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Mathematics
as a
Second Language
Arithmetic Revisited
© 2010 Herb I. Gross
Developed by
Herb I. Gross and Richard A. Medeiros
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Lesson 2 Part 1
Whole Number
Arithmetic
Addition
© 2010 Herb I. Gross
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Addition Through the Eyes
of Place Value
The idea of numbers being viewed as
adjectives not only provides a clear
conceptual foundation for addition, but
when combined with the ideas of place
value yields a powerful computational
technique. In fact, with only a knowledge
of the ordinary 0 through 9 addition tables
(i.e., addition of single digit numbers), our
“adjective/noun” theme allows us to easily
add any collection of whole numbers.
© 2010 Herb I. Gross
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The main idea is that
in our place value system,
numerals in the same column
modify the same noun.
Hence, we just add the
adjectives and “keep” the
noun that specifies the
column.
© 2010 Herb I. Gross
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To illustrate the idea, let’s carefully
analyze how we add the two numbers
342 and 517. According to our knowledge of
the place value representation of numbers,
we set up the problem as follows…
© 2010 Herb I. Gross
Hundreds
Tens
Ones
3
5
4
1
2
7
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In each column we use the addition
table for single digits. We then solve
the above problem by treating it as if it
were three single digit addition problems.
Namely…
adjective
noun
adjective
noun
adjective
noun
3
5
8
hundreds
4
1
5
tens
2
7
9
ones
hundreds
hundreds
© 2010 Herb I. Gross
ten
tens
ones
ones
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Of course, in everyday usage we do not
have to write out the names of the nouns
explicitly since the digits themselves hold
the place of the nouns.
Thus, instead of using the chart form below…
…we usually perform the addition in the
following succinct form…
Hundreds
Tens
Ones
3
5
4
1
2
7
342
+ 517
8
5
9
859
© 2010 Herb I. Gross
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Since the nouns are not visible in the
customary format for doing place value
addition, it is important for a student to
keep the nouns for each column in mind.
For example, in reading the leftmost
column of the above solution out loud (or
silently to oneself) a student should be
saying…
“3 hundred + 5 hundred = 8 hundred”
rather than just using the adjectives,
as in “3 + 5 = 8.”
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In that way, one reads 859, the answer
to 342 + 517, as…
“8 hundreds, 5 tens, and 9 ones.”1
In using place value to perform the above
addition problem, you may have missed
our subtle use of the associative and
commutative properties of addition.
note
1 Or in every-day terminology, we would read the solution as “eight hundred fifty-nine”.
© 2010 Herb I. Gross
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Note
The commutative property of
addition is a more formal way
of saying that the sum of two
numbers does not depend on
the order in which the two
numbers are written.
For example, 3 + 5 = 5 + 3.
Stated more generally, it says if a and b
denote any numbers, then a + b = b + a.
© 2010 Herb I. Gross
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The associative property of
addition is a more formal way of
saying that the sum of three
Note
(or more) numbers does not
depend on the how the numbers
are grouped.
For example, (3 + 4) + 5 = 3 + (4 + 5).
More generally, it says if a, b, and c denote
any numbers, then (a + b) + c = a + (b + c)2.
note
2
Mathematicians use parenthesis in the same way that hyphens are used in grammatical
expressions. That is, everything in parentheses is considered to be one number. Thus,
(3 + 4) + 5 tells us that we first add the 3 and 4 and then add 5; while 3 + (4 + 5) tells us
to add the sum of 4 and 5 to 3.
© 2010 Herb I. Gross
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Thus,
342 + 517
is an abbreviation for writing…
(3 hundreds + 4 tens + 2 ones) + (5 hundreds + 1 ten + 7 ones)
However, in using the
vertical form of addition,
we had actually used the
rearrangement…
342
+ 517
(3 hundreds + 5 hundreds) + (4 tens + 1 ten) + (2 ones + 7 ones)
© 2010 Herb I. Gross
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So whether or not we know the
formal terminology, the fact
remains that the vertical format for
doing addition of whole numbers is
justified by the associative and
commutative properties of addition.
© 2010 Herb I. Gross
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Classroom Application
Using “play money”, give each student
3 hundred dollar bills,
4 ten dollar bills, and
2 one dollar bills.
Then, give them…
5 more hundred dollar bills,
1 more ten dollar bill, and
7 more one dollar bills.
Then, ask them how much money each
of them has.
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Classroom Application
See how many of them simply combine the
bills the way we do in vertical addition;
that is…
the 3 hundred dollar bills with the 5 hundred dollar
bills; the 4 ten dollar bills with the 1 ten dollar bill;
and the 2 one dollar bills with the 7 one dollar bills.
If they do this, they are painlessly using the
commutative and associative properties
of addition.
© 2010 Herb I. Gross
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Note
There is a difference between
a job being “difficult” and
just being “tedious”.
For example, we see from the
illustration below that it is no more difficult
to add, say, twelve-digit numbers than
three-digit numbers. It is just more tedious
(actually, more repetitious).
2 3 4, 2 6 7, 5 8 0, 2 9 4
+ 3 5 2, 3 1 2, 2 1 9, 6 0 2
© 2010 Herb I. Gross
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Note
2 3 4, 2 6 7, 5 8 0, 2 9 4
+ 3 5 2, 3 1 2, 2 1 9, 6 0 2
That is, instead of carrying out
three simple single-digit
addition procedures we have to
carry out twelve.
© 2010 Herb I. Gross
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In general, no matter how many
digits there are in the numbers
that are being added, the process
remains the same. Namely…
Note
2 3 4, 2 6 7, 5 8 0, 2 9 4
+ 3 5 2, 3 1 2, 2 1 9, 6 0 2
5 8 6, 5 7 9, 7 9 9, 8 9 6
The problem is very easy, but requires
some patience.
© 2010 Herb I. Gross
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Practice Problem #1
In terms of the adjective/noun theme, how
would you correct a student who had
made the following error, namely…
to add 234 and 45, the student, believing
that numbers should be aligned from left
to right, writes…
and obtains the result…
© 2010 Herb I. Gross
234
+ 45
684
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Solution for Practice Problem #1
Place value addition is based on the fact
that numbers in the same column must
modify the same noun.
Notice that the 2 in 234 is modifying
hundreds while the 4 in 45 is modifying tens.
So in adding 234 and 45, when the student
wrote 2 + 4 = 6; in place value notation he
was saying that
2 hundreds + 4 tens = 6 hundreds
(and also that 3 tens + 5 ones = 8 tens).
© 2010 Herb I. Gross
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Notes
on Practice
Problem #1
This error couldn’t happen in
Roman numerals because the
nouns are visible.
In other words, if you wrote the
problem in the form…
CCC XXX IIII
XXXX IIIII
…the fact remains that X means ten no
matter where it is placed.
© 2010 Herb I. Gross
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Notes
on Practice
Problem #1
Even if the student is unaware
of the adjective/noun theme, a
little number sense should
warn the students that the
answer 684 can’t possibly be
correct.
Namely, since 234 + 100 = 334,
and 45 is less than 100…
234 + 45 must be less than 334.
Clearly 684 is not less than 334.
© 2010 Herb I. Gross
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“Trading In” or “Carrying”
Because the nouns are not visible in the
place value representation of a number,
certain ambiguities can occur that require
resolution.
Suppose,
for example,
that you have
3 $10-bills and
5 $1-bills.
© 2010 Herb I. Gross
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Someone then gives you 2 more
$10-bills and 9 more $1-bills. It is clear that
you now have a total of 5 $10-bills and 14
$1-bills.
© 2010 Herb I. Gross
If you want to (but you certainly don’t
have to) you may exchange ten of your
$1-bills for one $10-bill; thus leaving you
with six $10-bills and 4 $1-bills.
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After this exchange
you have $64,
just as before.
© 2010 Herb I. Gross
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The same reasoning applies to the use
of Roman numerals. Namely, since the
numerals are visible we do not have to
restrict ourselves to having no more than
nine of any denomination.
For example, we can write the sum of, say,
67 and 54 as XXXXXX I I I I I I I XXXXX I I I I.
If we wish to “economize” inXour use of
symbols, we exchange ten I’s for an X ten
and ten X’s forCone C to obtain…
XXXXXX I I I I I I I
© 2010 Herb I. Gross
XXXXX I I I I
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The point is that as long as the nouns
are visible it is okay to have more than 9 of
any denomination.
However, if we wish, we may exchange
10 $1-bills for 1 $10-bill. That is, Line 1
and Line 2 in the chart below provide two
different ways to represent the same
amount of money.
$10-Bills $1-Bills
© 2010 Herb I. Gross
Line 1
5
14
Line 2
6
4
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$10-Bills $1-Bills
Line 1
5
514
14
However, if the nouns are now omitted, all
we see is Line 1 in the form 514.
How can we tell whether we are naming
5 hundreds, 1 ten, and 4 ones or
5 tens and 14 ones [that is, 5(14)]?
This is a problem that many students
encounter when first learning to add.
© 2010 Herb I. Gross
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Namely, given an addition problem
such as 35 + 29, students will often write
the problem in vertical form and treat it as
if it involved two separate single digit
addition problems.
For example…
3 5
+ 2 9
3
5 (14 )
note
3
If we wanted to use grouping symbols we could write 5(14) to indicate that there are 14 ones
and 5 tens; but with numbers having a greater number of digits this would quickly become very
cumbersome.
© 2010 Herb I. Gross
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3 5
+ 2 9
5 14
To avoid such ambiguities as illustrated
above in which 5 tens and 14 ones can be
confused with 5 hundreds, 1 ten and
4 ones, we adopt the following convention
(or agreement) for writing a number
in place value.
We never use more than one digit per
place value column.
© 2010 Herb I. Gross
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By keeping this agreement in mind,
we avoid the type of confusion that
results in writing 514 dollars
when 64 dollars is meant.
The notion of trading in ten 1’s for one 10
is precisely the logic behind the concept
usually referred to, in the “traditional”
mathematics curriculum, as carrying and
in the “modern” mathematics curriculum,
as regrouping.
© 2010 Herb I. Gross
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Trading-in/Carrying/Regrouping
1
Thus, for example,
3 5
in computing the sum…
+ 2 9
4
we often start by saying something like
“5 plus 9 equals 14. Put down the 4 and
carry the 1”.
By placing the 1 over the 3 and noting that 3
is in the tens place, what we have said is
5 ones + 9 ones = 14 ones = 1 ten + 4 ones.
© 2010 Herb I. Gross
Continuing with this concept, one can
lead a student in a step-by-step fashion
through the process of “carrying” by initially
allowing the denominations to be visible.
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For example, to compute the sum,
5,286 + 2,959, we would first rewrite the
problem as…
thousands
5
2
+
hundreds
2
9
tens
8
5
ones
6
9
7
11
13
15
© 2010 Herb I. Gross
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Notice that at this stage of the process
there is no need to exchange ten of any
denomination for one of the next
denomination (unless one feels like doing
it) because the denominations are visible.
A more tangible way to see this is in
terms of our play money model.
© 2010 Herb I. Gross
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Namely, suppose you have “play
money” in the classroom, and you first
hand the student five $1,000-bills,
two $100- bills, eight $10-bills, and
six $1-bills.
Then you hand the student an additional
two $1,000-bills, nine $100-bills,
five $10-bills and nine $1-bills.
Altogether, the student sees that he/she
has seven $1,000 bills, eleven $100- bills,
thirteen $10-bills and fifteen $1-bills.
© 2010 Herb I. Gross
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Including all denominations,
the student now has a total of 46 bills, and
may wish to have a smaller stack but yet
have the same amount of money.
Thus, the student can systematically
proceed to exchange currency by
converting ten of one denomination into
one of the next denomination,
beginning with the lowest denomination
and proceeding step-by-step to
the higher denominations.
© 2010 Herb I. Gross
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The following chart shows each step
of the regrouping process.
$1,000 bills
5
2
+
$100 bills
2
9
7
11
Step 1
Step 2
Step 3
© 2010 Herb I. Gross
7
7
7
8
11
12
11
12
2
$10 bills $1 bills
8
6
5
9
13
14
13
14
4
4
15
15
5
5
5
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$1,000 bills
Step 1
Step 2
Step 3
7
7
7
8
$100 bills
11
12
11
12
2
$10 bills $1 bills
14
13
14
4
4
15
5
5
5
In Step 1, the student has traded in ten
$1-bills for one $10-dollar bill; in Step 2,
he/she has traded in ten $10-bills for one
$100-bill, and in Step 3, he/she has traded
in ten $100-bills for one $1,000-bill.
© 2010 Herb I. Gross
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The student knows from this chart
that at each step of the process the value
of the currency has not changed, but at
the end of this process, the total number of
bills has been reduced from 46 to 19, and
the following general principle has become
clear…
The process of exchanging ten of one
denomination for one of the next higher
denomination ends when the number
remaining in each denomination is less
than ten.
© 2010 Herb I. Gross
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In terms of currency, what we are
saying is that regardless of how much
money we want to have in our wallet, we
never have to have more than nine bills of
any denomination.
Once students see the above sequence of
steps in a logical and easy to understand
fashion, it is relatively simple to turn
from the concrete illustration using
currency to the abstract concept of
place value.
© 2010 Herb I. Gross
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They will then understand from a logical
point of view that since the denominations
are no longer visible, we have to write the
sum in the form of Step 3 (that is, as 8,245)
unless we want to run the risk of having
our answer misinterpreted.
In summary, the visible transition from
Step 1 through Step 3 should help the
student understand the concept of
“carrying”.
© 2010 Herb I. Gross
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A Classroom Note
It might be difficult for some students
to work with more than a single digit at
a time. Hence rather than write 14 as…
tens
ones
14
it might be easier for them if we wrote it as…
tens
ones
1
4
© 2010 Herb I. Gross
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A Classroom Note
In this way, an intermediate way
for solving the above problem
would be…
+
5
2
1
7
© 2010 Herb I. Gross
8 ,
2
9
1
1
2
8
5
1
3
4
6
9
5
5
(15 ones)
(13 tens)
(11 hundreds)
(7 thousands)
(8,245 ones)
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“Exploring the Counting on
Your Fingers Myth”
As teachers, we often tend to discourage
students from “counting on their fingers”.
We often say such things as,
“What would you do if you didn’t have
enough fingers?”
The point is that in place value we
always have enough fingers!
© 2010 Herb I. Gross
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Consider, for example, the following
addition problem…
5,
2,
+ 1,
9,
2
9
6
9
8
5
7
1
6
9
3
8
and notice that this result could be obtained
even if we had forgotten the simple addition
tables, provided that we understood
place value and knew how to count.
© 2010 Herb I. Gross
Remembering that numbers in the
same column modify the same noun and
using the associative property of addition4,
we could start with the 6 in the ones place
and on our fingers add on nine more to
obtain 15. Then starting with 15 we could
count three more to get 18; after which we
would exchange ten 1’s for one 10 by saying
“bring down the 8 and carry the 1”.
We may then continue in this way, column by
column, until the final sum is obtained.
note
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4Up
to now we've talked about the sum of two numbers. However, no matter how many numbers we're
adding, we never add more than two numbers at a time. For example, to form the sum 2 + 3 + 4, we
can first add 2 and 3 to obtain 5, and then add 5 and 4 to obtain 9. We would obtain the same result if
we had first added 3 and 4 to obtain 7, and then added 7 and 2 to obtain 9.
© 2010 Herb I. Gross
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More explicitly…
5, 2 8 6
2, 9 5 9
+ 1, 6 7 3
1 8
2 0
1 7
8
9 9 1 8
© 2010 Herb I. Gross
(6 + 9 + 3) ones = 18 ones = 1 ten 8 ones
(8 + 5 + 7) tens = 20 tens = 2 hundreds
(2 + 9 + 6) hundreds = 17 hundreds
= 1 thousand 7 hundreds
(5 + 2 + 1) thousands = 8 thousands
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However, the point we wanted to
illustrate in the above example is that even
though there is a tendency to tell youngsters
that “grown ups don’t count on their
fingers”, the fact remains that with a proper
understanding of place value and knowing
only how to count on our fingers we can
solve any whole number addition problem.
In particular at any stage of the addition
process we are always adding two numbers,
at least one of which is a single digit.
© 2010 Herb I. Gross
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An Application of
Number Sense
By using our adjective/noun theme,
we can paraphrase a problem like
35 + 29 into a more “user friendly”
addition problem.
Namely, suppose John has 35 marbles
and Bill has 29 marbles.
© 2010 Herb I. Gross
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John 35 marbles
Bill 29 marbles
John 34 marbles
Bill 30 marbles
64 marbles
Notice that the above addition would have
been “simpler” if Bill had 30 marbles
instead of 29. So let’s suppose John gives
one of his marbles to Bill.
By sight, 34 + 30 = 64. However, since the
total number of marbles hasn’t changed,
35 + 29 is also 64.
© 2010 Herb I. Gross
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More generally, the sum of two
numbers remains the same if we
subtract an amount from one of the
numbers and add it to the other.
So, for example, to find the sum of 998 and
277, we notice that 998 + 2 = 1,000. Hence,
we add 2 to 998 and subtract 2 from 277.
998
+277
© 2010 Herb I. Gross
+2
–2
1, 0 0 0
+ 275
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In this way we obtain the equivalent
addition problem 1,000 + 275 from which
we quickly see that this sum is 1,275.
Therefore, we also know that
998 + 277 = 1,275.
998
+277
© 2010 Herb I. Gross
+2
–2
1, 0 0 0
+ 275
1, 2 7 5
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Teaching students to use the
“add and subtract” theme gives them a
relatively painless way to practice whole
number addition. For example, they can
find the sum of 497 and 389 by rewriting
the sum in the equivalent form 500 + 386.
They rather easily see that the sum of 500
and 386 is 886; and they can then practice
“traditional” addition by adding 497 and
389 to verify that the obtain the same sum.
© 2010 Herb I. Gross
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One goal of critical thinking is to
reduce complicated problems to a
sequence of equivalent but
simpler ones.
Here we have a very nice example
of the genius that goes into
making things simple!
© 2010 Herb I. Gross