Problem solving modeling with theory of containerization
By Michael Vershima Atovigba PhD
Department of Curriculum and Teaching
Benue State University, Makurdi, Nigeria
[email protected]
Abstract
The work aimed at establishing the effect of use of theory of containerization on learners’
problem solving. Containerization entails doing mathematics using abstract containers and
objects. The work used a quasi-experimental design and it was a post test only study. A
sample of 101 Primary IV pupils in Nigeria was studied. The Arithmetic Achievement Test
was used to obtain data. The mean mark of 50 was set as cut off mark for comparing
achievement of pupils. The student t-test was used to test the hypothesis at 0.05 alpha. The
pupils achieved significantly higher than the set mean mark, after exposure to theory of
containerization. The work recommends that the theory of containerization be considered as
an addition to existing problem solving strategies.
Introduction
Problem solving is a major independent variable of intelligence. Pigott (2015) explains that
the very problem with problems, that they should result in the solver being stuck, is at the
heart of what problem-solving is about. Mathematics educators posit that the aim of
mathematics is problem solving and the right method of mathematics education is problem
solving.. Thus McClure (2014) writes: “What children should be doing is solving problems,
their own as well as those posed by others. Because the whole point of learning mathematics
is to be able to solve problems. Learning those rules and facts is of course important, but they
are the tools with which we learn to do mathematics fluently, they are not mathematics
itself”. Moreover, the Mathematics education content is prepared for learners; to enjoy
mathematics and develop patience and persistence when solving problems, to develop
mathematics curiosity and use inductive and deductive reasoning when solving problems, to
analyze and solve problems both in school and in real life situations; to develop in students
abstract, logical and critical thinking and ability to reflect critically upon their work and the
works of others (Senri International School Foundation, 2008).
Many scholars have devised problem solving models and strategies. Pólya (in McAllister,
1996) provides a four-step strategy of problem solving including: Understanding the
problem; devising a plan; carrying out the plan; and looking back. Cherry (2015) notes that
problem-solving refers to the mental process through which one discovers, analyzes and
solves problems, including discovery of the problem, deciding to tackle the issue,
understanding the problem, researching the available options and taking actions to achieve
goals. The solver then faces a set of psychological fixations which are firstly overcome
before the problem is solved finally. The Osborne-Parnes Creative Problem Solving Process
(in Hunt, 1998), provides a total of six stages of problem solving including: mess-finding
(Objective or goal finding), fact-finding (data gathering), problem-finding (clarifying the
problem), Idea-finding (generate ideas), solution finding (idea evaluation), and acceptancefinding (idea implementation). Also, Chapra and Canale (2002) looking at mathematical and
engineering problem solving modeling note that a mathematical model of problem solving is
a functional relationship of the form DV=f(IV, parameters, forcing functions). Here, DV is
for dependent variable, IV for independent variables, parameters for reflections of the
system’s properties or composition, and forcing functions for external influences acting upon
the system. The Singapore Maths Teacher (2005) on the other hand, recommends as a
strategy of problem solving that, to achieve proficiency in mathematics problem solving, it is
useful to (1) draw a picture, (2) look for a pattern, (3) guess and check, (4) make a systematic
list, (5) logical reasoning, (6) and work backwards.
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Cai (2008) says worthwhile problems should constitute the focus and stimulus for students’
learning and mathematics explorations while teachers provide guidance. Also Rusczyk
(2011) remarks that problem solving is an approach to mathematics as opposed to
‘memorize-use-forget’ approach and that a successful mathematician possesses a few tools
but knows how to apply them to a much broader range of problems.
The containerization model was earlier applied and found to have enhanced mathematics
problem solving, during the 2012 Millennium Development Goals project at Otukpo, Benue
State, Nigeria. The project involved retraining of unqualified mathematics teachers who,
owing to dearth of mathematics teachers, had been teaching mathematics at elementary
schools (Atovigba and O’Kwu, 2013).
Materials and Methods
This study was conducted in 2012 during the author’s doctoral research in effect of activitybased approach on pupils’ attitude and achievement in arithmetic. Materials used in the
research include containerized activities. The method of this research was quasi-experimental
as intact classes were used. Activity-based lesson plans were prepared and administered using
containers and objects as instructional materials. The Arithmetic Achievement Test was used
for data collection. Mean achievement was compared with the set mean mark of 50, to
answer the research question. The research question was: what will be the difference of
students’ mean achievement in mathematics problem solving from the set mean mark of 50
after exposure to containerized activities? The analysis of covariance was used to test the
hypothesis at 0.05 alpha. The hypothesis of the study was: exposure to theory of
containerization does not significantly enhance learners’ mathematics problem solving.
To conduct the study, pupils were exposed to the theory of containerization (Atovigba, 2013)
which states that nature expresses itself in containers of objects (or particles) and
mathematics and science must be studied from the point of view of containers and objects
from where basic binary operations are founded and abstracted and extended deductively to
higher order operations as relations of objects and the given containers. Using containers and
objects activity in doing mathematics is in line with the scientific method. According to
Edmund (2007), the scientific method consists in reasoning from evidence to conclusions that
can be tested or verified, and the scientific method has been developed for problem solving.
The theory specifically presents basic binary operations from an activity-based view point as
follows: Addition of x and y is the activity of having x count of objects in one container, and
y count of objects in another container, and then emptying contents of both containers into
one single container and then counting all its contents as sum. Example: 2 + 3 has container
A of 2 counts of an object or particle and container B of 3 counts of objects. The sum is
emptying A and B’s contents into Container C and counting the total (5 objects: Fig. 1).
+
=
Fig 1: showing 3 objects and 2 objects in respective containers
summing to 5 objects in one container
Subtraction as an activity has one container with a number of objects out of which some are
removed (Fig. 2, having 5 count of objects and 2 removed under white background: what is
left or the difference is 3).
5
-
3
2
=
F ig 2 : s h o w in g in a c o n ta in e r :
2 o b je c ts s u b tr a c te d fr o m 5 o b je c ts
h a v in g o n ly 3 le ft in th e c o n ta in e r
2
Multiplication of ‘x’ and ‘y’ (i.e. x.y) has x count of objects in each of y containers (Fig 3
having 2x3 as 3 containers each containing 2 counts of an object: the result is count of all
objects from each of the containers i.e. 6).
2x3
6
=
F ig 3 : s h o w in g in th r e e c o n ta in e r s e a c h w ith 2 o b je c ts :
R e s u ltin g in 6 o b je c ts w h e n a ll th e o b je c ts
fr o m e a c h c o n ta in e r a r e c o u n te d
Division of ‘x’ by ‘y’ is having ‘x’ count of objects distributed in each of ‘y’ containers (Fig.
3: read as 6  3  2 i.e. 6 objects distributed equally into 3 containers: Result: each container
has 2 objects).
Remark 1: Any other advanced form of binary operation is an extension or implication of
these four basic operations. For instance, the partial order a > b implies that a = b + c where
c > 0.
Remark 2: Using containers and objects, the operations of multiplication and division are the
same except that the results are interpreted differently: for multiplication the sum of contents
of all the containers is the result (product); for division the content of each of the containers
which is uniform is the result (quotient) as depicted in Fig. 3.
Remark 3: This containerized activity advances from arithmetic to algebra sense, with a
problem considered to be an abstract container of five sets of abstract particles of constant or
variable values (i.e. aim, subjects, objectives, knowledge and execution, abbreviated:
ASOKE) which are identified and manipulated in problem solving. Each of these particles or
sets there-of is of distinct identity and character, but all must be present in the abstract
container, else there is no problem. The attributes of particles in a container inform whether
the problem is framed within one or more of the basic strands of mathematics i.e. arithmetic
or geometry or algebra or change or a mix of at least two of the basic strands of mathematics.
Thus these particles converge into what is called a system or problem. Particle ‘A’ stands for
aim of the problem (the solution of the problem). Particle ‘S’ stands for the set of subjects
(Sn) of constant or variable values, involved in the problem and interacting in either direct or
inverse relationships (whether jointly as factors or partially as parts added up). Particle ‘O’
stands for objective (or object) in the problem. The objective (which is either of constant or
variable value) serves as the intersecting (or common) element in the container by virtue of
its being the basis for the interaction of the subjects. The absence of O nullifies existence of a
problem since the subjects will have nothing in common (i.e. nothing connecting them)
except their mere presence. An example is having a deaf and a vocalist in a container as
subjects and then the object being ‘hear’ which does not constitute an intersection between
the two sets of subjects. Thus the problem, as to whether communication took place by
speech or voice, is a nullity. The particle ‘K’ stands for knowledge of existence of A, S, O
and which of S or O are held constant or varying and whether O1,O2,…On, S1,S2 ,…,Sn are
inversely or directly proportional in the relation, what initial and final values of O1,O2,…On,
S1,S2 ,…,Sn are supplied as initial or final conditions, and knowledge of execution of the
defined operation(s). The particle ‘E’ stands for execution of A. This is sheer interpretation
and imputation of supplied values in the defined relation based on K and then manipulation
or rearrangement of values of the variables usually done with use of inverses. If E does not
take place, the problem remains unsolved.
There are two broad sets of problems: problems involving directly or inversely related
particles O and S in the container. Both directly and inversely related particles often take
place simultaneously in what is known as joint variations or partial variations (and normally
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every part or term of a partial variation is considered as a number made up of factors which
could be of constant or variable values). The direct and inverse relations are considered as
follows, with illustrations.
Problems involving Inversely Related Particles (i.e. increase in value of a subject, implies
decrease in value of another or other subjects; while Value of the object remains the same):
Here, Particle O (common object of activity or problem) is held constant in value while some
or all of Particles S n  S1 , S 2 ,... vary in value. In other words, if Object’s value is held
constant the problem is an inverse problem – or the term of the expression is an inverse term
such that increase in the value of one subject creates a decrease in the value of another or
other subject(s).
C
C
(1)
 S1 
 S2 
 C  S1 S 2 .
S2
S1
where C is the constant value obtainable whenever the subjects are mixed. For example: If 10
men can paint a house in 3 days. To find out how long the house could be painted if only 6
men are available: Here, we identify the subjects Sn=(S1,S2): S1  men, S2  days, while the
object O  painting a house. To solve the problem we note what final value(s) are given (S1 =
6 men); and further establish the problem’s aim: A = to establish the number of days (S2) for
which painting of the house could be done given 6 men (S1). We then establish: what values
vary or are constant among S1,S2 and O. Here, O is constant while S1,S2 vary. Since O is
constant, the relation is inversely, Using (1):  S1  C  S 2  C  C  S1S 2 . where C is
S2
S1
the proportionality constant. Initial values supplied are: S1=10, S2=3 which are used to define
C=S1S2=10(3)=30, thus making (1) to be
30 The final values supplied involve S1=6 which is imputed into (2) to execute A. That
S 
1
S2
is: E: S2  5 . Conclusion: It takes 5 days for 6 men to paint the same house which took 3
days for 10 men to paint.
Problems involving Directly Related Particles (i.e. increase in value of one implies increase
in value of the others): Here, Particle O varies as some or all of Particles S n  S1 , S 2 ,... vary.
Suppose that O varies with any of Sn, say S1. The aim (A) would then be to establish the
value of S1 when a new value of O is imputed, or vice versa. The variation of O indicates a
direct proportionality of O with S1 while other Sn are constant. i.e. the equation relating the
variables is:
(2).
S1  CO
For example, if it takes 10 men to paint 2 houses in a stated period of time. For 7 such
houses, to find how many men will do the job within the same period of time: Here, the
object (O) which is painting of houses varies in value from 2 to 7 and the subject (S1) which
is number of men involved also changes in value. The problem is thus a direct relation
because increase in value of object requires increase in value of subject. Given initial values:
S1=10, O =2, then C  s1  10  5 so that (2) becomes S1  (5)O Thus if O = 7 for 7 houses
O
2
to be painted, S1=5(7)=35. That is: it will take 35 men to paint 7 such houses which took 10
men to paint 2 of the houses within the same period of time.
Results, discussion and Recommendations
A total of 101 Primary IV pupils (average age: 10) of two separate schools in Benue State,
Nigeria were studied. The students were exposed to theory of containerization and a test
value of 50% set for acceptable mean mark of achievement. The scope of the test was limited
to arithmetic problem solving. The Arithmetic Achievement Test had 20 items out of which
11 were knowledge based, 5 were comprehension based and 4 were application based
4
according to recommended levels of the taxonomies of cognitive development for elementary
education. The scope of the test was limited to arithmetic problem solving. The Arithmetic
Achievement Test designed by the researcher was used as instrument for data collection. It
had a Kuder-Richardson 20 reliability coefficient of 0.88. The result of the study is shown in
Table 1 and Table 2.
Table 1: Mean Achievement of Sample after exposure to Theory of Containerization
___________________________________________________
N
Mean Std. Deviation Std Error Mean
___________________________________________________
Group
101
52.7228
7.70600
.76678
____________________________________________________
Data in Table 1 shows that the sampled students scored 52.7228 after exposure to theory of
containerization. This mean mark is above the cut off mark of 50 set for the study which
answers the research question.
Table 2: t-test result of Sample’s Achievement after exposure to Theory of containerization
_________________________________________________________________________
Test Value = 50
t
df
Sig. (2-tail) Mean Diff.
95% Confidence Int. of the Diff.
Lower
Upper
_________________________________________________________________________
Group
3.551
100
.001
2.72277
1.2015
4.2440
_________________________________________________________________________
Analysis in Table 2 shows a t value of 3.551 with significance of 0.001 which is below the set
alpha of 0.05 thus rendering the hypothesis rejected. This indicates that the difference
between students’ mean achievement and the set mean mark of 50 is significant. Thus based
on the evidence from the study, exposure to theory of containerization significantly enhances
learners’ mathematics problem solving.
The results may have confirmed that activity-based methods of teaching as problem oriented
enhance problem solving. The result also supports the findings by Atovigba & O’Kwu,
(2013) when the theory of containerization was exposed to primary school teachers in Benue
State, Nigeria at the Millennium Development Goals retraining programme.
Based on the results, the study recommends that the theory of containerization should be
considered as an addition to existing problem solving strategies.
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