Subtraction
Decimal Methods for Subtraction - how to justify
Reference: The site section Decimal Arith - Video Based includes a newer development.
Physical Concept
Put 15 objects in a bag. Ask your child to take away or subtract 7 of
them.
Then ask him or to count how many remain in the bag. (Also taking 15 steps
to right and then taking 7 steps to left yields the same result as 8
steps to
the right when all steps are equal sized.)
Three Different Column Methods for Subtraction
The rest of this lesson explains three column (vertical) methods for subtraction: (I) with borrows when necessary and (II) with a new two row column methods, (III) a complementary subtraction method. The second method is a site invention and curiosity perhaps. The third method is a re-invention
(I) Column Method with Conversions (Borrows) when needed
Subtraction with no conversion needed.
Now lets try 144 minus 31 without counting. On paper this can be written as _______________________________________\
/
244 (read 4 ones plus 4 tens plus 2 hundreds ) |
- 31 (read 1 ones plus 3 tens plus 0 hundreds) |
---------- subtraction gives |
213 3 ones plus 1 tens plus 2 hundreds \|/
---------- | /____________________________________
\
Subtraction with conversion (borrows)
First Example: Now consider 365 - 149 |
365 (read 5 ones plus 6 tens plus 3 hundreds ) - 149 (read 9 ones plus 4 tens plus 1 hundreds ) ---------- ----------
| Now taking 9 from 5 ones is not possible. But 5 ones and 6 tens is the same as 5 + 10 ones plus 6 - 1 tens |
10
3 6 5 (read 15 ones plus 5 tens plus 3 hundreds )
- 1 4 9 (read 9 ones plus 4 tens plus 1 hundreds )
---------- subtraction yields
6 1 2 6 ones plus 1 tens plus 2 hundreds
----------
-1 The -1 indicates a borrow and
The number above is obtained from
6 - 1 - 4
| The foregoing illustrates and justifies the borrowing method in a simple case. This is the method I met in school. An alternate method follows. Pick one that you, your child, or the child's teacher likes and appreciates. (If you have to battle over your child's education, this point of which method use for subtraction is too minor to argue over.) |
Second Example: Now consider 825 - 273
8 2 5 (read 8 hundreds plus 2 tens plus 5 ones )
- 2 7 3 (read 2 hundreds plus 7 tens plus 3 ones )
---------- This is the same as
8-1 hundreds plus 12 tens plus 5 ones
minus 2 hundreds plus 7 tens plus 3 ones. This yields
5 5 2 or 5 hundreds plus 5 tens plus 2 ones
----------
-1 This 1 below the bar indicates the conversion
of 8 hundred into
7 hundreds plus 10 tens -- the "borrow".
Third Example: Consider 8234 - 4816
8234 (8 thousand + 2 hundreds + 3 tens + 4 ones)
- 4816 - (4 thousand + 8 hundreds + 1 tens + 6 ones)
------- or (8-1 thousand + 12 hundreds + 3-1 tens + 14 ones)
- (4 thousand + 8 hundreds + 1 tens + 6 ones
3418
--------
1 1 Here is the shorthand indication of the borrows or
the conversions of one thousand into 10 hundreds and
one tens into ten ones.
Fourth Example: Another example (repeated borrows)
4823 (4 thousand + 8 hundreds + 2 tens + 3 ones)
- 3987 - (3 thousand + 9 hundreds + 8 tens + 7 ones)
------- or (4-1 thousand + 18-1 hundreds + 12-1 tens + 13 ones)
- ( 3 thousand + 9 hundreds + 8 tens + 7 ones
836
--------
111 Here is the shorthand indication of the -1s.
Note that 17 - 9 = 8
The pattern is as follows.
- 8 hundreds less than 9 hundred, so replace 4 thousand + 8 hundred by its equal 4 - 1 thousand + 18 hundred.
- 2 tens are less than 8 tens. So replace 18 hundred plus 2 tens by 18 -1 hundreds plus 12 tens.
- 3 ones is less than 7 ones. So replace 12 tens plus 3 ones by its equal 12-1 tens plus 13 ones.
These three replacements imply 4823 equals 4-1 thousands + 18-1 hundreds +
12-1 tens plus 13 ones,
that is, 3 thousands + 17 hundreds + 11 tens + 13 ones. Think of the conversion
of larger bills into
smaller ones -- the conversion can be done as convenient.
Fifth Example - A Case of Repeated Borrows or Conversions
Column: mlkjihgfedcba
Label the columns.
7881239562583
-234892345682
In the first two columns a and b, no borrowing is needed
as the lower digits, those being subtracted are less than
the upper digits.
Column: mlkjihgfedcba
7881239562583
-234892345682
01
Since it is difficult to type small, I am going insert a space between each
column. That gives
Column: m l k j i h g f e d c b
a
7 8 8 1 2 3 9 5 6 2 5 8
3
- 2 3 4 8 9 2 3 4 5 6 8 2
0 1
Now 5 < 6, 25 < 56, but 625 > 456. So we
convert or write or think
625 = 25 + 600
= 25 + 590 +10
So we strike through the 6 in column e, leave the 25 in place, and write 590 +10
above what was the top row.
That is a multicolumn conversion.
Column: m l k j i h g f e d c b a
5 9 10
<== Here is 590.
7 8 8 1 2 3 9 5 6
2 5 8 3
- 2 3 4 8 9 2 3 4 5 6 8 2
0 1
Now 10 + 5 - 6 = 4+5 = 9, | Aside:
Think 625 - 456
9 + 2 - 5 = 4+ 2 = 6 | =
25 + 600 - 456
and 5 - 4 = 1
| = 25 + 590 + 10 - 456
| = 25 + 590 -450 + 10 - 6
| = 25 + 140 + 4 = 169
|________________________________
So we fill in more digits:
Column: m l k j i h g f e d c b a
5 9 10
<== Here is 590.
7 8 8 1 2 3 9 5 6
2 5 8 3
- 2 3 4 8 9 2 3 4 5 6 8 2
1 6 9 0 1
We continue filling in more digits:
5 - 3 = 2; 9 - 2 = 7 but oops 3 < 9. So we arrive at:
Column: m l k j i h g f e d c b a
5 9 10
<== Here is 590.
7 8 8 1 2 3 9 5 6
2 5 8 3
- 2 3 4 8 9 2 3 4 5 6 8 2
7 2 1 6 9 0 1
and we have to do another conversion. Now
3 < 9 (setting the need for a borrow),
23 < 89 (continuing the need),
239 < 892 (still continuing the need)
BUT 8123 > 3489. So we write or think
8123 = 8000+ 123 = 7990 + 10 + 123 and above the top row
So we strike through the 8 in column k,
leave the
123 in place, and write
7990 +10 above what was the top row - a conversion:
Column: m l k j i h g f e d c b a
7 9 9
10 5 9 10
<== Here is 590.
7 8 8
1 2 3 9 5 6 2 5 8
3
- 2 3 4 8 9 2 3 4 5 6 8 2
7 2 1 6 9 0 1
Now no further conversions or borrowings are
required.
We use
10 + 3 - 9 = 1 + 3 = 4
9 + 2 - 8 = 1 + 2 = 3
9 + 1 4 = 5 + 1 = 6
7 - 3 = 5
78 - 2 = 76
to complete the calculation:
Column: m l k j i h g f e d c b a
7 9 9
10 5 9 10
<== Here is 590.
7 8 8
1 2 3 9 5 6 2 5 8
3
- 2 3 4 8 9 2 3 4 5 6 8 2
7 6 5 6 3 4 7 2 1 6 9 0
1
Exercise: Check the calululation:
Column: mlkjihgfedcba
7656347216901
+234892345682
7881239562583
111
11
Two More Examples:
Steps
- subtract units: 9 - 6 =3
- subtract tens: 9 - 5 =4
- subtract hundreds: 9 - 4 = 5
Conclusion: 999 - 456 = 543
Subtraction with Conversions:
Steps:
A more standard way to do this is to cross-out the 4000 and replace it by 3990 + 10 as follows.
(II) Column Method with Two Rows
|
58.65
17.44 _
41.25
You will have 41.25 left.
|
Imagine again that you have 5 ten dollar bills, 8 one dollar bills, 6 dimes and 5 pennies in a piggy bank. So again, the total amount in the piggy bank is 58.65 dollars. Suppose you owe another 29. 87 dollars. If you give 2 tens, 8 ones and 65 cents, you will have $ 30.00 left and still owe 1 one and 22 cents. The latter remains to take from the 30.00 -- we can write the following.
Here 7 from 5 pennies leaves 0 with 2 more owing or to subtract; 8 dimes from 6 dimes leaves 0 with 2 more owing; and 9 from 8 dollars leaves 0 with 1 more to be subtracted. To pay the debt completely, compute 30.00 - 1.22 as follows.
We can write all the foregoing at once:
So 58.65 = 29.87 = 28.78 Observe we subtract as much as we can in each column without borrowing (or converting). That gives two rows. The first row gives the amount that still remains. The second row shows what still needs to be subtracted. Examples follow.
|
Steps:
6 from 6 gives 0 and nothing more to subtract; 4 from 5 gives 1 with
nothing more to subtract; ...; 4 from 2 gives 0 with 2 more to subtract; 9
from 8 gives 0 with 1 more to subtract; and so on. The foregoing
leaves 4 200 002 110 with 220 120 000 to be subtracted. See the last
three rows of the calculation.
III: Column Method using ComplementsThere is another name for this that will return, or be found in one of my books. We introduce this complementary column method by solving for unknowns, and then re-arranging the rows in a way that hides the unknowns. Start With UnknownsFirst ExampleOne way to find or define 825 - 273 is to consider the missing number puzzle CBA The question here is what should the digits A, B and C equal given they belong to the set 0 to 9.
The foregoing gives C52 with a carry of 1 in the hundreds column. Now we find C so that the carry 1+ 2+ C = 8. By inspection, C = 5 Hence, we have or should have 552 That latter is easily checked by the column addition method.
|
Now 825 = 552+ 273 . Therefore 825 -273 =
(552+273) -273 = 552.
The foregoing gives an alternative method for finding the
difference 825 - 273
Second Example
Compute 8234 - 4816
Write
4816
DCBA
----- +
8234
-----
Want 6+A = 4 modulo 10. So A = 8 with a carry
of 1
DCB8
4816
----- +
8234
-----
1
Need 1+ 1 + B = 3 exactly or modulo 10. So B =
1 with no carry
Need 8+C = 2 exactly or modulo 10. So C = 4 with a carry of 1.
The foregoing gives
D418
4816
----- +
8234
-----
1 1
Now we need 1 + 4+ D = 8 exactly. So D = 3
5418
4816
----- +
8234
-----
1 1
Our conclusion is 5418 = 8234 - 4816.
Second Example Revisited - Row Swapping
By swapping the first and third row in the above
calculations,
we get a sequence of column method to do a subtraction via
complements rather than borrows.
Write
8234
DCBA
----- -
4816
-----
Want 6+A = 4 modulo 10. So A = 8 with a carry
of 1
8234
4816
----- -
DCB8
-----
1
Need 1+ 1 + B = 3 exactly or modulo 10.
So B =
1 with no carry
Need 8+C = 2 exactly or modulo 10.
So C = 4 with a carry of 1.
The foregoing gives
8234
4816
----- -
D418
-----
1 1
Now we need 1 + 4+ D = 8 exactly. So D = 3
8234
4816
----- -
5418
-----
1 1
Our conclusion is 5418 = 8234 - 4816.
Third example with and without letters.
|
Steps in the computation of 6855- 2985 follow - with letters 6855
6855
6855 5 + A = 5 modulo 10,
A = 0 |
|
Steps in the computation of 6855- 2985
follow - without letters 6855
6855
6855 5 + ? = 5 modulo 10,
? = 0 |
6855
2985
---- -
3870
-----
11
Number Theory & Practices
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