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YOU are better than YOU think. Show yourself how:
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-/[]\- Logic chapters 1 to 5 re- appear not in sequence, as is or longer, in Volume 1A, Pattern Based Reason, Bon Appetite. Logic
Mastery Logic mastery makes the hard, easier. Logic mastery leads to better, stronger and richer comprehension. Logic mastery improves reading and writing. Logic mastery ease learning difficulties. Logic mastery gives a headstart. In sum, logic mastery will develops critical thinking, improve reading and writing, and give a firmer base for work and studies at many levels. Good luck. After logic, (a) continue reading Three Skills for Algebra, chapters 8 to 14 and do so alongside site area on solving liinear Equations ; or (b) see this calculus starter lesson and Volume 3, Why Slopes & More Math, chapters 2 to 6;
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-/[]\- What may be learnt and when depends on how skills and concepts are developed. Making the hard easier and clearer will allow earlier & richer development of skills and concepts.
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Long Division Continuedcalculation of places after the decimal point Long Division method provides a sequence of decimal approximations to a fraction M/N. If the sequence is finite the fraction is decimal. So if the sequence does not terminate, the fraction is non-decimal.
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 The remainder 8 is smaller than 23, but 3 x 23 = 69. So 80 = 23 x 3 + 11 and therefore, dividing by 10, yields 8 = 23 x 0.3 + 1.1
--------------------
This gives
478155 = 20789 x 23 + 8
= 20789 x 23 + 23 x 0.3 + 1.1
= 20789.3 x 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 x 23 = 92. So
110 = 23 x 4 + 18
and therefore division by 100 gives
1.1 = 23 x .4 + .18
This yields again
478155 = 20789 x 23 + 8
= 20789.3 x 23 + 1.1
= 20789.34 x 23 + .18
The remainder has become smaller. This division process can be
recorded in the shorthand form as follows
_
20789.34
---------
23 | 478155.0000
- 455000 as 23 x 2 = 46 --> 20000
------
18155
-16100 23 x 7 = 161 --> 700
-----
2055
-1840 23 x 8 = 184 --> 80
-----
215 23 x 9 = 207 --> 9
-207
----
8.0
-6.9 23 x .3 = 6.9 --> .3
---
1.10
-.92 23 x .0492 = .92 --> 4
-----
.18
Therefore 478155 = 23 x 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. Alternative ViewpointIf we multiply by 100, or drop the decimal point in the above figuring, we see that 47815500 = 23 x 2078934 + 18 where the remainder r = 18 < 23. Thus 0 < 47815500 - 23 x 2078934 = 18 < 23. Division by 100 = 102 now implies 0 < 478155 - 23 x 20789.34 = .18 < 23/102. General Case of Division by a whole number d In general, to calculate p/d to k-decimal places when p is a decimal with k
decimal places or less after the decimal point, and d is a whole number, we
apply the long division method to divide the whole numbers 10kp by d.
The latter gives a quotient q (a whole no.) and remainder r with 0 < r
< d such that Then division by 10k gives p = d (q/10k) + r/10k. Here the decimal expansion of the new quotient (q/10k) and
the E = p - d (q/10k) = r/10k < d/10k. General Case of Division by a decimal d In general, to calculate p/d to k-decimal places when both p is a decimals with k decimal places or less after the decimal point, and d is a decimal with m decimals after the decimal point. we may apply the long division algorithm to divide p by 10md. to obtain a quotient q (a whole no.) and remainder r with 0 < r
< d such that Here the decimal expansion of the new quotient (q/10k) and
the E = p - d (q/10k) = r/10k < 10md/10k.
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