Mathematical competences 27 The degree of action relates to the number of areas within the subject of mathematics or in more general the number of different situations one can act out a given competence. For instance not all have the same level of symbol-formalism competence within higher algebra as within arithmetic.
The technical level of a competence refers to the level of mathematical complexity of a given situation one can handle. We would say it is easier to handle the representation of a situation of linearity in relation to buying potatoes 75 eurocents a kilo than the representation of the linearity of the situation presented above.
Below we will present a few of the eight competences in more details, to exemplify how the competence thinking can be used to challenge the more gifted students within the frame of the ordinary class work.
Chapter 2 28
Russian peasant algorithm for multiplication is difficult to understand and explain before grade 7, because it requires higher order reasoning. The initial multiplication1 problem is rewritten in new ways. But in some of the rewritings a factor is not transported to the next line:
22 × 35 = 11 × 70 5 × 140 2 × 280 1 × 560
Add the numbers in bold to get the result.
Sum of angles in a triangle
• One can cut and arrange the three corners of a paper triangle into a whole angle (180°). This concrete approach is a nice way to enter into a more systematic or formal investigation.
• One can investigate and experience the same connection with dynamic software, e.g. GeoMeter or GeoGebra. This will convince most people that angles in any triangle sum up to 180°. The visual demonstration is of course not a proof in the traditional formal understanding. But the immediate visual feedback will act like an enzym to the reasoning process when building up mental constructs on this in the brain. The computer may strengthen the conviction when formulating conjectures but the important message is that Reasoning from empirical evidence is not a mathematical proof” [2].
• The proof of Euclids is an example of a formal proof met in high school, still depending on some axioms and other conjectures.
Rules for calculating powers
• 52·54 = 56
• ab·ac = ab+c, a ∈ R; b, c ∈ N
• a0 = 1, a ∈ R
• a-n = (an)-1 = 1 / an, a ∈ R\{0}; n ∈ N
The rules show increasing abstraction from an example to a generalization i.e. the technical level increases in relation to both the reasoning and the representation.
1mathforum.org/dr.math/faq/faq.peasant.html (Eds.)
Mathematical competences 29 Gifted students may reason about the power being any real (or even a complex number) before this eventually is taught in class.
Pythagorean theorem
• Examples of Pythagorean triples such as (5, 12, 13), since 52 + 122 = 132
• Presenting this conjecture with visual support, and eventually suggesting rearrangement of cardboard triangles and squares.
Poems (verses) related to the Pythagorean theorem, e.g. a poem by H. C. Andersen (1831). The Danish title: Formens evige Magie means The Eternal Magic of the Form:
Excerpt in Danish:
Trianglen ABC er givet her, Retvinklet og paa Siderne Quadrater; Beviset er nu om de to Krabater, Det, at Quadraterne paa hvert Catheder AC, BC (jeg naevne disse Steder) Er' just i Eet og Alt, som den Krabat, Hypothenusen kalder sin Quadrat. Nu gaae vi da til vore Præparater. En lodret Linie maa man som De veed Her drage til den større Side ned, Og saa forlænge den endnu til K, Da vil man finde, ei det mindste mangler, AB-Quadraten ganske rigtig staae Delt (som AK BK) i to Rectangler. (Thi tvende Linier, man veed, Har just det generelle, Naar paa en tredie de staae lodret' ned, Saa er' de ogsaa ganske parallelle.) Nu drages en fra A til G, fra C til I, Og da Præparationen er forbi. Ei sandt, o Mester! — true dog ei med Riset! Nu gaae vi til Beviset.
In fact, this text reflects the proof given by Euclid as Proposition 47 in the Elements, Book 1 (see e.g. [3] in which 75 proofs of the Pythagorean theorem are presented).
It could be rendered by a simpler text than this poem displays; however, the poetic language at the same time requires and consolidates a profound understanding of the proof. It may also be fascinating to some students, that a world famous author in fact was also rather good at doing mathematics.
The examples may be from different subject areas and represent different levels than those of the KOM-report. The first two (dialogs) are examples of following and evaluating a mathematical reasoning:
Chapter 2 30
A: As you square a number, the result always becomes bigger. This is the case for the endless number of whole numbers. Therefore it must also be the case for every number.
B: No, firstly the assertion is wrong, as
= <
1 2 1 1
2 4 2. Secondly you cannot transfer all properties in the set of whole numbers to a more extensive set of numbers.
A: Every odd number is composite. Assume n is odd, then + −
= −
2 2
1 1
2 2
n n
n , where both
+ 1= 2
n k and − 1= 2
n m are whole numbers.
Since n=k2−m2=(k−m k)( +m , n has to be composite. )
B: The reasoning is wrong because k−m=1, so you only claim that n=1×n , which does not make n a composite number.
The next example is an illustration of what it may mean to know and understand what a proof is (is not):
A: If lim ( )f x =b, as x→a, and lim ( )g y =c, as y→b, then lim ( ( ))g f x =c, as x→a. Since x approaches a , then f x( ) by asumption must be approaching b. This means that
( ( ))
g f x approaches c. And that was the assertion.
B: This is not a valid proof because the handling of the concept of limit is loose and unprecise.
In fact, the assertion that is attempted to be proven is wrong, unless g satisfies more conditions.
The problem is that the image of f may be contained in the domain of g in such a way that the composite function cannot approach c.
For example let f and g be defined this way:
( )=0
f x for all x, and g(0)=1, but g y( )=0in every other instance.
Then for a=0, f x( )→b=0, as x→a. Also g y( )→c=0, as y→b=0. But g f x( ( ))=1 for every x.
Therefore it is not the case that g f x( ( )) approaches c=0, as x approaches a.
The final example shows the ideas in a correct proof:
Gauss’s proof of the formula 1 2+ ++n=n n( +1) / 2 rests on the idea to determine the sum as follows. By adding n++2 1+ to the left side we get the sum twice, which could be presented also as n pairs of numbers with sum n+1, i.e. n n( +1).
We encourage you to give and discuss some examples of this competence as it is demanded and developed in teaching in your school, college or university. You might reflect also on the following question:
Mathematical competences 31
How should mathematics teachers assess and cope with the possible and optimal progression in the
• degree of coverage
• radius of action
• technical level
of reasoning competence?