3表示S（z）
is different from Zelda itself. 与z本身不同
But notice that having S of S of Zelda turn out to be S of Zelda 但请注意S（S（z））=S（z）
would still be consistent with the axioms so far, 仍然可以依据目前几条公理成立（意味着N中除了Z和S（z）依然什么都不需要）
and that would bring our building process to a halt. 而这也会使得我们的推理进程停滞
So we need to prevent that possibility. 所以我们得要避免这种可能性
Enter axiom four. 因此引入公理4
If x and y are in N, and S of x equals S of y, then x equals y. 如果x和y都属于N且S（x）=S（y），则x=y
In other words, different inputs to S 换言之，向S输入不同值
will always give different outputs. 则始终会得到不同的输出值
Now, finally, we are guaranteed an infinite chain. 到现在我们终于得到了一个无尽的循环
Watch. 请看
Zelda is in our set by axiom one. Z因公理1属于我们创建的数集
S of Zelda is in our set axiom two, S（z）因公理2属于数集
and it's different from Zelda by axiom three. S(z)≠z因公理3成立
S of S of Zelda is in our set by axiom two, S（S（z））因公理2属于数集
and it's different from Zelda by axiom three, S（S（z））≠z因公理3成立
and different from S of Zelda by axiom four. S（S（z））≠S（z）因公理4成立
Now we keep using axioms two, three, and four over and over. 我们一直重复使用公理2、3和4
So s of s of s of Zelda will be in the set, S（S（S(z)））也会在数集中
and it's different from Zelda and from every element we've 它不等于Z也不等于其他的数集中的元素
already mentioned and so on and so on. 如此往复循环下去
This is starting to look like an infinite list of counting 这看起来就像一个无尽的计数
numbers. 就如同自然数一样
If we introduce some shorthand names 如果我们引入一些简短的名称
for these elements borrowing Arabic numerals, 赋予N中的这些元素，使用阿拉伯数字
like 0 for Zelda, and one for S of Zelda, 譬如0代表Z，1代S（z）
and two for S of S of Zelda, and so on, 2代表S（S（z）），按顺序命名下去
it will look exactly like the counting numbers. 那它就和数数一模一样了
Now, what about the function S? 那么函数S有是什么呢？
Notice that we never specified its internal mechanics. 请注意我们从未具体界定其内部算法
We just enforced some constraints 我们只是在数学上
on its behavior that apparently capture the intuitive 强制约束了其直观表现（即输入和输出）
notion of next, or successor, ergo the S. 其实也就是“下一个”或“继任者”的定义，因此有了S
And we used as ingredients only the abstract concept 我们只是使用了函数和相等
of a function and the idea of equals. 的抽象概念而已
So we're done, right? 那么一切搞定了么？
Well, not so fast. 别着急~
There's a subtle loophole in axioms one through four. 这套理论有一个微小的漏洞
Suppose that we throw two other characters into N. Let's call 假定我们将另外两个元素放到N中
them Mario and Luigi. 让我们称他们为Mario 和 Luigi
And suppose it turns out that S of Mario 并假定S(Mario)=Luigi
is Luigi and S of Luigi is Mario. 且S（luigi）=Mario
Tacking on these plumbers with this weird S behavior 将这俩管道工和奇特的S函数结合
doesn't violate any of the axioms so far, 并不违反四条公理中的任何一条
but it does leave us with a set that 但却会造成我们的数集
looks somehow bigger than the familiar natural numbers. 看起来比我们所熟知的自然数要大一点~~
So axioms two through four apparently 所以公理2 到公理4
do not quite capture the idea of next 这个过程并没有很好的捕捉到“下一个”的概念
because they would allow Mario and Luigi to be in the set, 因为它们允许了Mario 和 Luigi这俩货进入了数集~~
but detached from everybody else so to speak. 还和数集中的所有元素不相关
To prevent this, we used Peano's fifth axiom also known 为了避免这种情况的出现，我们引入皮亚诺的公理5
as the axiom of induction, which cleans up this mess 作为归纳总结的公理，它会理清一切
by fiat in a very subtle way. 以一种非常精妙的方式诠释出来
Here's axiom five. 接下来就是公理5
Suppose that T is any subset of N that 假定T是N的任意子集
satisfies two properties-- first that Zelda is in T and second 并满足两个条件--第一，Z属于T；
that if x is in T, then S of x is also in T. Then 第二，若x属于T则S(x)也属于T
by declaration, T is the set N. In other words, axiom five 这也正是宣布T就是N。换言之
declares N to be the minimal set that will satisfy 公理5宣布N应为满足公理一到四的
axioms one through four. 最小的数集
Now since the infinite chain we already 现在，因为这个无尽的循环
have fits that bill by axiom five, 满足了公理5
that set is N. We are done. 即为数集N。到此结束介绍。
And adjoining Mario and Luigi along with their weird S loop 而将Mario和Luigi和奇葩S循环
would give you something different from N. 结合得到的东西是不等于N的
So what does it mean to be a natural number? 所以什么是自然数呢？
It means membership in a set that 它是某个数集中的一员
is endowed with the notion of successor, 该数集被赋予了“继任者”“下一个”的概念
or next, and one of whose elements 且其中任意元素
has the privileged status of not being anybody else's successor. 满足不成为其它数字“继任者”（例如按顺序数数2在1之后，而不在3或4或任何其它数字的后面）
Now, the miracle of Peano's axioms, in my opinion, 我认为皮亚诺公理
is the revelation that next is not a fundamental concept, 揭示了“下一个”的概念并非基础逻辑概念
that it can be reduced to more basic logical ideas. 它是可以被分解成多种基本逻辑概念的
And to me, the demonstration that this can even 而对我而言这一演示
be done at all is one of the most beautiful arguments 能完成简直是数学史上
in all of mathematics. 最美丽的论证
Now, for some loose ends. 接下来，我们补上一些以上没讲的东西
First, what about arithmetic? 第一，S算法是什么呢？
It turns out that you can define both addition 最终结果显示它可以被界定成
and multiplication solely in terms of the successor function 加法和乘法的组合来实现“继任者”函数S
S. Each of those operations, in fact, can be viewed 每种算术都可以被看成是
as a function that takes two inputs from N 一个函数，即从N取两个输入值
and yields one output in N. And they can be defined recursively 则输出一个也属于N的输出值。
by the following rules. 而他们可以循环式的被以下法则界定
Now, I don't have time to do it here, 但现在我没时间去演示它们了
but it is a fun challenge to play around with these rules 但对于你来说会是一项有趣的挑战
in order to derive simple facts like one plus two equals three, 你可以自己尝试下这些运算法则来了解1+2为什么等于3
or that addition and multiplication are 又或者加法和乘法
commutative and associative, or that they 是交互相关的
obey a distributive rule. 或者它们遵守的分配法则
Super fun. 炒鸡好玩！
Second loose end. 第二个，未了结的问题
We arrived at a notion of successorship, 我们得到了一个“继任者”的概念
and we built a set that contains entities related to one another 并且我们建立了一个数集其中元素彼此相关
by iterated succession. 通过一个循环往复的规律
But we never specified what those entities actually 但我们从未具体界定这些元素是什么
were with Zelda, or zero, or any other members of N 也没界定Z或者0或者N的任何其它成员
actually are. 实际上是什么
We just asserted that here's a thing 我们只是插入一个东西
and here's some other things, and they're all in this set N. 而这又是另一个东西，它们都属于数集N
But we're leaving open the question, 但我们却留下了一个开放性的问题：
what are those things? 这些东西到底是什么？
Are they irreducible fundamental ingredients 它们是数学金字塔上不可再被拆解的
in the mathematical pyramid? 最基本的构成元素吗？
Or can even these entities, the natural numbers, 亦或者说这些元素、自然数
be built out of simpler entities in the same way 是由更加简单基本的元素构成的
that the integers get built from the naturals? 就像上述自然数被构成一样呢？
We'll answer that next time, and I'll see you then. 下次视频我们将会回答这一问题，我们到时见~
Some cliffhanger, huh? 吊吊胃口哈？~~
Don't worry. 别担心
You're not going to have to wait a full week for the resolution. 你不用等一整周再得到这一问题的解了
Our next episode about the natural numbers 我们的下一期关于自然数的视频
will be coming out in just a few days, 将在几天后放出
so be on the lookout for it. 所以敬请期待吧~~
And at the tail end of that episode, 在这期视频的尾巴上
I'll be responding to your comments 我将回复大家对
from the Telling Time on a Torus episode, Telling Time on a Torus那期视频的评论
and I'll be announcing the winners of the torus clock 并且宣布上期挑战的获奖名单
challenge, so tune in. 所以请耐心等待
But in the meantime, I did want to give you 同时我非常想给大家
a brief update on the status of the metallic ratio 一个有关本期我发起的
challenge, which I posed in this episode metallic ratio挑战的更新
that you should go watch if you don't 如果你不知道我在说什么
know what I'm talking about. 你真该去好好看看
Here's the update. 以下就是更新了
First of all, a lot of you have submitted code or results 首先，很多人分享了编码或者
of code that shows numerical checks 编码的结果
for very high sided polygons and very high end metallic ratios 显示了高阶多边形和高阶金属率
to see if the diagonal to side ratio on any polygon 来看斜边和直边的斜率是否与任何
ever comes out to be any of the metallic ratios. 金属率相同
I think some of you have checked up 我估计你们中的部分人
to 100 million sided polygons and very high end metallic 可能算到了百万边的多变形和非常高的金属率
ratios with no affirmative results. 但却没有得到肯定的答案
That strongly suggests that there 这强烈显示出
is no metallic ratio above the silver ratio that ever 没有任何金属率是可以超过银率的
occurs in any polygon at all. 任何多边形都不可能
But that's not the same thing as having a proof that either 但是证明这种情况是从来不可能发生
this never happens or finding that one that elusive metallic 和找到那个非常特殊隐秘、满足条件的金属率
ratio where it does. 并不能混为一谈
So, so far we're not there. 目前我们还没达到
Now, some of you have submitted ostensible proofs 某些人分享了以些表面证据
of why no metallic ratio above the silver ratio 有关于金属率无法高于银率
will ever be the diagonal to side 成为斜边与直边的斜率
ratio in any regular polygon. 在任何多边形当中
I'm still going through all these responses individually. 我仍然在看这些回复
I'm pretty backlogged over here. 积压了很多没看的回复
But once I've managed to check them all, 我将它们全部验证一遍的
before I'd name names and give away any t-shirts or something, 在我公布获奖名单送出T恤之前
we want to make sure these things are correct. 我们要确定这些答案是正确的
I'm going to look them all over. 我会仔细都看一遍的
I'm going to have Tai-Danae look them over. 我也会让Tai-Danae一起帮着看的
And once we know what's going on, you will know. 我们有任何发现你们都会知道的
We'll announce that in the comments 我们将会在评论中公布
of some future Infinite Series episode. 一些无限系列的视频
Stay tune. 敬请期待！