Transcript From Arithmetic to Algebraic Reasoning

Learning Arithmetic as a
Foundation for Learning Algebra
Developing relational thinking
Adapted from…
Thomas Carpenter
University of Wisconsin-Madison
Defining Algebra
Many adults equate school algebra with symbol
manipulation– solving complicated equations
and simplifying algebraic expressions. Indeed,
the algebraic symbols and the procedures for
working with them are a towering , historic
mathematical accomplishment and are critical in
mathematical work. But algebra is more than
moving symbols around. Students need to
understand the concepts of algebra, the
structures and principles that govern the
manipulation of the symbols, and how the
symbols themselves can be used for recording
ideas and gaining insights into situations.
(NCTM, 2000, p. 37)
Never the twain shall meet

The artificial separation of arithmetic and
algebra deprives students of powerful
ways of thinking about mathematics in the
early grades and makes it more difficult
for them to learn algebra in the later
grades.
Arithmetic vs Algebra
Arithmetic


Calculating
answers
= signifies the
answer is next
Algebra


Transforming
expressions
= as a relation
Arithmetic U Algebra
Arithmetic


Transforming
expressions
= as a relation
Algebra


Transforming
expressions
= as a relation
Developing Algebraic Reasoning in
Elementary School

Rather than teaching algebra procedures
to elementary school children, our goal is
to support them to develop ways of
thinking about arithmetic that are more
consistent with the ways that students
have to think to learn algebra successfully.
Developing Algebraic Reasoning in
Elementary School
Enhances the learning of arithmetic in
the elementary grades.
 Smoothes the transition to learning
algebra in middle school and high
school.

Relational Thinking






Focusing on relations rather than only on
calculating answers
Looking at expressions and equations in their
entirety rather than as procedures to be carried
out step by step
Engaging in anticipatory thinking
Using fundamental properties of arithmetic to
relate or transform quantities and expressions
Recomposing numbers and expressions
Flexible use of operations and relations
6+2 =
□
+3
6+2 =



□
+3
David: It’s 5.
Ms. F: How do you know it’s 5, David?
David: It’s 6 + 2 there. There’s a 3 there.
I couldn’t decide between 5 and 7. Three
was one more than 2, and 5 was one less
than 6. So it was 5
57 + 38 = 56 + 39





David: I know it’s true, because it’s like the other one I
did, 6 + 2 is the same as 5 + 3.
Ms. F. It’s the same. How is it the same?
David: 57 is right there, and 56 is there, and 6 is there
and 5 is there, and there is 38 there and 39 there.
Ms. F. I’m a little confused. You said the 57 is like the 5
and the 56 is like the 6. Why?
David: Because the 5 and the 56, they both are one
number lower than the other number. The one by the
higher number is lowest, and the one by the lowest
number up there would be more. So it’s true.
57 + 38 = 56 + 39
(56 + 1) + 38 = 56 + (1 + 38)
Recomposing numbers
8+7 =
□
□
=□
8 + (2 + 5) =
(8 + 2) + 5
Recomposing numbers
½+¾ =
□
½ + (½ + ¼) =
□
Using basic properties
Relating arithmetic and algebra
70 + 40 = 7 X 10 + 4 X 10
= (7 + 4) X 10
= 110
7/12 + 4/12 = 7(1/12) + 4(1/12)
= (7 + 4) X 1/12
7a + 4a = 7(a) + 4(a)
= (7 + 4)a
= 11a
Using basic properties
(not)
7a + 4 b = 11ab
X2 - X - 2 = 0
(X – 2)(X + 1) = 0
X+1 = 0
X–2 = 0
X = -1
X = 2
X2 - X - 2 = 0
(X + 1)(X - 2) = 0
X+1 = 0
X–2 = 0
X = -1
X = 2
(X + 1)(X - 2) = 6
X2 - X - 2 = 0
(X + 1)(X - 2) = 0
X+1 = 0
X–2 = 0
X = -1
X = 2
(X + 1)(X - 2) = 6
X+1 = 6
X–2 = 6
X = 5
X = 8
X2 - X - 2 = 0
(X + 1)(X - 2) = 0
X+1 = 0
X = -1
X–2 = 0
X = 2
(X + 1)(X - 2) = 6
X+1 = 6
X = 5
X–2 = 6
X = 8
(5 + 1)(5 - 2) = 18
(8 + 1)(8 -2) = 54
Multiplication properties of zero

ax0 = 0

axb = 0 implies a = 0 or b = 0
Equality as a relation
8+4 =
□+5
Percent of Students Offering Various
Solutions to 8 + 4 =  + 5
Response/
Grade
7
12
17
12 & 17
1 and 2
?
?
?
?
3 and 4
?
?
?
?
5 and 6
?
?
?
?
24
Percent of Students Offering Various
Solutions to 8 + 4 =  + 5
Response/
Grade
7
12
17
12 & 17
1 and 2
5
58
13
8
3 and 4
9
49
25
10
5 and 6
2
76
21
2
Challenge--Try this!

What are the different responses that
students may give to the following
open number sentence:
9 + 7 = + 8
26
Challenging students’ conceptions
of equality




9
9
9
9
+
+
+
+
5
5
5
5
=
=
=
=
14
14 + 0
0 + 14
13 + 1
Challenging students’ conceptions
of equality





7+4
11 =
11 =
7+4
7+4
= 11
7+4
11
=7+4
= 4+7
Correct solutions to 8 + 4 =  + 5
before and after instruction
Grade
Before Inst.
After Inst.
1 and 2
5
66
3 and 4
9
72
5 and 6
2
84
Learning to think relationally,
thinking relationally to learn

Using true/false and open number
sentences (equations) to engage students
in thinking more flexibly and more deeply
about arithmetic
Learning to think relationally,
thinking relationally to learn


Using true/false and open number
sentences (equations) to engage students
in thinking more flexibly and more deeply
about arithmetic
in ways that are consistent with the ways
that they need to think about algebra.
True and false number sentences





7 + 5 = 12
5 + 6 = 13
457 + 356 = 543
7 13/16 – 2 17/18 = 4 11/15
12÷0 = 0
Challenge--Try this!

Construct a series of true/false
sentences that might be used to elicit
one of the conjectures in Table 4.1
(on p. 54-55).
34
Learning to think relationally
 26
+ 18 - 18 = 
 17 - 9 + 8 = 
Learning to think relationally
 750
7
+ 387 +250 = 
+9+8+3+1 =
More challenging problems
A. 98 + 62 = 93 + 63 + 
B. 82 – 39 = 85 – 37 - 
C. 45 – 28 =  - 24
True or False

35 + 47 = 37 + 45

35 × 47 = 37 × 45
35 + 47 = 37 + 45
True
30 + 5 + 40 + 7 = 30 + 7 + 40 + 5
35 × 47 = 37 × 45
False
35 × 47 = 37 × 45
False
35 × 47 =
(30 + 5) × (40 + 7) =
(30 + 5) ×40 + (30 + 5) × 7 =
(30×40 + 5×40) + (30×7 + 5×7)
37 × 45 =
(30 + 7) × (40 + 5) =
(30 + 7) ×40 + (30 + 7) × 5 =
(30×40 + 7×40) + (30×5 + 7×5)
35 × 47 = 37 × 45
False
35 × 47 =
(30 + 5) × (40 + 7) =
(30 + 5) ×40 + (30 + 5) × 7 =
(30×40 + 5×40) + (30×7 + 5×7)
37 × 45 =
(30 + 7) × (40 + 5) =
(30 + 7) ×40 + (30 + 7) × 5 =
(30×40 + 7×40) + (30×5 + 7×5)
Parallels with multiplying binomials

(X + 7)(X + 5) =
(X + 7)X + (X + 7) 5 =
X2 + 7X +5X + 35 =
X2 +(7 +5)X + 35 =
X2 +12X + 35
Thinking relationally to learn


Learning number facts with understanding
Constructing algorithms and procedures
for operating on whole numbers and
fractions
Number sentences to develop
Relational Thinking









(Large numbers are used to discourage calculation)
Rank from easiest to most difficult
a) 73 + 56 = 71 + d
b) 92 – 57 = g – 56
c) 68 + b = 57 + 69
d) 56 – 23 = f – 25
e) 96 + 67 = 67 + p
f) 87 + 45 = y + 46
g) 74 – 37 = 75 - q
Learning Multiplication facts
using
relational thinking


Julie Koehler Zeringue
A learning trajectory for
thinking relationally



Starting to think relationally
The equal sign as a relational symbol
Using relational thinking to learn
multiplication




Multiplication as repeated addition
Beginning to use the distributive property
Recognizing relations involving doubles, fives,
and tens
Appropriating relational strategies to derive
Multiplication as repeated addition
3
4
6
2
7
7
+6
9
=
=
=
=
7+7+7
7+7+7+b
26
h+h
Beginning to use the distributive property






36
36
54
56
+ 6 = 46
+ 3 = 46
= 24 + 4 + 8
= 36 + g
67 = a7 + b7
67 = h7 + h7
Multiplication facts
x
1
2
3
4
5
6
7
8
9
1
1
2
3
4
5
6
7
8
9
2
2
4
6
8
10
12
14
16
18
3
3
6
9
12
15
18
21
24
27
4
4
8
12 16
20
24
28
32
36
5
5
10 15 20
25
30
35
40
45
6
6
12 18 24
30
36
42
48
54
7
7
14 21 28
35
42
49
56
63
8
8
16 24 32
40
48
56
64
72
9
9
18 27 36
45
54
63
72
81
Generating number facts based on
doubles
3×8 = 2×8+8
3 × 8 = 16 + 8
3×8 = 8×3
4×9 = 2×9+2×9
Multiplication facts
x
1
2
3
4
5
6
7
8
9
1
1
2
3
4
5
6
7
8
9
2
2
4
6
8
10
12
14
16
18
3
3
6
9
12
15
18
21
24
27
4
4
8
12 16
20
24
28
32
36
5
5
10 15 20
25
30
35
40
45
6
6
12 18 24
30
36
42
48
54
7
7
14 21 28
35
42
49
56
63
8
8
16 24 32
40
48
56
64
72
9
9
18 27 36
45
54
63
72
81
Generating number facts for nines
and fives
9 × 7 = 10 × 7 – 7
9 × 7 = 10 × 7 – 9
9 × 7 = 10 × 7 -
□
7 × 5 = 10 + 10 + 10 + 5
Multiplication facts
x
1
2
3
4
5
6
7
8
9
1
1
2
3
4
5
6
7
8
9
2
2
4
6
8
10
12
14
16
18
3
3
6
9
12
15
18
21
24
27
4
4
8
12 16
20
24
28
32
36
5
5
10 15 20
25
30
35
40
45
6
6
12 18 24
30
36
42
48
54
7
7
14 21 28
35
42
49
56
63
8
8
16 24 32
40
48
56
64
72
9
9
18 27 36
45
54
63
72
81
Appropriating relational strategies
to derive number facts
6×6 = 6×5+6
7×8 = 7×9–7
A learning trajectory for
thinking relationally
Starting to think relationally
The equal sign as a relational symbol
Using relational thinking to learn
multiplication







Multiplication as repeated addition
Beginning to use the distributive property
Recognizing relations involving doubles, fives,
and tens
Appropriating relational strategies to derive
number facts
37 = 7 + 7 + 7
Ms L: “Could you read that number
sentence for me and tell me if it is true or
false”?
Kelly: “Three times 7 is the same as 7 plus 7
plus 7. That’s true, because times means
groups of and there are 3 groups of 7, 3
times 7 just says it in a shorter way”.
Ms L: “Ok, nice explanation”.
37 = 14 + 7
Ms L: “How about this, 37 = 14 + 7, is that true or
false”?
Kelly: “It’s true”.
Ms L: “Wow, that was quick, how do you know that is
true”?
Kelly: “Can we go back up here [pointing to 37 = 7 +
7 + 7]”?
Ms L: “Sure”.
Kelly: “Seven and 7 is 14, that is right here [drawing a
line connecting two 7s in the first number sentence
and writing 14 under them]. Fourteen went right into
here [pointing to the 14 in the second number
sentence]. Then there is one 7 left pointing to the
third 7 in the first number sentence], and that went
right here [pointing to the last 7 in the second
number sentence]”.
46 = 12 + 12
Ms L: “Ok, I have another one for you 46 = 12 + 12, true
or false”?
Kelly: “That is true”.
Ms L: “Ok, how did you get that one so quickly”?
Kelly: “Six plus 6 is 12, in this case, there are 4 groups of
6, so it is like this [writing 6 + 6 + 6 + 6]. Six and 6 is
12, that leaves another 6 and 6, and that equals 12. So
one 12 is here and one 12 went here [indicating the two
12s in the problem]. What I’m trying to say is there are
four 6s and you broke them in half and made them into
two 12s”.
46 = 12 + 12 Continued
Ms L: “Nice! Kelly, do you know right away what 4 times 6 is”?
Kelly: “Yes”.
Ms L: “What is it?”
Kelly: “It’s [pause] thirty- [long pause] two.”
Ms L: “Ok, do you know what 12 plus 12 is”?
Kelly: “Yeah. That is the same thing, 32”.
Ms L:” Do you have a way of doing 12 plus 12, to check it”?
Kelly: “Well, there are two 10s, 20- oh wait, I was thinking of a
different one”!
Ms L: “You were thinking of a different multiplication problem”?
Kelly: “Yes. 4 times 6 is 24, because 10 and 10 is 20, and 2 and 2 is 4,
put those together and its 24”.
47 =
□
Ms L: “Ok, here is another one. Four times 7 equals box. I want you
tell me what you would put in the box to make this a true number
sentence”.
Kelly: “That would be [short pause] 28”.
Ms L: “Ok, how did you get 28”?
Kelly: “Well, I kinda had other problems… that went into this problem.
If you go up here [pointing to 37 = 7 + 7 + 7] 3 times 7 is the
same as 7 plus 7 plus 7. That problem helped me and I used it with
this problem, [pointing to 37 = 14 + 7] 3 sevens is the same as 14
and 7… You add one more seven and that goes right to here. [Then
she points to 46 = 12 + 12.] This problem also helped me because
47 is like… My mind went back up to here [pointing to 37 = 14 +
7], and I said, there is another 7 so I could put those two 7s
together, that’s 14, and there are two 14s, 10 and 10 is 20, 4 and 4
is 8, 28”.
Problem sequence
37 = 7 + 7 + 7
37 = 14 + 7
46 = 12 + 12
47 =
□
Inventing algorithms
612
-457
200
-40
-5
155
300
-299
1
292
-549
-300
50
-7
-257
87
-49.02
40.00
-2.00
-.02
37.98
300
-294 5/8
6
-5/8
5 3/8
5 ½ ÷ __ = 
Number choices ½, ¼, ¾, 3/8



Choose one of the numbers depending on
the level of your students.
Change the problem to add additional
challenge as needed.
Allow students to choose one to vary the
problem and allow choice.
5 ½ ÷ __ = 
Number choices ½, ¼, ¾, 3/8



Put the problem into a context.
It takes __ of a cup of sugar to make a batch of
cookies. I have 5 ½ cups of sugar. How many
batches of cookies can I make?
Solve for 3/8 cup of sugar for a batch.
It takes 3/8 of a cup of sugar to make a batch of cookies. I
have 5 ½ cups of sugar. How many batches of cookies can
I make?
It takes 3/8 of a cup of sugar to make a batch of
cookies. I have 5 ½ cups of sugar. How many
batches of cookies can I make?

5 ½ ÷ 3/8 = 

 × 3/8 = 5 ½
 × 3/8 = 5 ½







8 × 3/8 = 3
4 × 3/8 would be ½ of 3 or 1 ½
12 × 3/8 = 4 ½ => 4 ½ cups makes 12 batches
Need to use 1 more cup of sugar
Because 8 × 3/8 = 3, a third as much would be 1
i.e. 1/3 × (8 × 3/8) = 1
So you need 1/3 of 8, which is 8/3
i.e. 1 cup makes 8/3 batches.
So altogether you get a total of 12 + 8/3 or 14 2/3 batches
 × 3/8 = 5 ½

8 × 3/8 = 3

½(8 × 3/8) = ½ × 3

(½×8)×3/8 = 1 ½

4 ×3/8 = 1 ½
Next subgoal:
How many 3/8 cups to use the remaining cup?

8 × 3/8 = 3

1/3 × (8 × 3/8) = 1/3 × 3

(1/3×8)×3/8 = 1

8/3 ×3/8 = 1
Putting the parts together



(8 ×3/8) + (4 ×3/8) + (8/3 ×3/8) =
3+1½+1 =5½
And
(8 ×3/8) + (4 ×3/8) + (8/3 ×3/8) =
(8 + 4 + 8/3) ×3/8 = 14 2/3 ×3/8
So 5 ½ cups of sugar makes 14 2/3 batches of 3/8 cups of sugar
Solving equations
k
+ k + 13 = k + 20
From arithmetic to algebraic
reasoning




Attend to relations rather than teaching
only step by step procedures
Align the teaching of arithmetic with the
concepts and skills students need to learn
algebra
Enhance the learning of arithmetic
Provide a foundation for and smooth the
transition to learning algebra
Learning arithmetic and algebra with
understanding




Algebra for all
Not watering down algebra to teach isolated
procedures
Develop algebraic reasoning rather than
teaching meaningless algebraic procedures
Learning arithmetic and algebra grounded in
fundamental properties of number and number
operations
Assignment
What you will do before the next meeting
You will be writing a series of problems that
you might use with your students to
encourage them to begin to look for
relations.
 Write a problem to assess student thinking
 Predict student responses
 Write a series of problems to address your
students…back up…extend???
 Try these with your students
Assignment cont’d




You will be planning a lesson with your
colleagues for a lesson study cycle
As a team choose a topic to be taught on
December 6th
Choose a topic that is typically difficult for
students
Bring planning materials to the October
session
Challenge

a)
b)
Addition is associative, but
subtraction is not. How about the
following:
Is (a + b) - c = a + (b - c) true for
all numbers?
Is (a - b) + c = a - (b + c) true for
all numbers?
Thinking Mathematically p. 120 #4
81
Challenge
What kind of number do you get
when you add three odd numbers?
 Can you justify your response?


Thinking Mathematically p.103 #1
82
Challenge--Try this!


Design a sequence of true/false
and/or open sentences that you
might use to engage your students in
thinking about the equal sign.
Thinking Mathematically p. 24 #4
83
References




Carpenter, T.P., Franke, M.L., & Levi, L. (2003).
Thinking mathematically: Integrating arithmetic and
algebra in the elementary school. Portsmouth, NH:
Heinemann.
Carpenter, T. P., Franke, M.L., & Levi, L. (2005). Algebra
in Elementary School. ZDM. 37(1), 1-7.
Carpenter, T. P., Levi, L., Berman, P., & Pligge, M.
(2005). Developing algebraic reasoning in the
elementary school. In T. A. Romberg , T. P. Carpenter,
& F. Dremock (Eds). Understanding mathematics and
science matters. Mahwah, NJ: Erlbaum.
Jacobs, V.J., Franke, M.L., Carpenter, T. P., Levi, L., &
Battey, D. (2007) A large-scale study of professional
development focused on children’s algebraic reasoning in
elementary school. Journal for Research in Mathematics
Education, 38, 258-288.