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Old 2021-04-26, 12:17   #1
tuckerkao
 
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Minus What's the Name of This Calculation Method?

At first, I thought this was the Common Core Additions and Subtractions. I asked several neighbors and they said this was a totally different thing.

I'm wondering whether there's a given name for the specific calculation method showing on the screenshot below or is it my own way to count?
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Old 2021-04-26, 13:09   #2
firejuggler
 
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decimal base, octal base and a dozenal base? i don't know.
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Old 2021-04-26, 13:39   #3
Dr Sardonicus
 
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This is simply regrouping and reassociating. I admit I don't see much point in doing things this way. One could equally well say 6 + 7 + 5 = 6 + 2 + 5 + 5 = 8 + 10 rather than 6 + 4 + 3 + 5 = 10 + 8.

There are well-known methods of doing the equivalent in addition and subtraction. With addition, it is called "carrying," and with subtraction, it is called "borrowing."

EXERCISE: Let m and n be positive integers in base b (b > 1 integer).

The sum of the base-b digits of m + n is equal to the sum of the base-b digits of m and the base-b digits of n, minus (b-1) times the number of carries in the base-b addition of m + n.

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Old 2021-04-26, 14:01   #4
tuckerkao
 
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Common Core Math does try to get rid of the carrying and the borrowing, instead using the distance between the numbers to figure out the sum or the difference.

I admit the method is slower when calculating only in the decimal base, but when I calculate them in other bases with the larger numbers like the screenshot showing below, I no longer need the base conversion calculators, also no hard additions or subtractions from the alternative bases. When applying the carrying and the borrowing in other bases, it requires the person to memorize the entire addition, subtraction, maybe multiplication tables of other bases to avoid making frequent mistakes on the larger numbers.


With the method I use, a person only needs to memorize:

[Octal]
1 + 7 = 10
2 + 6 = 10
3 + 5 = 10
4 + 4 = 10

10 - 1 = 7
10 - 2 = 6
10 - 3 = 5
10 - 4 = 4
[/Octal]

[Dozenal]
1 + Ɛ = 10
2 + Ӿ = 10
3 + 9 = 10
4 + 8 = 10
5 + 7 = 10
6 + 6 = 10

10 - 1 = Ɛ
10 - 2 = Ӿ
10 - 3 = 9
10 - 4 = 8
10 - 5 = 7
10 - 6 = 6
[/Dozenal]

The total amount of the unique formula needs to be memorized equals to the base itself, so pretty much only involves those formula of the 10 of the given base or with the sum under its 10 of that base.
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Old 2021-04-26, 15:44   #5
Dr Sardonicus
 
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Quote:
Originally Posted by tuckerkao View Post
When applying the carrying and the borrowing in other bases, it requires the person to memorize the entire addition, subtraction, maybe multiplication tables of other bases to avoid making frequent mistakes on the larger numbers.
Who might actually be routinely having to do addition and subtraction (not to mention multiplication and division with quotient and remainder) problems to multiple bases by hand escapes me, other than using 2-power bases associated with machine computing. So I see little point in devising alternate methods for doing arithmetic in other bases.

Coupled with the alleged difficulty of memorizing arithmetic tables for the digits of other bases (which IMO you pretty much have to do anyway to use the regrouping and reassociating you describe), this makes an excellent argument for avoiding more than a cursory introduction to actually doing arithmetic by hand in bases other than decimal. Better to get a decent command of doing decimal arithmetic, and then realize that the same basic principles and methods apply to any base.

You might enjoy this Tom Lehrer song about the "New Math"
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Old 2021-04-26, 15:50   #6
slandrum
 
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Quote:
Originally Posted by Dr Sardonicus View Post
Who might actually be routinely having to do addition and subtraction (not to mention multiplication and division with quotient and remainder) problems to multiple bases by hand escapes me, other than using 2-power bases associated with machine computing.
Well, if the US had ever gone metric...
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Old 2021-04-26, 16:10   #7
retina
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Quote:
Originally Posted by tuckerkao View Post
... no hard additions or subtractions from ...
... anything.

Addition and subtraction aren't hard. They aren't difficult. They aren't actually a problem at all. They are easy.

The problem seems to be the perception a lot of people have.

"Eww, adding? That's maths, and therefore it's too hard. What's the latest nonsense on Facebook that I can get angry about?"
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Old 2021-04-26, 16:13   #8
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Quote:
Originally Posted by slandrum View Post
Well, if the US had ever gone metric...
Even though I grew up with teaspoons, tablespoons, ounces, cups, pints, gallons, inches, feet, yards, and miles, I did all the arithmetic in decimal.

When introduced to metric units, I was immediately impressed by the fact that a cc of water was very nearly a gram, and a liter was 1000 cc's. I liked to ask, "Quick - how many cubic inches in a pint?" I later learned that the US standard for liquid measure was the US gallon, which by definition is 231 cubic inches. So, a pint is 28 and 7/8 cubic inches.

I am duly grateful that the good ol' USA had a decimal money system long before the UK
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Old 2021-04-26, 16:38   #9
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Multiplication and division , in other base than 10 isn't very intuitive

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Old 2021-04-26, 17:40   #10
slandrum
 
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Quote:
Originally Posted by firejuggler View Post
Multiplication and division , in other base than 10 isn't very intuitive
Really? It's much simpler in Binary!
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Old 2021-04-26, 23:36   #11
tuckerkao
 
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Quote:
Originally Posted by Dr Sardonicus View Post
So I see little point in devising alternate methods for doing arithmetic in other bases
The major step to figure out the addition is to find the distance between 9 and 10 for bases larger than Decimal.

9 + ? = 10
[Dozenal]9 + 3 = 10
[Hex]9 + 7 = 10


For Bases smaller than Decimal -

10 - 1 = ?
[Octal]10 - 1 = 7
[Senary]10 - 1 = 5
[Binary]10 - 1 = 1

Using the alternative methods have the strength to avoid the instant to operate decimal base when calculating in other bases, thus people will try to figure out the distance between each number first. Like Feb 29 in the leap year, the additional numerical become the leap numbers, so 02/28/1996 + 2 days = 03/01/1996

Quote:
Originally Posted by Dr Sardonicus View Post
The sum of the base-b digits of m + n is equal to the sum of the base-b digits of m and the base-b digits of n, minus (b-1) times the number of carries in the base-b addition of m + n.
In this case, everyone will still have to use the decimal base, then get the answers from other bases. How do you calculate the additions of the numbers when the leap numerical are in the formula? Let's use [Dozenal]ӾƐ + 7Ӿ = ? and Ӿ7 + ƐӾ = ? as the example.

[Dozenal] ӾƐ + 7Ӿ = ӾƐ + 11 + 69 = 100 + 69 = 169
[Dozenal] Ӿ7 + ƐӾ = Ӿ7 + 15 + Ӿ5 = 100 + Ӿ5 = 1Ӿ5 or
[Dozenal] Ӿ7 + ƐӾ = Ӿ7 + 100 - 2 = 1Ӿ7 - 2 = 1Ӿ5


[Hex] 8 + 9 = 9 + 8 = 9 + 7 + 1 = 10 + 1 = 11, thus it has been calculated independently from the decimal base.
[Hex] 11 - 8 = ? ...... 8 + 1 = 9; 9 + 7 = 10, 10 + 1 = 11 ...... 1 + 7 + 1 = 9

Last fiddled with by tuckerkao on 2021-04-27 at 00:34
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