我建议向数学背景有限的人教授无穷层次的第一步是“为什么数学家说偶数集和整数集大小相同？” 这就引入了“如果你可以将集合 A 的每个成员与集合 B 的一个成员关联起来，数学家就说这些集合具有相同的大小。” 接下来表明，使用对角线方法，每个分数（每个有理数）都可以与恰好一个计数数相关联。 一旦他们对此感到满意，我就提出 π，每个人都知道它的十进制表达式中有无数个不重复的数字，这意味着它不能表示为分数，所以它会被留下来，这意味着无理数集合大于可数集合。 一些聪明人会反对说，如果你以 π 为基数，即 1π，那么 π 的位数是有限的，但你可以用“好吧，聪明的，写下一周中的天数，以 π 为基数。”
My recommended first step for teaching levels-of-infinity to people of limited mathematical background is "Why do mathematicians say that the set of even numbers and the set of whole numbers are the same size?" This introduces "if you can associate every member of set A with exactly one member of set B, mathematicians say the sets have the same size." Next comes showing that every fraction (every rational number) can be associated with exactly one counting number, using the diagonal method. Once they're satisfied with that, I bring up π, which everyone knows has an infinite number of non-repeating digits in its decimal expression, which means it cannot be expressed as a fraction, so it will be left over, and that means that the set of irrational numbers is larger than the set of counting numbers. Some wiseguys will object that π has a finite number of digits if you're working in base π, namely 1π, but you can come back at them with "okay, brainiac, write down the number of days in a week in base π."
Edit: OK, I've been writing code professionally for 13 years and I wouldn't call levels of infinity relevant to anything I've ever worked on.
And I guess I would draw a different conclusion from your theory. How is "we can only solve an infinitesimally small fraction of all problems" the limit of our craft?
Sounds to me like there are an infinite (countable or uncountable doesn't seem to make a difference) number of problems. Therefore our craft is unlimited -- we will never run out of problems to solve.
I think your explanation is the simplest, as that is what I learned. It's almost as if real numbers have multiple dimensions of infinity. It is infinite in one direction, but also in another.
Diagonalization is a very cool experiment, but I can see how it may go over beginners heads. It does make sense though, if it is demonstrated in a very deliberate way, going very slowly. Just throwing up numbers quick can be hard to follow I imagine.
I think the principle of Cardinality of the Continuum is also helpful, although perhaps can be simplified to a beginner level. Showing that there is more beyond simple real vs. integers can potentially help something to 'click'.
There are several tens of thousands of words in the English language. You can count the number of words in a book or the number of books in the universe. You cannot count the number of books that will ever be
I personally think of the countability/uncountability dichotomy as being very closely related to Zeno's paradox of the arrow.
The set of all natural numbers is countable, there is a specific method of generating the "next" integer, and it will get you a step forward. Countable sets are forward-moving in that sense. It's almost as if it has a velocity, it keeps moving forward.
The set of all real numbers is uncountable, like zeno's arrow.
If you have to move between the origin (0) and the destination (1 == 2-0), you must first go through the midpoint (1/2 == 2-1).
Now your destination is 1/2; If you must then go between the origin (0) and the (1/2), you must go through the midpoint (1/4 == 2-2)
So on and so forth, so to get between 0 and 1, you must first get between something inbetween, which you must first get between something inbetween. There is no finite method of calculating the "next" step, so the velocity (in contrast to the velocity of natural numbers) doesn't really exist, your next step is not going to take you anywhere.
Edit:
I realize now that this probably has to do with the total ordering and mapping of the set of natural numbers to any countable sets. If you can't totally order the items in a set, or you can't create a method to determine what the next item is in a set, chances are it's uncountable.
G) Therefore, if a language is infinite, it has an uncountably infinite number of subsets. Each of these represents a problem.
Citation needed. You can't merely assume that any (possibly infinite) set of Turing machines necessarily represents a distinct 'problem'. At the very least, you have to (separately) formalize the definition of 'problem' as much as Turing machines have been formalized.
Programmers (or at least, myself) don't often have to worry much about infinity in this way. When you place a finite box over any portion of the machine-representable real number line, you get a finite quantity of real numbers. =)
我们不计算不可计算的东西。 我们产生一个近似结果（就像我们计算和使用 pi 的近似值，另一个无法计算的数字）。 随着更多信息的出现，我们不断更新结果。这就是视错觉的全部内容。 当你看“一个花瓶，还是两张脸？”的图片时 你的视觉系统说“这是一个花瓶。不。等等。这是两张脸。不。等等。这是一个花瓶。” 你会看到它在两种解释之间来回切换。
仅仅因为某些事情不可计算并没有理由不去做。
Here's an example of a computable problem: at the start of a chess game, is it possible for white to force a win?
The number of possible moves and counter-moves is finite. All we have to do is build the trees and prune them. We haven't done this yet only because with current technology it would take billions of years.
Here is an example of a problem that is not computable: Given a two-dimensional view of a scene, construct a full three-dimensional model of the scene.
We do this all the time. (Make a room with a peephole in the door. Have someone furnish it. Look through the hole and describe everything you see.)
We do not compute the incomputable. We produce an approximate result (just like we compute and use an approximate value of pi, another incomputable number). We keep updating the result as more information comes in. That's what optical illusions are all about. When you look at the picture of "a vase, or is it two faces?" your visual system says "It's a vase. No. Wait. It's two faces. No. Wait. It's a vase." You see it switching back and forth between the two interpretations.
Just because something is not computable is no reason not to do it.
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我建议向数学背景有限的人教授无穷层次的第一步是“为什么数学家说偶数集和整数集大小相同？” 这就引入了“如果你可以将集合 A 的每个成员与集合 B 的一个成员关联起来，数学家就说这些集合具有相同的大小。” 接下来表明，使用对角线方法，每个分数（每个有理数）都可以与恰好一个计数数相关联。 一旦他们对此感到满意，我就提出 π，每个人都知道它的十进制表达式中有无数个不重复的数字，这意味着它不能表示为分数，所以它会被留下来，这意味着无理数集合大于可数集合。 一些聪明人会反对说，如果你以 π 为基数，即 1π，那么 π 的位数是有限的，但你可以用“好吧，聪明的，写下一周中的天数，以 π 为基数。”
My recommended first step for teaching levels-of-infinity to people of limited mathematical background is "Why do mathematicians say that the set of even numbers and the set of whole numbers are the same size?" This introduces "if you can associate every member of set A with exactly one member of set B, mathematicians say the sets have the same size." Next comes showing that every fraction (every rational number) can be associated with exactly one counting number, using the diagonal method. Once they're satisfied with that, I bring up π, which everyone knows has an infinite number of non-repeating digits in its decimal expression, which means it cannot be expressed as a fraction, so it will be left over, and that means that the set of irrational numbers is larger than the set of counting numbers. Some wiseguys will object that π has a finite number of digits if you're working in base π, namely 1π, but you can come back at them with "okay, brainiac, write down the number of days in a week in base π."
“非常相关”的部分在哪里？
编辑：好吧，我专业编写代码已经有 13 年了，我不会将我曾经从事过的任何事情称为无限级别。
我想我会从你的理论中得出不同的结论。 “我们只能解决所有问题中的一小部分”怎么会成为我们技术的极限呢？
在我看来，问题的数量是无限的（可数或不可数似乎没有什么区别）。 因此，我们的技术是无限的——我们永远不会用完要解决的问题。
Where's the "very relevant" part?
Edit: OK, I've been writing code professionally for 13 years and I wouldn't call levels of infinity relevant to anything I've ever worked on.
And I guess I would draw a different conclusion from your theory. How is "we can only solve an infinitesimally small fraction of all problems" the limit of our craft?
Sounds to me like there are an infinite (countable or uncountable doesn't seem to make a difference) number of problems. Therefore our craft is unlimited -- we will never run out of problems to solve.
我认为你的解释是最简单的，因为这就是我所学到的。 几乎就像实数具有无穷多个维度一样。 它在一个方向上是无限的，但在另一个方向上也是无限的。
对角化是一个非常酷的实验，但我可以想象它会如何超越初学者的头脑。 不过，如果以一种非常深思熟虑的方式、非常缓慢地进行展示，它确实是有意义的。 我想，仅仅快速给出数字可能很难理解。
我认为连续体基数的原则也很有帮助，尽管也许可以简化为初学者级。 表明除了简单的实数与整数之外还有更多的东西可以潜在地帮助某些东西“点击”。
I think your explanation is the simplest, as that is what I learned. It's almost as if real numbers have multiple dimensions of infinity. It is infinite in one direction, but also in another.
Diagonalization is a very cool experiment, but I can see how it may go over beginners heads. It does make sense though, if it is demonstrated in a very deliberate way, going very slowly. Just throwing up numbers quick can be hard to follow I imagine.
I think the principle of Cardinality of the Continuum is also helpful, although perhaps can be simplified to a beginner level. Showing that there is more beyond simple real vs. integers can potentially help something to 'click'.
英语中有数万个单词。 您可以计算一本书的字数或宇宙中的书籍数量。 你无法计算将会有多少本书
There are several tens of thousands of words in the English language. You can count the number of words in a book or the number of books in the universe. You cannot count the number of books that will ever be
请原谅下面写得不好的比喻。
我个人认为可数/不可数二分法与芝诺的箭头悖论密切相关。
所有自然数的集合都是可数的，有一种特定的方法来生成“下一个”整数，它会让你向前迈出一步。 从这个意义上说，可数集是向前发展的。 它几乎就像有一个速度，它不断向前移动。
所有实数的集合是不可数的，就像芝诺之箭。
如果您必须在起点 (0) 和目的地 (1 == 2-0) 之间移动，则必须首先经过中点 (1/2 == 2-1< /sup>）。
现在你的目的地是1/2； 如果必须在原点 (0) 和 (1/2) 之间移动，则必须经过中点 (1/4 == 2-2)
依此类推，所以要获得 0 和 1 之间的值，您必须首先获得介于两者之间的事物，而您必须首先获得介于两者之间的事物。 没有计算“下一步”的有限方法，因此速度（与自然数的速度相反）并不真正存在，您的下一步不会带您去任何地方。
编辑：
我现在意识到这可能与自然数集到任何可数集的总排序和映射有关。 如果您无法完全对集合中的项目进行排序，或者无法创建方法来确定集合中的下一个项目是什么，那么它很可能是不可数的。
Forgive the poorly written metaphors below.
I personally think of the countability/uncountability dichotomy as being very closely related to Zeno's paradox of the arrow.
The set of all natural numbers is countable, there is a specific method of generating the "next" integer, and it will get you a step forward. Countable sets are forward-moving in that sense. It's almost as if it has a velocity, it keeps moving forward.
The set of all real numbers is uncountable, like zeno's arrow.
If you have to move between the origin (0) and the destination (1 == 2-0), you must first go through the midpoint (1/2 == 2-1).
Now your destination is 1/2; If you must then go between the origin (0) and the (1/2), you must go through the midpoint (1/4 == 2-2)
So on and so forth, so to get between 0 and 1, you must first get between something inbetween, which you must first get between something inbetween. There is no finite method of calculating the "next" step, so the velocity (in contrast to the velocity of natural numbers) doesn't really exist, your next step is not going to take you anywhere.
Edit:
I realize now that this probably has to do with the total ordering and mapping of the set of natural numbers to any countable sets. If you can't totally order the items in a set, or you can't create a method to determine what the next item is in a set, chances are it's uncountable.
需要引用。 您不能仅仅假设任何（可能是无限的）图灵机集必然代表一个独特的“问题”。 至少，你必须（单独）形式化“问题”的定义，就像图灵机已经形式化一样。
Citation needed. You can't merely assume that any (possibly infinite) set of Turing machines necessarily represents a distinct 'problem'. At the very least, you have to (separately) formalize the definition of 'problem' as much as Turing machines have been formalized.
通过这种方式，程序员（或者至少我自己）通常不必太担心无穷大。 当您将有限框放置在机器可表示实数轴的任何部分时，你会得到有限数量的实数。 =)
例如，双精度变量具有有限数量的可能值：2^64 。
Programmers (or at least, myself) don't often have to worry much about infinity in this way. When you place a finite box over any portion of the machine-representable real number line, you get a finite quantity of real numbers. =)
For example, a double precision variable has a finite number of possible values: 2^64.
这是一个可计算问题的例子：在国际象棋比赛开始时，白棋有可能强行获胜吗？
可能的移动和反移动的数量是有限的。 我们所要做的就是建造树木并修剪它们。 我们还没有做到这一点，只是因为以目前的技术来说，这需要数十亿年的时间。
下面是一个不可计算问题的示例：给定场景的二维视图，构建场景的完整三维模型。
我们一直这样做。 （建造一个房间，门上有一个窥视孔。请有人提供它。通过这个洞看，并描述你看到的一切。）
我们不计算不可计算的东西。 我们产生一个近似结果（就像我们计算和使用 pi 的近似值，另一个无法计算的数字）。 随着更多信息的出现，我们不断更新结果。这就是视错觉的全部内容。 当你看“一个花瓶，还是两张脸？”的图片时 你的视觉系统说“这是一个花瓶。不。等等。这是两张脸。不。等等。这是一个花瓶。” 你会看到它在两种解释之间来回切换。
仅仅因为某些事情不可计算并没有理由不去做。
Here's an example of a computable problem: at the start of a chess game, is it possible for white to force a win?
The number of possible moves and counter-moves is finite. All we have to do is build the trees and prune them. We haven't done this yet only because with current technology it would take billions of years.
Here is an example of a problem that is not computable: Given a two-dimensional view of a scene, construct a full three-dimensional model of the scene.
We do this all the time. (Make a room with a peephole in the door. Have someone furnish it. Look through the hole and describe everything you see.)
We do not compute the incomputable. We produce an approximate result (just like we compute and use an approximate value of pi, another incomputable number). We keep updating the result as more information comes in. That's what optical illusions are all about. When you look at the picture of "a vase, or is it two faces?" your visual system says "It's a vase. No. Wait. It's two faces. No. Wait. It's a vase." You see it switching back and forth between the two interpretations.
Just because something is not computable is no reason not to do it.
您必须是网页设计师。
You must be web designer.