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Transcript
Brief ideas about teaching algebra in the Soviet Union
1. Algebra started in the 6th grade (12-year old pupils)
2. In the 6th, 7th and 8th grades, it covered: Algebraic expressions, equations, inequalities,
functions, sequences (arithmetical, geometrical, oscillating and Fibonacci)
3. There were two main approaches to teaching algebra (and not only algebra):
 Calculative approach (“For the broad population”)
 Much attention paid to “calculative techniques”, not restricted only on simple
cases:
 Substituting
ab
a b
Examples:

 ... ; substitute a  2, b  1  3
2
( a  b) ( a  b) 2
a
x
a b
b
; substitute x 
a
ab
x
b
By substituting y  4 x , solve the equation 5 4 x  x  0 .

Manipulating expressions, often based on formulas for:
( a  b) 2 , ( a  b)3 , a 2  b 2 , a 3  b 3 , a 4  b 4 , ( a  b ) 2 , …

Finding ranges of definition of algebraic expressions and functions
c 9
Examples: Find the range of the expression
c 1
5 x
Find the range of the function y  2
x  2 x  80
 Use of prescribed, “ready-made” algorithms
 Underestimating of pupils ability to work independently and creatively
 Scientific approach (“For the elite”)
 Less drill
 Introducing complicated concepts and relationships
 Big attention paid to the connections between algebra and geometry (e.g. via
parametric situations)
b
c
 Work with parametric situations
Example: Find all relationships among various values:
V
a
d
U
a
 Presentation of several solving strategies
 Preparing several advanced mathematical ideas (e.g. limits, reducible and
irreducible polynomials, …)
 Required rather big amount of home work
Example: Solving the equation |1 |1  x ||
1
2
Calculative approach:
Only one solving strategy - (a) the use of intervals
1
1
In  ,0  : |1 |1  x ||  x,  x  , x  
2
2
1
In (0, 1): |1 |1  x || x, x 
2
1
3
In (1, 2): |1 |1  x || 2  x, 2  x  , x 
2
2
1
5
In (2,  ): |1 |1  x || x  2, x  2  , x 
2
2
Scientific approach:
Three solving strategies – (a) the use of intervals, (b) graphs of functions, (c) geometrical
(b)
y  1 x
y  1 x
y |1  x |
y  1 x
y  1 |1  x |
y   |1  x |
y   |1  x |
y  1 x
y |1 |1  x ||
y |1 |1  x ||
1
½
0
1
1
3
y y
2
2
2
1
1
3
|1  x |  x   x 
2
2
2
3
1
5
|1  x |  x    x 
2
2
2
(c) | a  b | is the distance of a and b; |1  y |
1
2
1
3
2
Results
 Entrance exams for important Universities based on complicated manipulations with
algebraic expressions and functions
Examples :
1977
2 y 3  2 x 2  3x  3  0,

Solve the system of equations
2 z 3  2 y 2  3 y  3  0,
2 x3  2 z 2  3 z  3  0.

1978

1 a
,
Find all a for which the system of inequalities
a  1 has a solution.
3x 2  10 xy  5 y 2  2
x 2  2 xy  7 y 2 

Solve all whole numbers solving the equation cos( (3 x  9 x 2  160 x  800))  1.
8
1983


Solve the equation cos( (3 x  9 x 2  160 x  800))  1.
8
 The idea of “a big nation” can be clearly seen – only about 1.5 % of students understood
algebra, but these were excellent