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Home < Algebra Starter Lessons < A Origins of Counting and Figuring Methods << D Long Division Methods
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D. Long Division MethodsDivision in general - How times does one object (measure) fit into another.When we have a whole number of objects, we may decide to form or try to form equi-sized or equipollent groups. The division of a whole number N by another whole number d counts the maximum number of times q that groups of size d can be formed and yields a remained less the divisor d. What is left-over may be zero or a group of count r < M. Division in the first instance may be appear as a physical operation. We say d divides N when and only when the remainder after division is zero. Here N = qd + r The foregoing discussion of division may be repeated or recalled later when needed. The concept of multiplication is needed next. When fractions are known and allowed as multipliers, the division of a whole number N by another whole number d is given by the proper or improper fraction N/d as is or expressed in lowest terms or expressed as a mixed number. In the latter case, the number of times that d goes into N is N/d exactly, with no remainder. That being said if N/d is not a whole number, it equals (qd + r) = q + r/d where r = remainder for the number of wholes times q that d goes in N. Decimal Methods for DivisionMotivating Questions.
Example 1. What number added to itself 3 times, gives The answer is 3 hundred + 2 tens + 1 ones =321. (Related question: How many times must 3 be added to itself to give 963?) Example 2. How many times does 7 go into 963 = 9 hundred + 6 tens + 3 ones? The Euclidean Division Method gives the answer.
9 = 7×1 + 2.
Thus 9 hundred = (7×1 + 2) hundred
Thus 963 = (7x1 + 2) hundred + 63 =7×1 hundred + 263 Thus
963 = 7×(1 hundred) + 263.
--------------------------
2 is smaller than 7.
But 7×3 = 21 and 7×4 =28 > 26.
and 26 = 7×3 + 5. Therefore
263 = (7x3 + 5) tens + 3 = 7×3 tens + 53
Thus 263 = 7×(3 tens) + 53
-------------------------------
5 is smaller than 7. But 7×7 = 49 and 7×8 = 56 > 53.
Therefore 53 = 7×7 ones + 4 ones.
Thus 53 = 7×(7 ones) + 4 leftover.
------------------------------------
4 is "too small to be divided by 7"
without the use of fractions.
Now
963 = 7×(1 hundred) + 263
= 7×(1 hundred) + 7×(3 tens) + 53
= 7×(1 hundred) + 7×(3 tens) + 7×(7 ones) + 4 left-
over
= 7×(1 hundred + 3 tens + 7 ones) + 4 leftover
= 7×137 + 4.
= 959 + 4
Our conclusion is that 7 goes into 963, 137 times completely, with
4 leftover = remainder.
In shorthand notation, we write the above calculation more compactly and briefly as follows.
137
----
7 | 963
-700
---
263
-210
---
53
49
--
4 <--- the remainder.
Conclusion: The Euclidean Division Method justifies the long division algorithm.Algorithm is just another word for method. If you know about fractions:
4
963 = 7×(137 + --- )
7
Example 3. How many times can 23 go into 478155? We will apply the Euclidean Division Method
47 = 23×2 + 1 leftover
Thus 470000= 23×20000 + 10000.
Adding 8155 gives
478155 = 23×20000 + 18155
---------------------------
18 is smaller than 23
but 7×23 = 161 and 181 - 171 = 20 is
smaller than 23. Thus
181 = 23×7 + 20 leftover
Thus 18100 = 23×700 + 2000 leftover
Adding 55 gives
18155 = 23×700 + 2055 leftover
------------------------
20 is smaller than 23. But 8×23 = 160 + 24 = 184
Thus 200 = 23×8 + 16 leftover
2000 = 23×80 + 160 leftover.
So adding 55 gives
2055 = 23×80 + 215 left-
over
---------------------
21 is smaller than 23. But
9×23 = (10-1)x23 = 230 - 23 =207.
Thus
210 = 9×23 + 7 or
215 = 23×9 + 8 left-
over
-----------------
Here the remainder 12 is less than 23. Now we substitute:
478155 = 23×20000 + 18155
= 23×20000 + 23×700 + 2055
= 23×20000 + 23×700 + 23×80 + 215
= 23×20000 + 23×700 + 23×80 + 23×9 + 12
left-over
= 23×(20000 + 700 + 80 + 9) + 12 leftover.
We conclude 23 goes into 478115,
20789 times completely with 12 leftover
The foregoing calculations can be written more compactly in the left column.Read it first. Consider it as summary of the above.
The rightmost column shows the usual long division algorithm, one that was taught in elementary school in the 1960s. Variations of it may be found. You should be able to see how the steps on the right correspond to those on the left. In the rightmost column, you will see rows with * in them. The row in-between in them is usually omitted to lessen the amount of writing. The row is included here to help in the comparison of the Euclidean Division Method and that which I met in elementary school in the 1960s. The right hand column is a more cryptic implementation and variation of the Euclidean Division Method. 8 Conclusion
478155 = 20789×23 + 8 = (20789 + -- )×23.
23
Continuing the Division Process
The remainder 8 is smaller than 23, but 3×23 = 69. So 80 = 23×3 + 11 and therefore, dividing by 10, yields
8 = 23×0.3 + 1.1
--------------------
This gives
478155 = 20789×23 + 8
= 20789×23 + 23×0.3 + 1.1
= 20789.3×23 + 1.1
The remainder has become 1.1 instead of 8. It is much smaller. We can do this again, and again. For example: 4×23 = 92. So those on the left. In the rightmost column, you will see rows with * in them. The row in-between in them is usually omitted to lessen the amount of writing. The row is included here to help in the comparison of the Euclidean Division Method and that which I met in elementary school in the 1960s. The right hand column is a more cryptic implementation and variation of the Euclidean Division Method.
8 80 = 23×3 + 11 Therefore, dividing by 10, yields
8 = 23×0.3 + 1.1
--------------------
This gives
478155 = 20789×23 + 8
= 20789×23 + 23×0.3 + 1.1
= 20789.3×23 + 1.1
The remainder has become 1.1 instead of 8.
It is much smaller. We can do this again,
and again. For example: 4×23 = 92. So
110 = 23×4 + 18
and therefore division by 100 gives
1.1 = 23×.4 + .18
This yields again
478155 = 20789×23 + 8
= 20789.3×23 + 1.1
= 20789.34×23 + .18
The remainder has become smaller. This division process can be recorded in shorthand form as follows
_
20789.34
---------
23 | 478155.0000
- 455000 as 23×2 = 46 --> 20000
------
18155
-16100 23×7 = 161 --> 700
-----
2055
-1840 23×8 = 184 --> 80
-----
215 23×9 = 207 --> 9
-207
----
8.0
-6.9 23×.3 = 6.9 --> .3
---
1.10
-.92 23×.0492 = .92 --> 4
-----
.18
Therefore
478155 = 23×20789.34 + .18 where 0.18 = the remainder. The remainder approaches zero as more and more decimal digits after the decimal point are computed via the long division method. |
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Home < Algebra Starter Lessons < A Origins of Counting and Figuring Methods << D Long Division Methods
[1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12][13]
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