School Mathematics Newsletter（SMN）

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SMN

ISSUE 23

學校數學通訊

第二十三期

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School Mathematics Newsletter  Issue No. 23

版權

，不得用任何方式抄襲、節錄或翻印作商業用途，亦不得以任何方式透過互聯網發放。

ISBN 978-988-8370-89-4

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School Mathematics Newsletter（SMN）

Foreword

The School Mathematics Newsletter（SMN）is for mathematics teachers. SMN aims at serving as a channel of communication for mathematics education in Hong Kong. This issue includes various articles written by academics and teachers, related to the current hot topics in education, e.g. STEM education, e- learning and the practical application of artificial intelligence, etc. Other articles involve different areas, including suggestions of effective strategies in learning and teaching of mathematics on specific topics; strategies to facilitate the teaching of Compulsory Part with Module One and story about the number e. I hope all the readers can get some fascinating insights in mathematics education.

We would like to take this opportunity to thank all the authors in this issue. Without your support, it would not be possible to publish SMN issue 23.

SMN provides an open forum for mathematics teachers and professionals to express their views learning and teaching in mathematics. We welcome contributions in the form of articles on all aspects of mathematics education. Please send all correspondence to:

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The Editor, School Mathematics Newsletter, Mathematics Education Section

Curriculum Development Institute

Room 403, Kowloon Government Offices 405 Nathan Road

Yau Ma Tei, Kowloon

email: schmathsnewsletter@gmail.com

We extend our thanks to all who have contributed to this issue.

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10.

由零開始的電子學習

孔令仁 ... 91

11. 電子教學於數學科的推行與反思

林澔基老師、關子雋老師 ... 95 12. When Probability meets STEM

WONG See Yan ... 102

13. 小一 STEM 教育初探

聖公會聖十架小學：陳寶儀主任、潘慧妍老師 ... 114 14. Exploration and Development of Effective Strategies for

Promoting and Implementing STEM Education in Secondary Mathematics

CHENG Po Chun, Kitty ... 122

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1. What ‘Teachers’ Can Do to Foster STEM Awareness through the Learning of Mathematics

LAW Huk Yuen

Department of Curriculum and Instruction The Chinese University of Hong Kong

With the ever-growing pace of technological advancement, STEM (Science + Technology + Engineering + Mathematics) education comes into the focus of attention in our revised elementary (both primary and secondary) Mathematics curriculum. From the perspective of curriculum design, the STEM initiative invites both constraints and challenges (see Herschbach, 2011) as we attempted to incorporate STEM education into our existing curriculum framework. There is as we can see a great variety of how STEM is and would be highlighting (in terms of educational resources) in schools such as through General Studies in the Primary and through Technology Education in the Secondary. And yet, from the epistemological point of view, mathematics as a long- established discipline remains as ever the highly significant core subject in the elementary education. Ontologically speaking, mathematics itself has a unique role to play in our undertaking of meta-ontological theorizing of being through which we may understand better of what is existing in the world we are living in. Thus, I argue that mathematics itself is both the key with which we open and the keyhole through which we vision the world of STEM (see Figure 1 as I used in my writing

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for SMN, Issue 21). The vision of seeing the world of STEM through the keyhole of mathematics requires teachers to have dedicated mission of reinventing their own teaching (Law, 2013) through researching their practices.

Figure 1. Making sense of STEM in Maths (HY Law, School

Mathematics Newsletter, 2017, Issue 21, p. 8)

M: Awareness of using mathematics as a language to interpret life experience E: Awareness of designing model for solving problems

T: Awareness of making tools for solving tasks S: Awareness of making inquiry for natural phenomena

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The three-teacher model

To foster students’ STEM awareness through the learning of mathematics, we need to have three ‘teachers’ – the learner-

as-teacher (the First Teacher), the school teacher (the Second

Teacher), and the environment-as-teacher (the Third Teacher).

Loris Malaguzzi (1994) argues strongly that “We need to define the role of the adult, not as a transmitter but as a creator of relationships — relationships not only between people but also between things, between thoughts, with the environment.”

With the adoption of such a three-teacher model in the STEM- in-Maths (SIM) classroom (see Figure 2), we need to reinvent the notions of curriculum and assessment – curriculum as

learning and assessment as learning. Such kind of reinvention

requires school teachers to see education in general and STEM education in particular as relationship – relating teaching to learning as self-inquiring process (by asking “what do I learn and why I want to learn it?”, “how do I know that I have learned?”) and in turn to the environment itself as ontological experiment with which we create space for learning.

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Figure 2. The three-teacher model for fostering STEM-in- Maths awareness

The First Teacher – Student as Agent

All human beings are born into the world as a learner to be. As natural born learners (see Beard, 2018), we are the first teacher who should know best to teach ourselves the way we learn. It is the human rather than the technology that teaches us through social interactions. With exploratory actions, we as self-

Environment as affordance

(The Third Teacher)

Teacher as agent

(The Second Teacher)

Task in Context

Student as agent

(The First Teacher)

Rich

Task

Affordance for engagement in task activity

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directed learners develop our agency for monitoring and

assessing the task as encountered in the environment. Tversky (2019, p.288) argues that “actions in space create abstractions”.

Yet, making sense of abstractions that we create requires us to engage in voicing through the dialogic space whilst doing the action-oriented task. Hand-on activity is the natural way of developing our spatial thinking (abstract thought of the first kind) that underlies our ability of making sense of STEM world we live in. Engaging in the hand-on activity will develop the kind of emotion or feeling that urges us to assess what kind of learning experience that we have gone through. Such kind of assessment is of vital importance for building up the motivation behind not just the learning of mathematics itself but the fostering of the disposition of self-directed learning (Guglielmino, 2013) for STEM awareness.

The Second Teacher – Teacher as Agent

Learning begins to take place by the moment when the teacher opens the action possibility by creating the task in the classroom and invites the students to engage in it. The teacher as agent for making the difference wants to foster students’

response-ability (capability to respond) by creating dialogic space through which she would see the chance of making a collaborative effort at making sense of the classroom discourse.

The teacher is well aware that making mathematics meaningful to the students can be possible by having a quality classroom interaction upon the task as engaged rather than by a sheer

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telling or saying of what to do with the task itself. Knowing how to respond the students’ responses enable the teacher to develop the milieu in the classroom for the adoption of

assessment as learning (Earl, 2003) as she knows well that the

kind of feedback to the learners is aimed to develop the thinking about thinking of what kind of learning experience that they would have. As a guide for teachers to design the task for the students, we may make sense of ‘rich task’ as an equation such as ‘Rich Task = Good Problems + Quality Learning Discourse’. In other words, teachers need to adopt various teaching strategies such as questioning and group work to bring in the meaning for the students’ learning through quality mathematical communication of the problem task as designed. In the making of the problem task, the teacher needs to have a dialogue with the environment as the third teacher who tells what is going on in our world.

The Third Teacher – Environment as Affordance

Reggio Emilia, the city in southern Italy, is where Loris Malaguzzi’s idea of the environment as the third teacher comes about. The Reggio Emilia philosophy values children as central to their own learning via negotiating what to learn with their surroundings (see Strong-Wilson & Ellis, 2007). If we as teachers are to adopt such a philosophy, we would see all kinds of action possibility of designing the task for fostering schoolchildren’s STEM awareness. As we see the emergence of what the students want to learn, we come to understand what

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to-be-designed task constitutes the notion of ‘curriculum as learning’. By the very moment we capture the image of our students as a curious learner with inquiring mind, we realise that it is the beginning of teaching something that fits what they are interested to learn (see Malaguzzi, 1994). In other words, environment affords learning if we can design the classroom task by adopting the learners’ perspectives of how they make sense of the world they are living in – task in context. I would argue that the task can be ‘rich’ only when the classroom teacher can articulate three kinds of affordance – ‘affordance for task design’, ‘affordance for engagement in task activity’, and ‘affordance for dialogue’ (see Figure 2). The concept of

‘affordance’ that comes from James Gibson (1979) empowers us with the awareness of developing possible actions for undertaking our STEM practices. In the STEM context, I would rather adopt the ideas of ‘semiotic affordance’ as “responses to a conceivable practical action made possible by habits”

(Morgagni, 2012) with which hopefully the teacher can provide her students the kind of semiotic mediation for developing their STEM awareness through the learning of mathematics.

Coda

Fostering STEM awareness through learning mathematics as confined in the classroom context is indeed a challenging mission as it involves a complex interplay among the three entities -- agent, task and environment. And yet, it is worth pursuing if we as teachers do not just want to make sense of

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STEM education but to make sense of learning mathematics itself for our schoolchildren. With the design of rich task, it is hoped that the hand-on activity as a way of undertaking the enactive explorations can help learners to make meaning out of what they would have learned in mathematics classrooms.

References

[1] Beard, A. (2018). Natural born learners: Our incredible

capacity to learn and how we can harness it. London:

Weidenfeld & Nicolson.

[2] Earl, L.M. (2003). Assessment as learning using classroom

assessment to maximise student learning. Thousand Oaks, CA:

Corwin Press.

[3] Herschbach, D.R. (2011). The STEM initiative: Constraints and challenges. Journal of STEM Teacher Education, 48(1), 96-122.

[4] Gibson, J.J. (1979). The ecological approach to visual

perception. Boston: Hougton Mifflin.

[5] Guglielmino, L.M. (2013). The case for promoting self- directed learning in formal educational institutions. SA-eDUC

Journal, 10(2), 1-18.

[6] Law, H.Y. (2013). Reinventing teaching in mathematics

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classrooms: Lesson study after a pragmatic perspective.

International Journal for Lesson and Learning Studies, 2(2),

101-114.

[7] Law, H.Y. (2017). STEM education: The challenge of using mathematics as the initiation for promoting STEM education (in Chinese). School Mathematics Newsletter (Education Bureau, Hong Kong), Issue 21, 6-11.

[8] Malaguzzi, L. (1994). Your image of the child: Where teaching begins. Early Childhood Educational Exchange, No.

96, 52-61.

[9] Morgagni, S. (2012). Affordances as possible actions:

Elements for a semiotic approach. Proceedings of the 10^th

World Congress of the International Association for Semiotic Studies. Universidade da Coruña (Spain), 867-878.

[10] Strong-Wilson, T., & Ellis, J. (2007). Children and place:

Reggio Emilia’s environment as third teacher. Theory into

Practice, 46(1), 40-47.

[11] Tversky, B. (2019). Mind in motion: How action shapes

thought. New York: Basic Books.

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2. 數學科差異化教學的理念和設計

張僑平

香港教育大學數學與資訊科技學系一、引言

香港自1970 年代推行強迫教育之後，學校教育開始普及，

更多不同背景的學生有機會入學。1990 年代推行融合教育，課室裡學生的學習差異加大。這種差異主要表現在學生的不同性向，不同學習能力、學習興趣和學習風格。當然不同的學校文化和課堂文化也是造成學習差異的原因之一。其實不單是香港，在世界範圍來看，如何在學校有效地照顧學生的學習差異也都是一個難題。

跟處理學習差異類似的說法大致有差異化教學

(differentiated instruction)、分層教學（tiered instruction）以及適應性教學(adaptive teaching/instruction)。雖然說法不同

，他們的內涵其實也都相近。在本文中我們主要使用差異化教學這一提法。目前在學校中的一些常見作法有：抽離式學習、加速學習、按學生能力分組甚至分班，或者借助校外資源進行補救或拔尖的教學。這些方法均或多或少受到特殊教育中的三層級學習支援模式（Response to Intervention Model）（Bender & Crane, 2010; Pierce & Adams, 2004）或資優教育中增潤三分層模式（Enrichment Triad Model）（Renzulli, 1977）的影響。增潤分級支援模式也為目前香港教育局推行資優教育所採用。如果說差異化教學在學校層面主要是上述的分層結構，那麼在課堂層面則主要是教學內容和材料的分層策略，如根據問題的難易程度

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劃分淺、中、難或者挑戰題等層級。無庸置疑，最有效和最理想的個別差異處理方式，自然是面對面的個別輔導，

或是針對全班不同學生設計出不同教學方式，或是給8 組

學生發展出8 種不同的學習路徑。但這些做法十分消耗資

源，亦難以實施。如何在學校的同一個常規班級（學生能力是混合的）中進行差異化教學，要更為貼近教師的日常。

這也是值得我們去深入探討的。

二、差異化教學

差異化教學的專家Carol Ann Tomlinson（1999）指出，差異化的教學尤其需要教師能從課程的三個要素即學習內容

（content）、學習過程（process）及學習成果（product）入手，根據學習個體在學習特徵（learning profile）、學習準備度（readiness）以及學習興趣（interest）等層面的不同，透過多元的教學設計，達到學生的學習參與最大化的目的^[2]。簡單來說，跟傳統意義上教師主導或學科內容為本的教學相比，差異化的教學設計更注重不同能力、興趣和風格的學生在課堂中的參與和投入，教師須要儘可能給更多學生提供學習的機會。學習的內容、過程和結果構成了學生在課堂上的完整學習體驗。這些體驗包括作為學習內容的知識和技能，作為過程的思維方式和問題解決的方法，以及通過口頭表達、動手操作、視覺觀察、書寫作答等表現出來的學習結果。

現有的差異化教學理論中，已經總結了不少處理學習差異的策略和方法，比如改變學習內容的深度、調整學習內容的抽象度、改變問題的複雜度、改變舉例的數目（增加或

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減少）、改變學習的方式（獨立或合作）等等，Tomlinson 稱為教學設計的調適均衡器（Equalizer Model）（見圖 1）。

圖1. 差異化教學設計均衡器

上述列出的均是一般的策略，結合具體學科特點和課程要求的教學策略並不多。差異化教學不是要消除差異。基於課堂的現實，務實而有效的差異化教學策略是能為同一班級絕大多數不同學習需求、學習方式、學習興趣及學習程度的學生，都能提供學習的機會，在同一時間能參與課堂活動，發展自己的思維，達至其最近發展區(The zone of

proximal development, ZPD)

(Vygotsky, 1978)。在數學課堂上，問題解決是學習數學的核心。不同學生的數學學習差異真正就體現在解題思路、解題策略和方法上。如何佈置

問題、如何啟動學生的數學思維，達成他們的ZPD，正是

數學科進行差異化課堂教學的重點。

基礎知識明確、實在事物

單一事件/意思單一層面

與已有知識範疇聯繫

多指引、少變化要求直接資料

教師主導循序漸進

延展、轉化抽象、深層意義

多重事件/意思多種層面

與已有知識範疇有聯繫少

少指引、多變化

要求複雜資料、沒有特定答案

學生主導刪減淺易部分

基礎性轉換性

具體簡單窄角度

小躍結構性界定清晰指導性慢

抽象複雜多角度大躍開放性界定模擬獨立性快

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美國Marina Small 教授基於長期試驗研究和教師專業發展課程的經驗，總結出《基於課程標準的數學差異化教學法》

（Great ways to differentiate mathematics instruction in the standard-based classroom）（Small，2017）。她將數學開放題

（open questions）結合平行任務題(parallel tasks)（以下簡

稱O-P 策略）融入到日常的數學教學中，結合美國數學課

程標準諸多學習領域，包括數和運算、數系、運算和代數思維、運算式和方程、幾何、測量和資料、統計和概率等，

在共同的架構下設計差異化的數學教學。當前香港已經推出了新修訂的數學課程文件，儘管和美國的數學課程標準不盡相同，但我們一樣面對處理學生數學學習差異的問題。

這種基於課程標準的教學設計，與我們教師日常教學的方式還是比較貼近，主要涉及的是教學任務的設計和內容安排。教師在運用上不會覺得完全陌生。因此，本文對這此差異化數學教學的做法進行一些介紹，希望能給前線教師處理學生數學學習差異提供借鑒，以及帶出一些啟示和思考。

三、《數學差異化教學法》的教學策略和實施模式

學生在數學學習上會有怎樣的差異？在《數學差異化教學法》中作者舉了這樣一個問三年級學生的問題：

在一個櫥櫃中，您有三個架子，每個架子上有五個盒子。

這個房間裡有三個櫥櫃。共有多少盒子存放在三個櫥櫃裡？

這個問題涉及的計算關係還是比較複雜的。不同的學生會

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有怎樣的回應呢？比如：

 有的學生可能會畫一幅圖畫，把盒子數畫出來，來描

述上面的文字問題；

 有學生會直接用加法，將所有盒子數相加 5+5+5+5+5+5+5+5+5；

 有學生可能快一點，先算出一個櫥櫃中的盒子數 5+5+5＝15，然後算出房間裡盒子的總數 15+15+15=45；

 有的同學可能也熟悉乘法，先計算35=15，然後再加 15+15+15=45；

 有的學生可能對題意和乘法運算很熟練，直接用乘法

353=45。

上面學生的這些回答未必全部發生在我們每一位教師的數學課堂，也可能有的課堂中出現更多不同方案，要知道還

有一些學生是不出聲的（這不代表他們沒有想法！）。不過，

這不正是學生多樣化的數學學習嗎？假如這樣的情境真的發生在我們的課堂，我們該怎麼回應呢？

學生的回答反映出他們不同程度的理解，我們是一個一個去解釋，還是告訴全班同學一個「標準」的理解和計算方法？還是在幫助學生在解決問題的同時繼續挑戰他們的思維呢？在前面的討論中，我們指出，要促進學生的最近發

展區（ZPD），透過提供適當的學習任務，讓所有學生都能

在老師的指導下或同學的協助下，參與課堂活動，獲得有意義的數學學習經歷。基於此，在《數學差異化教學法》

的教學設計中主要凸顯三個重要目標：

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 重視學生預測診斷(Pre-assessment): 預先的診斷評估以確定不同學生的需求，瞭解學生的學習準備度。

 圍繞課標重要觀點(Big ideas)：教學的焦點集中於重要觀點且確保都被提出；

 提供學生選擇機會(Choice)：無論是在課程內容、實施過程和學習結果都能在某些面向提供學生機會去選擇；

在每一堂課的教學中，教師有時會局限在每一個學習單元或者課題的教學目標，往往忽視了整個學習領域的重要觀點。比如，對數的認識，表示一個數可以有多種表達方式，

即便是進行加、減、乘、除這樣的運算，也可以有多種方法。幾何圖形可以有不同的表徵方式。這樣的一些重要想法不只是拘泥在一個課題、一個單元，而是貫穿各個年級的課程。

相信我們的教師都認同，教學須要提供學生參與的機會，

但真正實施起來並不容易。不少教師習慣於掌控整個教學的進程和節奏，害怕課堂鬆散和失控（這裡不只是紀律問題，還有學生理解很可能出乎教師的意料以致教師回答不

上來）。然而，沒有學生聲音的課堂，如何真正啟動他們的

主動思考呢？當我們提供學生表達的機會時，不僅幫助教師瞭解學生，也有助於讓學生建立學習的自信和投入學習。

上述三個目標並非是彼此分隔的，如何讓不同能力的學生都有機會參與課堂的學習活動，Small 提出了兩種核心的教學策略—開放問題和平行任務將上述目標有效地結合起來

（見圖2）。

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四、開放問題和平行任務

我們經常看到，教師在教學中往往提出一個問題，發現學生不會或少有人回答，於是改為問一個簡單的問題，降低難度或者給學生一些提示。這種策略也能起到一定的效果，

但對學生的要求還是比較高，降低到哪個程度也不容易掌握。開放的問題即是在同一個問題中可讓不同的同學依據他們自己的瞭解層次做出回應。比如，如果問學生如何比較4

5和8

9的大小，對全班學生來說，能回答的學生可能不

會太多。如果教師的問題改為問，你認為4

5, 4 10,2

9, 1 12,8

9哪兩個分數較易進行比較？這樣問的效果就會不同。

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重要觀點 Big ideas

需要關注

實施差異化的教學 Differentiated instruction

學習經驗

Learning Experience

學習內容

Content

教學過程 Process

學習結果 Product

教師根據學生的

準備程度 Readiness

興趣

Process 學習風格

Learning style

靈活運用教學策略

開放問題

圖2. 差異化教學的設計模式平行任務

教師帶動課堂討論

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將開放的問題引入教學，學生容易獲得學習的信心，而且能開始思考和回答問題。開放性問題提供學習者從不同角度思考的機會，讓學生能學會從多種觀點來思考數學概念，

絕大多數的同學均能完全的參與課堂，而且從針對數學問題的課堂討論中獲取知識。Silver（1995）曾簡要地將數學開放性問題化為三大類別：1）給出的問題條件是開放的；

2）問題的最終結果是開放的；3）問題的解題過程是開放的。當然，開放題也可以是這三種類型的綜合情況。不少開放題會提供多種不同的解法，有助於學生創造力的培養。

教師從學生的回答中，能診斷出學生的問題和學習需求，

從而組織下一步的教學。

儘管開放題的設計有很多方式，但不能漫無目的地放寬，

需要有數學意義，跟所教授的主題相關（重要觀點 big

ideas）。如問學生，看到2 你會想到甚麼？這當然是一個開放性的問題，但這會導致許多沒有甚麼數學意義的回答。

而比如問學生16 和 18 有甚麼相似的地方？這也是開放問題，有學生可能答都是偶數，有學生可能說都在 10 到 20 之間。這樣的問題就更有數學意義，每一個學生的回答都很有價值，教師跟進討論也能回到授課重點上去。開放題的例子我們前線教師並不陌生，這裏不贅述了。

至於平行任務的問題，其實是有相同的重要觀點而且有比較接近的問題情境的問題。其想法類似於變式教學設計的理念（黃毅英、林智中、孫旭花，2006；黃毅英、林智中、

陳美恩、王豔玲，2008），通常是兩個或三個問題作為一組，

用來滿足處於不同發展水平的學生需要，讓學生同時練習

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和討論。這些任務具有相同的知識要點，並且能在類似情境中討論。透過對問題的討論，能帶出問題情境中重要的數學知識。比如，這樣一組平行任務：

任務1: 如果你將兩個數相乘，結果是 24000，你會選怎樣的兩個數相乘?

任務 2: 如果你將兩個數相乘，結果是 24.00，你會選怎樣的兩個數相乘?

這兩個任務很相近，都涉及到乘法，也跟十進位有關係。

但還是有細微的分別，一個數字相對較大，一個有小數點。

有的學生可能選擇做24000，有的可能選擇做 24.00。學生根據自己的選擇解決任務後，教師需要有跟進的問題。因為是平行任務，教師跟進的問題在兩類學生中都可以作答:

 你選擇這兩個數位中的任何一個數都可以很小嗎?

 你選擇這些數中最大的是甚麼數?

 你知道任務中的數字24 有甚麼用嗎?

平行任務的佈置可以讓學生基於自己的理解進行自由選擇

，也可以分配不同組別。教師結合學生的回答進行綜合講解，學生通過對比不同的情境的問題解決，在變化中掌握其中的不變的主要觀點。下面是一些平行任務的例子。

任務1: 一個圖形的周長是 30 厘米，它可以是甚麼圖形?

任務 2: 一個圖形的面積是 30 平方厘米，它可以是甚麼圖形?

任務1: 試舉兩個相同分子的分數，解釋哪一個分數較大。

任務2: 試舉兩個相同分母的分數，解釋哪一個分數較大。

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五、結論與啟示

Tomlinson 在給《數學差異化教學法》寫的序言中，描述了一段自己學習數學不愉快的經歷，小學階段的數學學習已經讓她覺得害怕，以致長大後認為自己就是不擅長數學的人。相信不少人都會有類似的學習經歷，甚至有人在小學已經覺得自己不擅長數學，不會在數學上成功了。這裏我們並不準備展開討論情意因素對個人數學學習的影響。反而，我們想大家思考：是什麼造成學生學習數學的失敗，

或者說，怎樣才能幫助不同的學生學好數學。

《數學差異化教學法》儘管給出了一些清晰的、可操作的教學設計模式，但它是否適合每一個教師的課堂，需要實踐檢驗，更需要我們教師專業的判斷。我們想著重指出的是，這套差異化教學設計背後的理念值得我們借鑒。教師給每一個學生充分的學習數學機會，對學生的回應，教師能給予引領和回饋，讓弱的學生獲得成功的體驗，讓能力強的學生接受進一步的挑戰！在整個課堂教學中，將與課題相關的重要觀點貫穿其中，無論是提出開放問題，還是佈置平行任務，亦或跟進的問題和討論，無一不是在構建和促進學生的數學思考。這真正是當前我們不少數學課堂欠缺的。

每個人的學習進度有快有慢，學習方式有不同：有的喜歡符號，有的喜歡圖形，有的喜歡小組活動，有的喜歡個人練習等等。在這些活動中，我們都希望學生還能有數學的思考。2000 年課改以來，各種教學法不斷出現，一時喧鬧非常。無論怎樣，其實都是希望促進學生課堂參與，提升

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學習動機，最好還能獲得好的學習結果。差異化的數學教學，正是需要教師搭建起一個促進數學思考的教學環境。

開放題的特徵和功能，既重視了過程，也關注了結果。讓學生學會學習，掌握思維的方法。處理數學學習差異，既不是只選用簡單問題將就低程度學生，也不是一味挑選難題給能力高的學生。教師須要先掌握每一個教學單元的核心概念和觀點，為教學目標設定不同層次並提供相應的學習活動。平行任務既能讓普通程度與低程度的學生關注核心知識與技能，有達成學習目標的機會；同時也兼顧能力高的學生促進其高層次思維的發展，在問題變化中發現規律，在不同情境中學會應用。

參考文獻

[1]黃毅英、林智中、孫旭花（2006）。《變式教學課程設計

原理：數學課程改革的可能出路》。香港：香港中文大學教

育學院香港教育研究所。

[2]黃毅英、林智中、陳美恩、王豔玲（2008）。數學變式課程設計——以小學三個課題爲例。《教育學報》35 卷 2 期，

1-28。

[3] Bender, W. N., & Crane, D. N. (2010). RTI in math:

Practical guidelines for elementary teachers. Bloomington, IN:

Solution Tree Press.

[4] Renzulli, J. S. (1977). The enrichment triad model: A guide for developing defensible programs for the gifted and talented.

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Mansfield Center, CT: Creative Learning Press.

[5] Pierce, R. L., & Adams, C. M. (2004). Tiered lessons: One way to differentiate mathematics instruction. Gifted Child Today, 27(2), 58–66.

[6] Small, M. (2017). Good Questions: Great ways to differentiate mathematics instruction in the standards-based classroom (3^rd Edition). New York, NY, U.S.A.: Teachers College Press.

[7] Silver, E. (1995). The nature and use of open problems in mathematics education: Mathematical and pedagogical perspectives. ZDM-International Reviews on Mathematical Education, 27 (2), 67-72.

[8] Tomlinson, C. A. (1999). The differentiated classroom:

Responding to the needs of all learners. Alexandria, VA, U.S.A.: Association for Supervision and Curriculum Development.

[9] Tomlinson, C. A. (2001). How to differentiate instruction in mixed ability classrooms (2nd ed). Alexandria, VA: ASCD.

[10] Vygotsky L. S. (1978). Mind in Society: The development of higher psychological processes. Cambridge, MA, U.S.A.:

Harvard University Press.

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3. 解方程要驗算嗎？

羅家豪香港大學數學系

在本地的數學課程中，學生一般在高小開始接觸方程。以下是一個簡單例子：

例一：解方程 3

x

 4 22。

解： 3 4 22

3 4 4 22 4 3 18 3 18 3 3

6

x

x x x x

 

   



驗算：當

x

 時，左方 6 3(6) 4 22   右方。

因此

x

 是正確答案。 6

解方程後可以「驗算」，就是把答案¹代入原方程看看兩邊

是否相等。驗算的定位對於大部分老師和學生來說似乎都

是「額外的一步」，也就是說驗算並不是解方程的一部分，

只是如果題目要求，或是題目沒有要求但有時間剩下來，

我們可以這樣做來檢驗答案是否正確。事實上，連課程文件²也是這樣寫的：

1 何謂方程的「答案」或「解答」？何謂「解」方程？這些問題和本

文的主旨環環相扣，暫不在此扯得太遠，稍後會再提及。

2 《數學教育學習領域課程指引補充文件》小學數學科學習內容，課

程發展議會，2017。

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「學生須認識如何在解方程或解應用題後作驗算」

由此可見，驗算是在解方程「後」進行的，它並不被視為解方程的一部分。

現在看看另一例子（這種「牽涉根號且可變換為二次方程

的方程」在現行高中課程中已被刪去）：

例二：解方程

x

   。 2

x

4 解：

2

2 4

2 8 16

9 18 0 ( 3)( 6) 0

x x

x x x

x x

  

   

  

  

x = 3 或 6

完成了嗎？還沒有！先來驗算一下：

當

x

 時，左方 6  6 4 2 6 4     右方。

當

x

 時，左方 3  3 2 1  ，可是右方 3 4    ！ 1 因此

x

 須捨去，從而 3

x

 是方程的唯一解。 6

之前不是說驗算是事後額外的一步嗎？為何在上例中驗算突然變得必要（否則會多出一個錯誤的解）？筆者遇過的老師一般的解釋都是「因為曾經對方程兩邊平方，所以必

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須驗算」。這個解釋有點似是而非：我們當然明白「對方程

兩邊平方」會出現的問題，可是我們很容易構作出「沒有對方程兩邊平方，但仍必須驗算」的例子，因此這個解釋並沒有觸及問題的核心。

讓我們回到一些現行高中課程會出現的方程。以下例子取自某出版社的高中教科書：

例三：解方程 log(3x  1) 1 log(x2)。解： log(3 1) 1 log( 2)

log(3 1) log10 log( 2) log(3 1) log10( 2)

3 1 10( 2) 7 21

3

x x

   

  

 

 

當

x

  時， 33

x

 和 1

x

 都是負數，即 2 log(3x1)和 log(x2) 都是沒有意義的，所以原方程沒有實數解。

該書在題解後還有一個溫馨提示：

「注意解對數方程後必須進行驗算，以檢查方程中的每一個對數是否有意義。」

可是由此至終還是沒有解釋為何「必須」進行驗算，而跟

例二一樣，有關解釋亦只針對特定的「對數方程」，仍是沒

有解答在甚麼情況下「驗算」的角色會突然有所不同（我們總不可能把方程分成根式方程、對數方程、三角方程、…… ，再逐一判斷每種方程中驗算是額外還是必須的吧）。

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事實上這樣的例子有很多，除了牽涉根號的方程和對數方程外，另一常見的類別是分式方程，例如：

例四：解方程 ₂⁴ ¹ 1 ²

4 2 2

x

x  x   x

   。

解方程的步驟從略，反正結果就是其中一個解出來的根須

捨去，這次的原因是它會使原方程的最少一個分母變成0。

然而各位讀者不妨思考一下：如果你只有紙筆而要構作一

道這樣的試題（即解出來的根中有最少一個必須捨去），有

甚麼好的方法？

在說下去之前，先看一個似乎不太相關的例子。

例五：蛋糕每件售價 5 元，現在特價八折。買 5 打蛋糕，

付款1000 元，應找回多少？

解：應找回的金額 1000 (5 80%) (5 12) 1000 4 60

1000 240 760

    

  

 



上例與解方程無關，但大家可以注意一下題解的表達方式。

在上例中，整個題解基本上只包含了一道等式，我們可以把它在同一行寫出來：

應找回的金額

1000 (5 80%) (5 12) 1000 4 60 1000 240 760

           。

只是習慣上一般會分成幾行，每行一道算式，這樣比較清

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晰易讀。再比較一下之前「解方程」的例子，例如例一，

不難發現題解的結構是不同的：它每一行都是一道等式（由

於等式中含未知數，所以這等式又稱為「方程」），而我們

也無法把整個題解放在同一行表示。

筆者在小學階段曾遇過幾位數學老師教「解方程」，記得其

中一位（只有一位）有提及過可以在題解的每行之間加上

「」符號，這樣的話例一的題解會變成這樣：

3 4 22 3 4 4 22 4

3 18 3 18 3 3

6

x

x x x x

 

    

 

這樣做的話，「無法把整個題解放在同一行表示」的問題也

解決了：

3 18

3 4 22 3 4 4 22 4 3 18

3 3

6

x x x x

x

          

 

上面提到，例一的題解中「每一行都是一道方程」。特別地，

第一行是原來的「題目」，最後一行是「答案」，為甚麼一

道方程可以成為另一道方程的「答案」？

所謂「解方程」，基本上就是「求未知數的所有可能值，使

等式成立」。在上述「加了  號的例一題解」中，整個邏輯就變成「如果 3

x

 4 22 成立，那麼

x

 成立」，也就6 是說「除了6 以外，其他 x 值都無法滿足方程 3

x

 4 22」。

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但這樣並不保證

x

 滿足方程，因此我們應該要驗算！6

可是我們也知道，每個「」牽涉的逆命題也成立（這當

然須要小心逐一檢查！），於是題解可重寫成

3 18

3 4 22 3 4 4 22 4 3 18

3 3

6

x x x x

x

          

 

也就是說「3

x

 4 22 成立當且僅當

x

 成立」，這樣6 我們就可以確定

x

 是原方程的唯一解而不需驗算了。 6 這樣，隨後幾個例子的迷思也就水落石出了。在例二的題解中

2 4 2 2 8 16

x  x  x x  x

的逆命題並非真確（一般來說

a b

 

a

² 

b

² ，但

2 2

a



b

 

a b

），因此題解基本上證明了

「 x   2 x 4 x3或 6」，也就是說「除了 3 和 6 以外，其他

x 值都無法滿足方程 x

   」，於是只要2

x

4

（而且須要！）逐一檢驗3 和 6 是否滿足方程，便能得出方程的所有解。同樣，大家自然也可以立即找到例三和例四「需要驗算」的原因，這裡不再多花篇幅。

以下例子取自2018 年香港中學文憑試的延伸部分（單元二）：

例六: (a) 若 cot

A

3cot

B

，證明sin(

A B

 ) 2sin(

B A

 )。

(b) 利用 (a)，解方程 4 5

cot 3cot

9 18

x  x 

     

   

   ，其中

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School Mathematics Newsletter  Issue No. 23 0_{ }x 2

。

解：(a) 從略

(b) 利用 (a) 部的結果可解方程如下：

4 5

cot 3cot

9 18

4 5 5 4

sin 2sin

9 18 18 9

sin 2 13 2sin

18 6

sin 2 13 1

18 13 3

2 18 2

7 18

x x

x x x x

x

 

   

 



 



     

   

   

             

       

   

    

   

   

   

 

 

 



看到這裡，大家當然立即會想到「需要驗算嗎」的問題。

筆者跟一些中學老師討論過，大部分都覺得不需要。在考

評局出版的題解和評卷參考³中，雖然有驗算的一步，可是

不佔分數（也就是以上「沒有驗算」的題解也可得滿分）。

題解中的每一步都是等價（即可用「 」表示）的嗎？的

確如此（這也得小心驗證，例如從第四行到第五行牽涉到

3 《香港中學文憑考試 2018 試題專輯：數學（延伸部分）》，香港考試

及評核局，2018。

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x 的取值範圍）。可是從第一行到第二行我們採用了 (a) 部

的結果，而 (a) 部只證明了「若 cot

A

3cot

B

，則

sin(

A B

 ) 2sin(

B A

 )」而沒有證明其逆命題（儘管其逆命題也是真確的，事實上只要懂得證明原命題，則證明逆命題基本上不用多花太多功夫），因此如果使用 (a) 部的結果，從第一行到第二行只有「」而非「  」。也就是說，

除非在 (a) 部額外證明了「若 sin(

A B

 ) 2sin(

B A

 )，則 cot

A

3cot

B

」這個逆命題，否則以上的「滿分題解」中存在一個重大的邏輯漏洞！

筆者認為，姑且撇開驗算步驟不佔分數的問題（畢竟評分

準則有其局限性和很多考慮因素），但讓學生清楚知道本題

中驗算的必要性是很重要的。以上「沒有驗算的題解」即使得到滿分，也只是「幸運」而已（因為在本題最終得出的解毋須被捨去，而評分準則中亦未有把分數分配到驗算的步驟）。

更一般地，我們應培養學生判斷是否需要驗算的能力，而非只記著「解對數方程必須驗算」這類指引。事實上，在很多情況下，這種「每行寫一個命題（命題可以是等式、

方程或其他形式）」而不搞清楚命題之間關係的論證方式，

很容易會掉進邏輯陷阱。因此解方程時能養成「思考是否需要驗算」的習慣，對學習其他課題也會有所裨益。

以上談到的例子涵蓋根式方程、對數方程、分式方程、三角方程等。大家不妨思考一下一些相關的問題，例如：

解聯立方程要驗算嗎？

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承上題，唯一解和無窮多個解的情況是否有分別？

在中學課程中，除解方程外，有沒有類似「需要驗算」的情況？

最後，以一個笑話式的例子作結，藉以說明弄清邏輯關係的重要性。當中所證明的結論固然是不正確的，大家可以細想一下問題的所在。

例七：證明 2 4 。解：考慮方程 ^x^x^x ^²



，其中

x

 。解方程0 ⁴如下：

 

2

2 2 2

xx

x x

x





再考慮方程

y

^y^y^ 4，其中

y

 。解方程如下： 0

 

4

2

4 4 2 2

yy

y y y y





4 當有「多於一層」的指數時，應從「上至下」計算，舉例說，

  ^{ }

5 5 5

4 4 4

3 3 3

2 2 2 。

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由於

x  2

滿足第一道方程，故此

2 2

2 2



。由於 y 2 滿足第二道方程，故此

2 2

2 4



。結合上述兩條等式，可得 2 4 。

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4. 一類分數的小數表示

梁景信

香港教育大學數學與資訊科技學系

考慮分數，其中，且、互質，把化成

小數，有以下三種情況：

（1） ( 、是非負整數)：有盡(terminating)小

數，如；

（2）和互質：純循環(pure recurring)小數，如（橫綫下的數字不斷重複，即

）;

（3） ( 、是非負整數，和互質)：混循

環(mixed recurring)小數，如。

但有盡小數也可以表示成循環小數，如。有關

表示分數為循環小數的論述，可參考[1]和[2]。

以下介紹一個表示分數為循環小數的重要定理：

定理1 假設

(一般寫作， ) 和

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School Mathematics Newsletter  Issue No. 23 個

，則。

證明： ^೙ ^షభ ^೙షభ ^షమ _ష೙^మ ^ష೙శభ ^భ ^ష೙

應用定理 1，我們可以用簡單筆算，把分母數位是同一數

字的分數化為小數。

例 1（分母的數位全是 9）把化為小數。

解：把寫成，應用定理1 得。

以下例 2 和 3 中的分母可以很容易地倍大成

個

。

例2 （分母的數位全是 1）把化為小數。

解：＝。

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解：＝。

在以下例 4－8 中的分數，雖然分母皆不能倍大成

個

，但可以倍大成

個個

，然後把分數表示成 _೘ _೙ _೘，其中

和，只要找到和，便可

把原來的分數表示成混循環小數。

解：＝，其中和

，解

得和，所以

。

解：，其中和

，解得和，

所以。

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解：，其中

和，解

得，，所以

。

解：，其中和

，解

得，，所以

。

解：，其中和

，解得，，所以

。

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最有趣的情況是分母的數位全是 7，以下例 9－11 都是把

分母倍大為。

例9 （分母是 7）把化為小數。

解：。

解：

。

注意：10101 是 7 的倍數，所以第一次擴分後的分子可以被7 整除。

解：

。

注意：1001 是 7 的倍數，所以第一次擴分後的分子可以被 7 整除。

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習題1 把化為小數。

提示：把分母倍大為

個

。

習題2 把化為小數。

提示：把分母倍大為

個

，

習題3 讀者自擬。

提示：不需要。

參考文獻

[1] Hardy, G. H. & Wright, E. M. (1979). An introduction to

the theory of numbers (5

^th ed.). Oxford: Clarendon Press.

[2] Niven, I. M. (1961). Numbers: Rational and irrational.

Providence: American Mathematical Society.

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5. The e Story

WONG Hang Chi

International Christian Quality Music Secondary and Primary School CHEUNG Ka Luen

The Education University of Hong Kong

1 Introduction

The Euler's number plays a very special role in Mathematics, especially in Calculus. In the Mathematics Curriculum in Hong Kong, for instance, students may select the Extended Part Module 1 (Calculus and Statistics) or Module 2 (Algebra and Calculus), in which this concept is introduced.

This learning objective is stated clearly in the Mathematics Curriculum and Assessment Guide (Secondary 4–6) provided by the Education Bureau.

For Module 1 students: “recognise the definition of the number and the exponential series ^మ ^య ” and

For Module 2 students: “recognise the definitions and notations of and the natural logarithm.”

However, the introduction of requires tactful planning. Some students found it difficult to believe in the existence of the limit

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of the sequence ; some struggled with the sum to infinity of the series and wonder why this could possibly be finite; while some just could not imagine how it works. This article aims to discuss a different pedagogical approach to introduce this special number to secondary students, so that students could have a deeper understanding of the definition of and the exponential function.

In this article, for the sake of simplicity, the sets of natural numbers, rational numbers and real numbers are denoted by ,

and respectively throughout.

2 A Problem Solving Activity

2.1 A Function which Equals its Derivative

The lesson planning we would like to present here reverses the order of the traditional approach, which suggests the existence of first and then goes on with the differentiation of polynomials, exponential functions, etc. We shall, however, delay the introduction of after the discussion of the differential calculus of polynomials and the basic differentiation theorems such as Addition Rule, Product Rule, Quotient Rule and Chain Rule.

In this activity, students are first asked to find the derivatives of certain algebraic expressions like polynomials , , ,

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and rational functions , _మ, _య, . Then, the teacher may guide students to summarize these results as the Power Rule:

where is a non-zero integer. Or, equivalently stated, we have . The interesting part is that this result is not valid if , otherwise the right side of the equality results in the division by zero. Let us also investigate the left side by putting . It equals . So, the question is: “the derivative of which function gives ?” From the above discussion, we know that if a differentiable function satisfies , it must not be of the form , where is any integer. We need to think about other possibilities.

The teacher may then motivate the students to investigate the properties of this sought for function. If such a function exists, then it must satisfy the equation .

According to the Chain Rule, we then obtain , which implies that . In other

words, if can also be expressed as a function of , then the

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derivative of equals itself! From the observation before, the function cannot be a polynomial of with any degree , or otherwise, will be another polynomial of degree , which is not equal to . Intuitively, in order that the equality to hold true, the function

should grow more rapidly than any polynomial. At this stage, students might start to be curious to look for something of exponential growth!

2.2 A Limit as the Slope of Tangent

Let us investigate the exponential function to see whether it is indeed the solution to our problem. Let be a positive constant such that . Students had already learned, in the Compulsory Part, that the exponential function is well-defined for all . The teacher might explain to students that is dense in , and so this function could, somehow, be extended to be defined for all .

We aim to find the value of such that , that is, . Because if this “idealistic” value was found, then the function would satisfy the property that its derivative equals itself, which in turn solves our original problem.

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Putting , in particular, we have . By the first principles, let us calculate the limit of

೓ as .

If this limit does exist, then ^೓

In general, assume that , we have

೤శ೓ ೤ ೓

.

Taking , we obtain

provided that the limit exists.

In other words, we may “localize” the study of the derivative at . The solution of the equation will come down to the understanding of finding the value of such that

೓ .

In the process of problem solving, it is beneficial to us if we can look at the problem in a different perspective. In fact, and

are merely dummy variables. We shall also write to represent the same relation. What is the

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geometric meaning of ? We may introduce a rectangular coordinate system to represent the graph.

We know that equals the slope of tangent of the curve at . Using the slope-intercept form, the equation of this tangent line can be expressed as .

3 Exploration of the Exponential Function 3.1 The Graph of the Exponential Function 3.1.1 The Relation of and

One way to illustrate this is to use a Dynamic Geometry Software (DGS), such as GeoGebra, to demonstrate the situation. In the figure below, the slide bar controls the value of , which is initially set at . As varies, the software can immediately display the behaviour of the graph of

so that it can be compared with the fixed straight line . This helps students to visualize the shape of the

function in the investigation.

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As we can see from the graph, when , the graph of

intersects the straight line at two distinct points and

.

On the other hand, if we drag the slide bar to, say, , the curve also intersects the straight line

at two distinct points and .

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Combining these two observations, it is natural to guess that there must be an “idealistic” position between and

, at which the curve and the straight line has exactly

one point of intersection. By adjusting the slide bar carefully, it could be found that the sought for value occurs at about , as shown in the following figure. We shall call this idealistic value .

3.1.2 The Nature of the Points of Intersection

Let us formulate the above results more rigorously. We know that in general, for any constant , the curve cuts the straight line at the point , and then they intersect again at another point for some , except when , where the curve and the straight line

touch each other at exactly one point .

In the DGS, it could be observed that the point will move towards when the value of approaches . Intuitively, the point serves as the “bending centre” of

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the curve , which is concave upward for any with .

The behaviour of the curve, however, changes at . Intuitively, when , the rate of increase of the curve is less than at the point , and so the other point of intersection is at the right side of the -axis.

Therefore, we have . On the contrary, vice versa for the situation of , we know that the other point of intersection is at the left side of the -axis, which implies that . These observations can help us establish the relation between and (Nelsen, 2015). Thinking backward, we can consider as an independent variable and express as a function of . Our aim is to take to be really small, so that can be made very close to .

If , we have . At the point of intersection , we know that , and so ^భ^೟. Now, as approaches indefinitely from above, the value of can be taken as close to as we please from the below. The limiting case is that the two points of intersections coincide with one another, and the secant line, with slope , becomes the tangent line at the point . That is, the right hand limit exists.

శ

భ

೟

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Similarly, for , we have . In this case, we have also

భ

೟. If approaches indefinitely from below, the value of can also be taken as close to as we please from the above. Therefore, the left hand limit also exists.

ష

భ

೟

Summarizing these two results, we have

భ

೟

3.2 The Study of the Limit

3.2.1 Expressing as the Limit of a Sequence

In particular, if we substitute , for , we have

భ

೟ . On the other hand, if we substitute , for , then ^భ^೟

. Hence, we have established the inequality

for all .

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Note that when , the value of ^భ^೟ increases towards when decreases, as shown in the DGS. Since decreases towards as increases, we know that the sequence is monotonic increasing. By using similar

arguments, we can also deduce that the sequence is monotonic decreasing. As a result, we

have for all

. Both of the sequences and are monotonic and bounded. By the Monotone Convergence Theorem for sequences, and must also be convergent! Furthermore,

since when

, by using the Sandwich Theorem, we know that the limiting values of and ought to coincide at the constant . This ultimately explains why the definition of the Euler's number can be expressed in this peculiar form of

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3.2.2 Estimating the Value of

Once the students are convinced by the definition of , the teacher may introduce

some numerical values by

using to

get a taste for the students.

This sequence, however, converges quite slowly as increases. Therefore, it would be beneficial to them if a spreadsheet software, such as Excel, is used instead of a scientific calculator. For instance, by putting , for

and calculating the corresponding value of in the software, one could obtain the following approximations, as shown in the table below.

Unfortunately, the drawback of this method to estimate is that the accumulation error is very large when exceeds . It is because the precision of the software or calculator is not enough to handle values of such a large scale. The first way to solve this problem is to use a software that support high precision arithmetic. The second way is to apply the

10 2.59374246 100 2.704813829 1000 2.716923932 10000 2.718145927 100000 2.718268237 1000000 2.718280469 10000000 2.718281694 100000000 2.718281798 1000000000 2.718282052 10000000000 2.718282053

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exponential series ^మ ^య at

(Dörrie, 1956). Nevertheless, the technical details are out of the scope of this article.

4 Conclusion

Now, we have found the answer to our original problem.

Firstly, as we saw earlier in the previous section, for the exponential function to satisfy the equation , the only possible value of is (Maor, 1994). Secondly, since the natural logarithm is the inverse function of , we conclude that is a solution to the equation . Moreover, the growth rate of is faster than that of , whereas the growth rate of is slower than that of , for any .

As a side product of the above investigations, we also discover the reason why the Euler's number is chosen as the base of the natural logarithm. Consider , then

, where and . Consequently, we have , and hence . As a result, we would have a strange constant multiplied to .

School Mathematics Newsletter（SMN）

SMN

ISSUE 23

ISBN 978-988-8370-89-4

School Mathematics Newsletter（SMN）

Foreword

Contents

10.

1. What ‘Teachers’ Can Do to Foster STEM Awareness through the Learning of Mathematics

Mathematics Newsletter, 2017, Issue 21, p. 8)

The three-teacher model

as-teacher (the First Teacher), the school teacher (the Second

learning and assessment as learning. Such kind of reinvention

The First Teacher – Student as Agent

(The Third Teacher)

(The Second Teacher)

Task in Context

(The First Teacher)

directed learners develop our agency for monitoring and

The Second Teacher – Teacher as Agent

assessment as learning (Earl, 2003) as she knows well that the

The Third Teacher – Environment as Affordance

Coda

References

capacity to learn and how we can harness it. London:

assessment to maximise student learning. Thousand Oaks, CA:

perception. Boston: Hougton Mifflin.

Journal, 10(2), 1-18.

International Journal for Lesson and Learning Studies, 2(2),

World Congress of the International Association for Semiotic Studies. Universidade da Coruña (Spain), 867-878.

Practice, 46(1), 40-47.

thought. New York: Basic Books.

2. 數學科差異化教學的理念和設計

proximal development, ZPD)

重要觀點 Big ideas

Learning Experience

3. 解方程要驗算嗎？

x

x

x x x x

x

x

x

x

x

x

x

x

x x

x x

x x

x x

x x

x

x

x

x

x x x x

x

x

x

x

x

x

x

a b

a

b

a

b

a b

x 值都無法滿足方程 x

x

A

B

A B

B A

x  x 

 

   