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Logic
Mastery
Amazing, Amusing, Amorous, Delicious, Delightful,
Edifying, Strengthening Elixir.
It eases work & learning difficulties Makes the hard easier. Opens
eyes. Leads to greater precision.
in reading and writing
Do not leave here without it - Logic
mastery will develops critical thinking, improve reading and
writing, and give a firmer base for work and studies at many levels.
Good luck.
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Caution: Site advice is
approximately correct, for some circumstances, not all. Site How-TOs
are logically developed, but not tried and tested. That leaves
room for thought and refinement.. |
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After logic,
(a) continue reading Three
Skills for Algebra, chapters 8 to 14 and do so alongside site
area on solving
linear2007 Equations ; or (b) see this calculus
starter lesson and Volume 3, Why
Slopes & More Math, chapters 2 to 6;
For online automated help in senior
high school maths & calculus, visit quickmath.com
For Automatic Calculus and Algebra Help with derivatives, integrals,
graphs, linear equations, matrix algebra, visit calc101.com
With overlap, each site quickmath
& calc101offers a different
range of services, some free, some not, all based on webmathematica.
Good luck.
|
Explore collaborative whiteboards from groupboard,
twiddla or
scriblink.
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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 Column Methods for Subtraction
The rest of this lesson explains three column 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
- subtract units: 9 - 6 =3
- subtract tens: 9 - 5 =4
- subtract hundreds: 9 - 4 = 5
Conclusion: 999 - 456 = 543
Subtraction with Conversions:

Steps:
- Cannot subtract 835 from 000. So Convert 4000 into 3990 + 10. That
is, erase the 4000 and replace it by 3990 + 10.
- 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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Number Theory
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Start of Number Theory
Origins of Counting or Tallying
Adding Wholes
Multipling Wholes
Distributive Law Preamble
Distributive Law for Wholes
Consequences
More Consequences
What is a Fraction
Compound Fractions
Number Theory
Continued
Decimal Place Value Place Value Reinforcement Comparison Method Addition Method Subtraction Methods Multiplication Methods Division Methods Remainder Arithmetic I Primes & Composites Primes Factorization Theorem Primes & Composites Prime Factorization Aids Prime Factorization Examples Counting Whole No. Factors Arithmetic Videos Square Roots & Primes Long Division Continued Fractions & Decimals Fractions as Decimals 1 = 0.999 Recurring Infinite Decimals Expansion Arithmetic Ratio of Simple Fractions Ratio of Decimal Fractions Unsigned Reals Numbers Signed Coordinates Plane Vectors Horizontal Vectors Adding Vector Multiplies Adding Signed Numbers Multiplying Signed Numbers Distributive Law for Reals Real Numbers Axioms Remainder Arithmetic II
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