想談談『數學理解』這件事，先想到 □□、○○ 『理解』是何意？那『理解』一詞的『意義』將要如何『定義』呢？？中文維基百科詞條這麼講︰

理解

理解（Understanding），又稱為領會、了解、懂得、思維作用（intellection），是指一種心理過程，與諸如人、情形或訊息之類的某種抽象的或有形的對象相關，籍此一個人能夠對其加以思考，並且運用概念對該對象加以適當的處理。

理解乃是概念表達（又稱為概念化）的界線。理解某一事物，也就是已經對該事物實現了一定程度的概念表達或者說概念化。

有關理解的例子

一個人如果能夠做到對天氣加以預測並對天氣的一些特點加以解釋等等之類的話，那麼，就說這個人理解了天氣。
如果一位精神病醫生知道某位病人的焦慮及其原因，並且能夠針對如何應對焦慮給予有益的忠告，那麼，就說這位精神病醫生理解了該病人的焦慮。
就某條命令來說，如果某個人知道這究竟是誰下達的命令，下令者的期望究竟是什麼以及該項命令是否合法或者說正當等等，那麼，就說這個人理解了這條命令。
如果一個人能夠清醒地某條消息所傳達的信息內容，那麼，就說這個人理解了其中的修辭、論據或某種語言。
如果一個人能夠利用某一數學概念解決問題，尤其是那些此前未曾見過的問題，那麼，就說這個人理解了這個數學概念。

彷彿可意會的多，能言傳的少。就像解釋『如人飲水』，然後說其『自知冷暖』一般。那麼換個語境，由不同的人解釋︰

Understanding

Understanding is a psychological process related to an abstract or physical object, such as a person, situation, or message whereby one is able to think about it and use concepts to deal adequately with that object. Understanding is a relation between the knower and an object of understanding. Understanding implies abilities and dispositions with respect to an object of knowledge sufficient to support intelligent behavior.^[1]

Understanding is often, though not always, related to learning concepts, and sometimes also the theory or theories associated with those concepts. However, a person may have a good ability to predict the behaviour of an object, animal or system — and therefore may, in some sense, understand it — without necessarily being familiar with the concepts or theories associated with that object, animal or system in their culture. They may, indeed, have developed their own distinct concepts and theories, which may be equivalent, better or worse than the recognised standard concepts and theories of their culture.

是否就能促進『理解』之『深、淺』乎︰

Shallow and deep

Someone who has a more sophisticated understanding, more predictively accurate understanding, and/or an understanding that allows them to make explanations that others commonly judge to be better, of something, is said to understand that thing “deeply”. Conversely, someone who has a more limited understanding of a thing is said to have a “shallow” understanding. However, the depth of understanding required to usefully participate in an occupation or activity may vary greatly.

For example, consider multiplication of integers. Starting from the most shallow level of understanding, we have (at least) the following possibilities:

A small child may not understand what multiplication is, but may understand that it is a type of mathematics that they will learn when they are older at school. This is “understanding of context”; being able to put an as-yet not-understood concept into some kind of context. Even understanding that a concept is not part of one’s current knowledge is, in itself, a type of understanding (see the Dunning-Kruger effect), which is about people who do not have a good understanding of what they do not know.
A slightly older child may understand that multiplication of two integers can be done, at least when the numbers are between 1 and 12, by looking up the two numbers in a times table. They may also be able to memorise and recall the relevant times table in order to answer a multiplication question such as “2 times 4 is what?”. This is a simple form of operational understanding; understanding a question well enough to be able to do the operations necessary to be able to find an answer.
A yet older child may understand that multiplication of larger numbers can be done using a different method, such as long multiplication, or using a calculator. This is a more advanced form of operational understanding because it supports answering a wider range of questions of the same type.
A teenager may understand that multiplication is repeated addition, but not understand the broader implications of this. For example, when their teacher refers to multiplying 6 by 3 as “adding 6 to itself 3 times”, they may understand that the teacher is talking about two entirely equivalent things. However, they might not understand how to apply this knowledge to implement multiplication as an algorithm on a computer using only addition and looping as basic constructs. This level of understanding is “understanding a definition” (or “understanding the definition” when a concept only has one definition).
An teenager may also understand the mathematical idea of abstracting over individual whole numbers as variables, and how to efficiently (i.e. not via trial-and-error) solve algebraic equations involving multiplication by such variables, such as $2x=6$ . This is “relational understanding”; understanding how multiplication relates to division.
An undergraduate studying mathematics may come to learn that “the integers equipped with multiplication” is merely one example of a range of mathematical structures called monoids, and that theorems about monoids apply equally well to multiplication and other types of monoids.

For the purpose of operating a cash register at McDonald’s, a person does not need a very deep understanding of the multiplication involved in calculating the total price of two Big Macs. However, for the purpose of contributing to number theory research, a person would need to have a relatively deep understanding of multiplication — along with other relevant arithmetical concepts such as division and prime numbers.

『了解』耶？若是強要朔本追源，終究會遇到『天性』、『本能』這一類的『基元概念』哩！！所以作者假設『理解』 understand ，就是『理解中』 understanding ，就是『理解』加深『持續中』 understand-ing… ？？宛如『領會』羅伯特‧里德『畫作』一樣︰

Robert Lewis Reid (1862-1939) at the Library of Congress

Robert Lewis Reid alternated between the easel & painting murals in public buildings. These at the Thomas Jefferson Building of the Library of Congress are worth note.

【智慧】

【理解】

【知識】

【哲學】

人人都有條『不言自明』之『思路』，只待自己發現走上矣！！

如斯者當能『領悟』他人『思路』之『根』夫☆

一個係數是整數的『多項式』 $f(x) = \sum \limits_{i=0}^{n} c_i x^i$ 是一個在任何閉區間 $[a, b]$ 裡『連續』而且於開區間 $[a, b]$ 中『可微分』的『函數』，如果用『均值定理』來看所對應的 $n$ 次『方程式』 $f(x) = 0$ 的『根』 $\alpha$ ── 稱之為『代數數』 ──，『劉維爾』證明了

如果『無理數』 $\alpha$ 是一個 $n$ 次『多項式』之根的『代數數』，那麼存在一個『實數』 $A > 0$ ，對於所有的『有理數』 $\frac{p}{q}, \ p, q \in \mathbb{Z}, \ \wedge \ q > 0$ 都有

$\left\vert \alpha - \frac{p}{q} \right\vert > \frac{A}{q^n}$

，現今這叫做『劉維爾定理』。

既然 $\alpha$ 是 $f(x) = 0$ 的一個『解』 $f(\alpha) = 0$ ，假設除此之外它還有 $\{ \alpha_1, \alpha_2, ..., \alpha_m \}$ 個與 $\alpha$ 值不同的『解』，考慮一個由 $\alpha$ 構造的『閉區間』 $[\alpha - 1, \alpha +1]$ ，由於 $f^{\prime}(x) = \sum \limits_{i=1}^{n} c_i \cdot i \cdot x^{i-1}$ 存在且連續，因此 $| f^{\prime}(x) |$ 存在且連續，從『極值定理』可以知道 $| f^{\prime}(x) |$ 在任何『閉區間』裡都有『極大值』，將 $| f^{\prime}(x) |$ 在 $[\alpha - 1, \alpha +1]$ 中的『最大值』記作 $M, | f^{\prime}(x) | \leq M, \ x \in [\alpha - 1, \alpha +1]$ 。讓我們選擇一個滿足 $0 < A < \min \left(1, \frac{1}{M}, \left\vert \alpha - \alpha_1 \right\vert, \left\vert \alpha - \alpha_2 \right\vert, \ldots , \left\vert \alpha-\alpha_m \right\vert \right)$ 的 $A$ ，『假使』有一個『有理數』 $\frac{p}{q}$ 違背『劉維爾定理』，將會有 $\left\vert \alpha - \frac{p}{q} \right\vert \le \frac{A}{q^n} \le A< \min\left(1, \frac{1}{M}, \left\vert \alpha - \alpha_1 \right\vert, \left\vert \alpha - \alpha_2 \right\vert, \ldots , \left\vert \alpha-\alpha_m \right\vert \right)$ ，此處 $\frac{A}{q^n} \leq A$ 是因為『不等於零』的『正整數』至少是一。由於 $A < 1$ 和 $A < | \alpha - \alpha_{i} |, i=1 \cdots m$ ，因此 $\frac{p}{q} \in [\alpha - 1, \alpha +1]$ 而且 $\frac{p}{q} \notin \{ \alpha_1, \alpha_2, ..., \alpha_m \}$ ，也就是說 $\frac{p}{q}$ 不是 $f(x)$ 的『根』，而且 $f(x)$ 在 $\alpha$ 與 $\frac{p}{q}$ 的『閉區間』內沒有『根』，按照『均值定理』一定有一個 $x_0$ 界於 $\alpha$ 與 $\frac{p}{q}$ 之間，使得 $f(\alpha)-f(\frac{p}{q}) = (\alpha - \frac{p}{q}) \cdot f^{\prime}(x_0)$ 。因為 $f(\alpha) = 0$ 與 $f(\frac{p}{q}) \neq 0$ ，所以可以將之改寫成 $\left|\alpha -\frac{p}{q}\right |= \frac{\left | f(\alpha)- f(\tfrac{p}{q})\right |}{|f^{\prime}(x_0)|} = \left | \frac{f(\tfrac{p}{q})}{f^{\prime}(x_0)} \right |$ 。

由於 $\left|f \left (\frac{p}{q} \right) \right| = \left| \sum \limits_{i=0}^n c_i p^i q^{-i} \right| = \frac{1}{q^n} \left| \sum \limits_{i=0}^n c_i p^i q^{n-i} \right | \ge \frac {1}{q^n}$ ，此處 $\left| \sum \limits_{i=0}^n c_i p^i q^{n-i} \right | \ge 1$ 是因為 $f(\frac{p}{q}) \neq 0$ 而『不等於零』的『正整數』至少是一。然而 $\left| f^{\prime}(x_0) \right| \le M$ 以及 $1/M > A$ ，因此 $\left | \alpha - \frac{p}{q} \right | = \left|\frac{f(\tfrac{p}{q})}{f'(x_0)}\right| \ge \frac{1}{Mq^n} > \frac{A}{q^n} \ge \left| \alpha - \frac{p}{q} \right|$ ，產生了 $\left | \alpha - \frac{p}{q} \right | > \left | \alpha - \frac{p}{q} \right |$ 的『矛盾』，所以『假使』有一個『有理數』 $\frac{p}{q}$ 違背『劉維爾定理』的『假設』不成立。

在此回顧一下『劉維爾數』的定義

如果一個實數 $w$ 滿足，對任何正整數 $n$ ，都存在著整數 $p, q$ ，其中 $q > 1$ 而且『定然』的會有 $0 < \left| w - \frac{p}{q} \right| < \frac{1}{q^{n}}$ ，如此我們就將此數 $w$ 叫做『劉維爾數』。

假使一個『劉維爾數』 $w$ 是一個『代數數』，那麼一定會有 $\exists n \in \mathbb Z, \ A > 0 \ \forall p, q \ q > 0, \ \left( \left\vert w - \frac{p}{q} \right\vert > \frac{A}{q^{n}} \right)$ 。但因為它也是『劉維爾數』，所以當取滿足 $\frac{1}{2^r} \le A$ 的正整數 $r$ ，並使 $m = r + n$ ，一定存在整數 $a, b$ 其中 $b > 1$ 使得