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Home < Arithmetic and Number Theory Skills < 2 Arithmetic with Decimals < B Decimal Comparing and Subtracting Methods << Appendix 2 Three Decimal Subtraction Methods

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Subtraction

Decimal Methods for Subtraction - how to justify

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)

Note (January 25th, 2008): The repeated borrowing method below  in the fitfh example represents
advances my understanding of the multicolumn grouping and borrowing case.
I was never sure how to explain it before.

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

  1. subtract units: 9 - 6 =3
  2. subtract tens: 9 - 5 =4
  3. subtract hundreds: 9 - 4 = 5

Conclusion: 999 - 456 = 543

Subtraction with Conversions:

Steps:
  1. Cannot subtract 835 from 000. So Convert 4000 into 3990 + 10. That is,  erase the 4000 and replace it by 3990 + 10.
  2. Now subtract 5 units from 10 units, 3 tens from 9 tens, 8 hundreds from 9 hundreds and 2 thousands from 3 thousands to get 5 units, 6 tens, 1 hundred and one thousand. 

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 
       Reinventing(?)  a Two Row Method

Imagine you have 5 ten dollar bills,  8 one dollar bills, 6 dimes and 5 pennies in a piggy bank. Then the total amount in the piggy bank is 58.65 dollars. 

Now suppose you owe  another 17. 44 dollars.  Then you can give the other one of the five tens, 7 of the eight ones, 4 of the six dimes and 4 of the five pennies. There is nothing else to do.

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.

58.65
29.87 _
30.00   Amt left
  1.22   Amt still to be subtracted (owed)

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.

29.9
30.0010    
  1.22   _
28.78

We can write all the foregoing at once:

58.65
29.87 _
30.00     
  1.22  _ 
28.78

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 Complements

There 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 Unknowns

First Example

One way to find or define  825 - 273 is to consider the missing number puzzle

  CBA
 273
 ------ +
 825
------

The question here is what should the digits A, B and C equal given they belong to the set 0 to 9.

  • Here  A = 2 works as 3+2 = 5. 
  • 7+5 = 12 = 2 modulo 10.  So we take B =5.  The latter is the only digit 0 to 9 that satisfies 7+B = 2 modulo 10.

The foregoing gives

   C52
 273 
------ +
 825
------
 1

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
 273
------ +
 825
------
 1

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
2985       2985      2985         8 + B = 5 modulo 10, B = 7, Carry 1
---- -     ---- -    ---- -   1 + 9 + C = 8 modulo 10, C = 8, carry 1.
DCBA        970      3870     1 + 2 + D = 6 modulo 10, D = 3
----       ----      ----
            1         1

Steps in the computation of  6855- 2985 follow - without letters
--- letters have been removed

6855       6855      6855         5 + ? = 5 modulo 10, ? = 0
2985       2985      2985         8 + ? = 5 modulo 10, ? = 7, Carry 1
---- -     ---- -    ---- -   1 + 9 + ? = 8 modulo 10, ? = 8, carry 1.
            970      3870     1 + 2 + ? = 6 modulo 10, ? = 3
----       ----      ----
            1        11

6855
2985
---- -
3870
-----
11

The above defines a complementary method for subtraction, one free of borrows.

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Home < Arithmetic and Number Theory Skills < 2 Arithmetic with Decimals < B Decimal Comparing and Subtracting Methods << Appendix 2 Three Decimal Subtraction Methods

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