Online resources for
Affecting the flow of a proof by creating presence
a case study in Number Theory
By Mika Gabel and Tommy Dreyfus
Online Resource 1 The theorem and its proof
Lemma: Let a, b, c  Z . If c | b and c | a then k , m  Z : c | (ka  mb)
Claim 1: If the equation ax  by  d has a solution x, y  Z then gcd(a, b) | d
Proof of Claim 1: By the Lemma each common divisor of a , b is a divisor of d . In
particular gcd(a, b) | d
Corollary: d  gcd(a, b)
Theorem (Claim 2):
The greatest common divisor (gcd) of two integers a , b , at least one of which is not 0,
equals the smallest natural number of the form ma  nb , where m, n are integers:
gcd(a, b)  min{ma  nb  0 : m, n  Z } .
Proof:
Define a set A  {ma  nb  0 : m, n  Z }
Suppose a  0 . Then if a  0 then 1 a  0  b  A and if a  0 then 1 a  0  b  A .1
Therefore A is not empty  A has a minimal element (well-ordering principal).
Let d  min A  m, n  Z such that d  ma  nb
We will show that d is a common divisor of a , b :
By division with reminder theorem: ! q, r such that a  qd  r ; 0  r  d
 r  a  qd  a  q( ma  nb)  a (1  qm)  qnb  d
If r  0 then r  A and r  d  contradiction  r  0  d | a.
Similarly can show that d | b so d is a common divisor of a , b , therefore d  gcd(a, b) .
In addition, by the corollary of Claim 1 d  gcd(a, b) . Therefore d  gcd(a, b) .
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In Lesson1 the lecturer considered only the case a, b  N therefore it was sufficient to argue that
1 a  0  b  A in order to show that the set is not empty.
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Online Resource 2 Students’ questionnaires
1. Write a short summary of the proof as it was taught in the lesson.
2. Given a  36 ; b  64 . Let the set A be A  {36m  64n  0 : m, n  Z }.
a. It is known that 2  36  3  64  120 . Is it true that 120  A ?_______________
Explain your answer_________________________________________________
b. Write 3 elements in A and explain how you got them:
c. It is known that gcd(36, 64)  4 . Is it true that 4  A ?______________________
Explain your answer_________________________________________________
d. Is it possible to find m, n  Z such that 36m  64n  2 ?_____________________
Explain your answer_________________________________________________
e. Is there a solution to the equation 36 x  64 y  8 , x, y  Z ?__________________
Explain your answer_____________________________________________________
3. The proof shows that gcd(a, b)  min{ma  nb  0 : m, n  Z } . Explain in your own
words the relation between the proof and between finding a solution to the
Diophantine equation ax  by  d
4. In the beginning of the proof it is demonstrated that the set A  {ax  by  0 : x, y  Z }
is not empty. Why is it important to show that?
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
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5. The proof concluded that d  gcd(a, b) . Please explain in your own words:
a) why d cannot be smaller than gcd(a, b) :_______________________________
___________________________________________________________________
b) why d cannot be greater than gcd(a, b) :_______________________________
___________________________________________________________________
6. Please state true/false for the following statements:
Statement
True
False
The proof describes a method of finding a solution to the
Diophantine equation ax  by  d
According to the proof, if d  gcd(a, b) then there exists a solution
to the Diophantine equation ax  by  d
According to the proof, if d  gcd(a, b) then there exists a unique
solution to the Diophantine equation ax  by  d
If gcd(a, b)  1 then the proof "doesn't work”
d is the minimal element of the set A and this leads to a
contradiction during the proof
The conclusion r  d comes from the calculation
r  a  q(ma  nb)  a(1  qm)  qnb  d
7. Do you have any comments (positive/ negative) regarding the way the proof was
presented in class?
8. Which part of the lesson was the most interesting and which part was more boring?
Please explain.
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9. In your opinion, your understanding level of the theorem is:
a. Very high
b. High
c. Reasonable
d. None
10. In your opinion, your understanding level of the theorem’s proof is:
a. Very high
b. High
c. Reasonable
d. None
11. What was the contribution of the proof to your level of interest in the lesson?
a. Greatly increased
b. Somewhat increased
c. Not interesting but important
d. Redundant part of the lesson
12. What was the effect of the proof on your understanding of the theorem?
a. Greatly increased
b. Somewhat increased
c. No influence
d. Only caused confusion
* In year 2, two questions were added:
13. Please complete the following sentence:
“The proof of the theorem is________________________________”
14. Please circle the adjectives that you feel represent the proof:
Discouraging
Enriching
Cumbersome
Fascinating
Enjoyable
Charming
Clear
Complicated
Vague
Challenging
Important
Technical
Beautiful
Tiresome
Redundant
Boring
Trivial
Annoying
Interesting
Hard to understand
Fun
Clear
Elegant
Deep
Awesome
Amazing
Casual
Convincing
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Online Resource 3 Rhetorical figures used to create presence and suppress presence
Creating presence
The lecturer selects certain elements on which he focuses attention by endowing them
with ‘presence’. The elements left out are not ignored but ‘pushed to the background’.
The rhetorical figures used to endow presence for elements are:
• Hypotyposis: vivid picturesque description
• Synonymy or anaphora: using repetition of words or phrases for emphasis
• Metabole: repeating the same ideas in different words to accent another aspect
• Amplification: the division of the whole into its parts
• Aggregation: enumerating the parts and end with a synthesis
In addition to the above rhetorical figures other possible methods for creating presence
have emerged from data:
• Multiple warranted claims: This can be demonstrated by using Toulmin’s scheme.
• Type of discourse: An argument that is built through debate with the students or
an argument that is achieved as a shared effort by the lecturer and the students
may have more presence than an argument that is merely presented to the
students.
• Using humor: Perelman did not relate directly to humor as a rhetorical figure and
did not think of it as pertinent to persuasive discourse but he did state that “humor
is a very important factor in winning over the audience or, more generally, in
establishing a communion between the speaker and his hearers” (The new
rhetoric, p. 188).
Suppressing presence
Elements are coded as having suppressed presence if the lecturer is actively trying to
divert students’ attention from them by using rhetorical figures.
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For example, during the introduction, four examples of Diophantine equations were
presented to the students. Some of the examples require mathematical skills beyond the
scope of the students and were presented very briefly and partially. One of the examples
was Fermat’s last theorem. Clearly the lecturer did not expect the students to prove or
even read the proof of Fermat’s theorem, as reflected in the following excerpt from the
lesson:
Lecturer:
Andrew Wiles... proved the theorem with extraordinary efforts and the
use of advanced mathematics that Fermat was not even aware of its
existence…
Student:
What are such advanced mathematical means?
Lecturer:
Oh… it will take me a full seminar to teach this to people who are
very knowledgeable in mathematics; it will take me about a year to
clarify these mathematical means…
Student:
Can it be summarized?
Lecturer:
…it’s
called
elliptical
curves
and…modular…
the
required
mathematics is beyond the scope of understanding for most of
humanity… so let’s not get into that…
By using these repeated exaggerated phrases (or maybe not so exaggerated in the case of
Fermat’s theorem) the lecturer is really emphasizing that the focus of the students should
not be on the proof of the theorem but rather on other aspects of the example.
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Online Resource 4 Schematic global analysis of the flow of the proof in Lesson 1
I. Introduction: Diophantine equations
Ia. Historical background – Diophantus
Ib. Examples of various Diophantine problems:
i. Linear Diophantine equation
ii. Pythagorean triples
iii. Fermat's last theorem
iv. 2m-3n=±1
(40’)
(7')
(33')
Central topic: Linear Diophantine equations
Given a,bϵN, what is the condition on d such that the equation
ax+by=d will have a solution with x, yϵZ?
II. Numerical example:
Discussing why the equation 8x+12y=3 has no solution x,yϵZ
(5’)
A necessary condition for the existence of a solution was obtained
III. Stating and explaining Claim 1:
If the equation ax+by=d has a solution x,yϵZ then gcd(a,b)|d
(2’)
Is that really
equivalent?
Is that sufficient? i.e. is it true that if gcd(a,b)|d then a solution exists?
IV. Stating and proving Claim 2: gcd(a,b)=min{ax+by>0, x,yϵZ}
IVa. Stating Claim2, describing the set A={ax+by>0, x,yϵZ}
IVb. Proving that the set is not empty and the existence of a minimal
element denoted d
IVc. Proving that d|a (division with remainder)
(30')
(6')
(5')
(16')
So far we have shown that d≤gcd(a,b)
IVd. Finalizing the proof: d≤gcd(a,b) and d≥gcd(a,b) therefore d=gcd(a,b)
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(3’)
d|b
Online Resource 5 A detailed Toulmin scheme representation of Lesson-1, Sub-module IVb
1. Abduction
C1: The set
A={ax+by>0 , x,yϵZ}
is not empty
D: given a,bϵN
C2: There exists a
minimal element in
the set A, denoted d
W3: There exist x,yϵZ such
that ax+by>0
W1: Induction?
Infinite regression?
W4: take x=1, y=0 then
a*1+b*0=a>0 so aϵA
(Abduction)
W5: Can do the same for b
to show it belongs to A
W2: The well
ordering principle
2. Deduction
W6: take x=0, y=1
W7: at least a=a*1+b*0 belongs to
A, and b=b*1+b*0 belongs to A
Deduction
repeated twice
W8: can find many more elements,
an infinite number of elements in A
Legend:
Wrong student C, W, B, R
Lecturer C, W, B, R
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Online Resource 6 Examples of students’ difficulties as reflected in Year-1 questionnaires
Number of filled questionnaires in Year-1: 10 (out of 38 registered to the course)
Question 1
Writing a short summary of the proof
Main idea
No. of students
7 (4 of 7 relate explicitly to
division with remainder
theorem)
5 (2 of 5 gave satisfactory
explanation)
6
(1 of 6 gave no warrant)
1 (no warrant)
2
4
3 (1 quoted Claim 1)
1
Using division with remainder
Using proof by contradiction
The set A is not empty therefore by the well ordering principle
it has a minimal element d
The set A has a minimal element d
d|a (did not mention that d|b)
d is a common divisor of a,b
gcd(a,b) is a divisor of d
d≤gcd(a,b) and d≥gcd(a,b)
minL[a,b]=gcd(a,b)
* remarks:
(1) Shaded cells represent the central ideas as perceived by the lecturer
(2) The number of students that mentioned all 3 ideas stated by the lecturer: 2
Question 3
Explain the relation between the proof and finding a solution to the Diophantine equation
ax  by  d
Satisfactory answer
Wrong answer
No answer
2
3
5
Question 5
a) The reason d cannot be smaller than gcd(a, b) :
3 students answered correctly, 7 students gave a wrong reason or no reason at all.
b) The reason d cannot be greater than gcd(a, b) :
1 student answered correctly, 9 students gave a wrong reason or no reason at all.
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Online Resource 7 Schematic global analysis of the flow of the proof in lesson 2
I.
Introduction: Linear Diophantine equations
Ia. Formulating the problem and historical background
Ib. Why linear? A geometrical interpretation (a line in the plane)
(13’)
(8')
(5')
It’s not difficult to find a linear Diophantine equation that has no solution x,yϵZ
II. Numerical example: Discussing why the equation 12x+18y=68 has no solution x,yϵZ
(4’)
A necessary condition for the existence of a solution was obtained
III. Stating and explaining (without proof) Claim 1:
If the equation ax+by=d has a solution x,yϵZ then d(a,b)|d
(2’)
Actually, we can assume without loss of generality that d>0
IV. Explaining (without proof) that if ax+by=d has a solution then ax+by=d* has a
solution if d|d*
(4’)
If the equation ax+by=d has a solution x,yϵZ then gcd(a,b)|d
So what would the smallest d be such that ax+by=d has a solution?
V. Class discussion, concluding that this “fundamental” d would be gcd(a,b)
(12’)
We will prove this by two ways: The first is an existence proof and the second is
a constructive proof. But before we prove it, let me define the set L[a,b]
VI. Defining
and giving examples
(8’)
By the examples I showed you can see what we are going to prove next
VII. Stating and proving theorem 2:
.
(34')
VIIa. Explaining the theorem
(8')
VIIb. Proving that L[a,b] is not empty and the existence of a minimal element
denoted d
(8')
VIIc. Non-formal explanation of the proof’s structure, giving motivations for the next
steps of the proof
(10')
VIId. Proving that d|a (division with remainder)
(6')
So far we have shown that d≤gcd(a,b)
VIIe. Finalizing the proof: d≤gcd(a,b) and d≥gcd(a,b) therefore d=gcd(a,b)
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d|b
(2’)