Зборник на трудови од IV конгрес
на математичарите на Македонија
Струга, Македонија, 19-22.10. 2008
Стр. 41-50
Proceedings of IV congress
of mathematicians of Republic of Macedonia
Struga, Macedonia, 19-22.10. 2008
Pages 41-50
IDEAS ON FINDING SUM AND DIFFERENCE UP TO 20 IN
TEACHING PRIMARY SCHOOL MATHEMATICS
Ivanka Mincheva Georgieva
Abstract. Aim of the paper is to explore some didactic ideas on ways to solve
arithmetic problems in teaching primary school mathematics. Some representative
groups of problems are used and different ways, including practical ones, of solving
them are proposed.
1. Introduction
The operations addition and subtraction within the set of natural
numbers are introduced in first grade mathematics teaching and are one of the
basic arithmetic pieces of knowledge taught in this period. The skills, connected
with addition and subtraction include not only finding sum and difference but
also searching for and applying different ways of solving problems of type
mentioned above. In this way generalized skills for solving arithmetic problems
in next primary school years as well as in next stages of mathematics teaching
are built. The aim of the paper is to propose ideas which pupils need to solve
groups of problems on addition and subtraction up to 20.
2. Motivation
Finding sum and difference of natural numbers is a basic activity in 1st
grade mathematics teaching. However most of the problems usually taught are
solved by one or two ways which fact leads to mechanic search for the answer of
a given problem counting primarily on memorized problems and cases of
addition and subtraction. That is why here representative groups of problems and
various ways of their accessible solving in 1st grade are presented. By doing that
development of flexible thinking of pupils is reached to and generalized skills
for solving groups of arithmetic problems are built.
3. Basic methodical ideas for solving addition and subtraction
problems with up to 20 natural numbers.
Theoretical basis for introduction the operations of adding and
subtracting in 1st grade teaching are respectively union and difference of sets (in
the case of difference of sets the subtrahend set is a subset of the minuend set).
We do not consider problems dealing with introduction a given case of addition
or subtraction. Aims of the ideas below are to clarify logical structure of a given
problem solution and to suggest various accessible ways for solving the
problem. Two systems of basic representative problems of addition and
41
subtraction up to 20 are chosen. The systemizing is based on the connection
between addition and subtraction. The cases chosen are presented by two groups
of problems in order to differentiate between addition “to” 10, subtraction
“down to” 10 and subtraction “of” 10. Systemizing on these principles provides
comprehensiveness, accessibility and consecutiveness after all the cases from
addition and subtraction up to 20 are assimilated.
3.1. Ideas for solving problems of the type: 10 + 5 = ?, 15 – 5 = ?, 15 – 10 = ?
10 + 5 = ?
Writing down
Ways of solving – logical
Basic component problems
the solution in
structure of the solution
the real teaching
practice
Practical activities connected
with union of sets (adding
********** → 15
objects)
10 + 5 = 15
*****
10 + 5 = 1ten + 5units = 15
10 = ?tens
10 + 5 = 15
1tens + 5units = ?
10 + 1 = ?
5 1+1+1+1++1+1
10 + 5 = 10 + 1 + 1 + 1 + 1 + 1 Counting from 10 by units in
10 + 5 = 15
= 15
right order
Inductive reasoning through
which a conclusion for the way
10 + 2 = 12
10 + 5 = 15
of adding to 10 of one unit
10 + 4 = 14
number is reached to
10 + 7 = 17
Therefore 10 + 5 = 15
15 – 5 = ?
Ways of solving – logical
structure of the solution
**********
*****
15 – 5 = (1ten + 5units) –
5units = 1ten + (5units –
5units) = 1ten + 0units = 10
Basic component problems
Practical activities connected
with difference of sets (taking
away objects)
15 = ?tens + ?units
(а + в) – в = а + (в – в) = а +
0=а
1ten + 0units = ?
Remark 1: The second component problem is actually applying
42
Writing down
the solution in
the real teaching
practice
15 – 5 = 10
15 – 5 = €
15 = 10 + 5
5–5=0
15 – 5 = 10
the associative property but its writing down in 1st grade is nonaccessible. The property however can be presented pictorially
(look at the practical solution above).
15 = 10 + ?
15 – 5 = (10 + 5) – 5 = 10 + (5
– 5) = 10 + 0 = 10
(а + в) – в = а + (в – в) = а +
0=а
5–5=?
15 – 5 = €
15 = 10 + 5
10 + 0 = ?
5–5=0
Remark 2: This way is equivalent to the one before but is 15 – 5 = 10
didactically different from it because an explicit writing the
decimal form of 15, 5 and 10 is not given. Known problems of
adding and subtracting are used here (given here as component
problems).
Counting from 15 by units in
15 – 5 = 15 – 1 – 1 – 1 – 1 – 1 reverse order
15 – 5 = 10
= 10
Relationship between addition
We know:
and subtraction in direction
10 + 5 = 15.
Then:
10 + 5 = 15, therefore 15 – 5 = from addition to subtraction:
15 – 5 = 10
10
(а + в) – в = а
Relationship between addition
and subtraction in direction
from subtraction to addition
15 – 5 = 10,
(definition of difference)
because
10 + 5 = 15
? + 5 = 15 – finding the
15 – 5 = ?, so that ? + 5 = 15
unknown number through a
known sum (10 + 5) or
through adding to 5 up to 15
14 – 4 = 10
11 – 1 = 10
19 – 9 = 10
Therefore 15 – 5 = 10
Inductive reasoning through
which a conclusion for the way
of subtracting the units of a
two digit number is reached to
15 – 5 = 10
43
15 – 10 = ?
Ways of solving – logical
structure of the solution
**********
*****
Basic component problems
Practical activities connected
with difference of sets (taking
away objects)
15 = ?tens + ?units
15 – 10 = (1ten + 5units) –
1ten = (1ten – 1ten) + 5units =
0tens + 5units = 5
(а + в) – а =( а – а) + в = 0
+в=в
0tens + 5units = ?
Writing down
the solution in
the real teaching
practice
15 – 10 = 5
15 – 10 = €
15 = 1ten + 5units
1ten – 1ten =
0tens
0tens+ 5units = 5
Remark 3: Analogous to remark 1.
15 = 10 + ?
15 – 10 = (10 + 5) – 10 = (10
– 10) + 5 = 0 + 5 = 5
(а + в) – а =( а – а) + в = 0
+в=в
10 – 10 = ?
0+5=?
Remark 4: Analogous to remark 2.
Counting from 15 by fives in
15 – 5 = 15 – 5 – 5 = 5
reverse order
Relationship between addition
and subtraction in direction
10 + 5 = 15, therefore 15 – 10 from addition to subtraction:
=5
(а + в) – а = в
Relationship between addition
and subtraction in direction
from subtraction to addition
(definition of difference)
15 – 10 = ?, so that ? + 10 = 15
? + 10 = 15 – finding the
unknown number through a
known sum (5 + 10) or
through adding to 10 up to 15
14 – 10 = 4
11 – 10 = 1
16 – 10 = 6
Therefore 15 – 10 = 5
44
Inductive reasoning through
which a conclusion for the way
of subtracting the ten from a
two digit number is reached to
15 – 10 = €
15 = 10 + 5
10 – 10 = 0
0+5=5
15 – 10 = 5
15 – 10 = 5
We know:
10 + 5 = 15.
Then:
15 – 10 = 5
15 – 10 = 5,
because
5 + 10 = 15
15 – 10 = 5
3.2. Ideas for solving problems of the type: 12 + 3 = ?, 15 – 3 = ?, 15 – 12 = ?
12 + 3 = ?
Ways of solving – logical
structure of the solution
Practical activities
connected with union of
sets (adding objects)
12 = ?tens + ?units
********** → 15
** ***
12 + 3 = (1ten + 2units) + 3units =
1ten + (2units + 3units) = 1ten +
5units = 15
12 + 3 = (10 + 2) + 3 = 10 + (2 + 3)
= 10 + 5 = 15
Remark 5: Analogous to remark 2.
12 + 3 = 12 + 1 + 1 + 1 = 15
+3
10
11
12
Basic component
problems
15
11 + 2 = 13
14 + 3 = 17
15 + 1 = 16
Therefore 12 + 3 = 15
(а + в) + с = а + (в + с)
1ten + 5units = ?
12 = 10 + ?
(а + в) + с = а + (в + с)
10 + 5 = ?
12 + 1 = ?
3 1+1+1
Counting from 12 by units
in right order
Practical activities for
calculating sum by using
numerical line (adding 3
segments to 12 segments)
Inductive reasoning
through which a
conclusion for the way of
adding to a two digit
number one digit number
with no passing over 10 is
reached to
Writing down
the solution in
the real
teaching
practice
12 + 3 = 15
12 + 3 = €
12 = 1ten +
2units
2units + 3units =
5units
1ten + 5units =
15
12 + 3 = €
12 = 10 + 2
2+3=5
10 + 5 = 15
12 + 3 = 15
12 + 3 = 15
12 + 3 = 15
45
15 – 3 = ?
Ways of solving – logical
structure of the solution
**********
** ***
15 – 3 = (1ten + 5units) – 3units =
1ten + (5units – 3units) = 1ten +
2units = 12
Basic component
problems
Practical activities
connected with difference
of sets (taking away
objects)
15 = ?tens + ?units
(а + в) – с = а + (в – с)
1ten + 2units = ?
15 = 10 + ?
15 – 3 = (10 + 5) – 3 = 10 + (5 – 3)
= 10 + 2 = 12
(а + в) – с = а + (в – с)
5–3=?
10 + 2 = ?
Remark 6: Analogous to remark 4 and remark 5.
15 – 1 = ?
3 1+1+1
15 – 3 = 15 – 1 – 1 – 1 = 12
Counting from 15 by units
in reverse order
–3
Practical activities for
10 11 12
15
finding difference using
numerical line (taking away
3 segments from 15
segments)
Relationship between
addition and subtraction in
12 + 3 = 15, therefore 15 – 3 = 12
direction from addition to
subtraction:
(а + в) – в = а
Relationship between
addition and subtraction in
direction from subtraction
to addition (definition of
difference)
15 – 3 = ?, therefore ? + 3 = 15
46
Writing down
the solution in
the real
teaching
practice
15 – 3 = 12
15 – 3 = €
15 = 1ten +
5units
5units – 3units =
2units
1ten + 2units =
12
15 – 3 = €
15 = 10 + 5
5–3=2
10 + 2 = 12
15 – 3 = 12
15 – 3 = 12
We know:
12 + 3 = 15.
Then:
15 – 3 = 12
15 – 3 = 12,
because
12 + 3 = 15
? + 3 = 15 – finding the
unknown number through a
known sum (12 + 3) or
through adding to 3 up to
15
14 – 3 = 11
18 – 5 = 13
16 – 2 = 14
Therefore 15 – 3 = 12
Inductive reasoning
hrough which a conclusion
or the way of subtracting
ne digit number from a two
digit number with no passing
ver 10 is reached to
15 – 3 = 12
15 – 12 = ?
Ways of solving – logical structure
of the solution
**********
*** **
15 – 12 = (1ten + 5units) – (1ten +
2units) = (1ten – 1ten) + (5units –
2units) = 0tens + 3units = 3
Basic component
problems
Practical activities
connected with difference
of sets (taking away
objects)
15 = ?tens + ?units
12 = ?tens + ?units
(а + в) – (с + d) =(а – с) +
(в – d)
0tens + 3units = ?
15 = 10 + ?
12 = 10 + ?
15 – 12 = (10 + 5) – (10 + 2) =
(10 – 10) + (5 – 2) = 0 + 3 = 3
(а + в) – (с + d) =(а – с) +
(в – d)
Writing down
the solution in
the real
teaching
practice
15 – 12 = 3
15 – 12 = €
15 = 1ten +
5units
12 = 1ten +
2units
1ten – 1ten =
0tens
0tens + 3units =
3
15 – 12 = €
15 = 10 + 5
12 = 10 + 2
10 – 10 = 0
0+3=3
10 – 10 = ?
5 – 2= ?
0+3=?
47
Remark 7: The second and third ways of solving are
mathematically equivalent but they are didactically different.
Remark 8: The rule for subtracting sum from sum is not taught in
primary school. However it may be accessibly presented by the
practical solution (first way of solution above).
15 – 12 = 15 – (10 + 2) = (15 – 10)
–2=5–2=3
12 = 10 + ?
15 – 10 = ?
5–2=?
15 – 12 = €
12 = 10 + 2
15 – 10 = 5
5–2=3
15 – 12 = 15 – (10 + 2) = (15 – 2)
– 10 = 13 – 10 = 3
12 = 10 + ?
15 – 2 = ?
13 – 10 = ?
15 – 12 = €
12 = 10 + 2
15 – 2 = 13
13 – 10 = 3
–2
3
– 10
5
15
Practical activities for finding difference using numerical line
(consecutively taking away 10 and 2 segments from 15 segments)
– 10
3
–2
13 15
Practical activities for finding difference using numerical line
(consecutively taking away 2 and 10 segments from 15 segments)
Relationship between
addition and subtraction in
12 + 3 = 15, therefore 15 – 12 = 3 direction from addition to
subtraction:
(а + в) – а = в
Relationship between
addition and subtraction in
direction from subtraction
to addition (definition of
difference)
15 – 12 = ?, so that ? + 12 = 15
? + 12 = 15 – finding the
48
15 – 12 = €
12 = 10 + 2
15 – 10 = 5
5–2=3
15 – 12 = €
12 = 10 + 2
15 – 2 = 13
13 – 10 = 3
We know:
12 + 3 = 15.
Then:
15 – 12 = 3
15 – 12 = 3,
because
3 + 12 = 15
unknown number through a
known sum (3 + 12) or
through adding to 12 up to
15
14 – 13 = 1
18 – 15 = 3
16 – 12 = 4
Therefore 15 – 12 = 3
Inductive reasoning through
which a conclusion for the
way of subtracting two
digit number from a two
digit number with no
passing over 10 is reached
to
15 – 12 = 3
4. General conclusions
1. The two groups of problems considered above are representative
examples of problems of given type and they are a model of a randomly
chosen concrete problem of addition or subtraction up to 20.
2. Logical structure of a given way of solution includes component
problems in a respective order. Taking it in consideration teachers can
be oriented to which problems to be reminded prior to introduction the
new problem as well as they can systemize the teaching material and
order problems so that each problem is followed by one using it in its
solution. In this way the basic didactic principle of accessibility of
teaching is fulfilled.
3. Including three problems in each group is connected with the
relationship between addition and subtraction which leads to connection
between problems and their solutions respectively.
4. Calculations while solving problems of the type 12 + 3 = ?, 15 – 3 = ?,
15 – 12 = ? could be written down vertically but the logical structure of
their solutions remains the same. A horizontal writing is chosen here
because of its association with mental calculations within numbers up to
20.
5. Most of the ways of solution suggested are mathematically equivalent.
However it is better they to be taught as relatively independent because
they are didactically different – by different component problems and
correspondent pieces of knowledge used or by different level of
abstraction of the solution.
6. The explicit use of associative property in some ways of solution is non
accessible but its accessible presentation is provided by the
correspondent practical solution.
Conclusion
Exploring the peculiarities of cases of addition and subtraction up to 20 is
useful due to the fact that these cases are component problems of all the rest
49
problems with whole numbers and fractions. This phenomenon makes necessary
understanding, assimilation and long term memorizing all the problems of this
group. Different ways of solution proposed here are accessible to 1st grade
pupils. However one or two of them are present in textbooks in mathematics as
well as in the real teaching practice. This usually leads to mechanical (without
understanding) and partial solving of a problem. That is why it is useful more
ways of solution to be learned. Doing so the logical thinking of pupils is
developed, it becomes flexible and combinatorial. In result when facing an
unknown problem the pupil finds out faster and rational way of its solution.
Taking into account all mentioned above we believe that the ideas of the
paper contribute to enrichment of theory and practice of 1st grade mathematics
teaching as well as the next grades and stages of mathematics teaching.
[1]
[2]
[3]
[4]
[5]
[6]
REFERENCES
Bogdanova, M., K. Nikova, N. Dimitrova. Mathematics for 1st grade.
Bulvest 2000. Sofia, 2003
Manova, A., R. Rangelova. Манова, А., Mathematics for 1st grade.
Prosveta. Sofia, 2002
Mincheva, I. Relationship addition – subtraction and multiplication –
division and preliminary learning the concept of equation in teaching primary
school mathematics. May readings 2007 – pending
Mincheva, I. Relationship addition – subtraction and multiplication –
division and the analogy between them in teaching primary school
mathematics. In (Ed. Tomislav Diakov) Continuity and perspectives in the
development of pedagogic theory and practice. 125 years preschool
education in Bulgaria (277 – 282). Veda Slovena – JG. Sofia, 2007
Mincheva, I. Relationship between the arithmetic operations addition and
subtraction in teaching primary school mathematics. In (Ed. Milka
Atanassova, Atanass Diakov) Training teachers and social pedagogues on the
eve of the European integration. (261 - 267). Veda Slovena – JG. Sofia, 2006
Mincheva, I. Methods of teaching primary school mathematics – special
issues. Astarta. Veliko Tarnovo, 2002
Ivanka Mincheva Georgieva
Veliko Tarnovo University
Department of Algebra and Geometry
Bulgaria
vmincheva2002@yahoo.com
The paper is sponsored by a Fund “Scientific research” of the University of Veliko
Tarnovo “St. Cyril and St. Methodius”, project № РД- 491- 06, 2008
50