Damo and Water need Schooling Again on 1/3

Here is a good example of the breakdown of communication. I don't think I ever argued the point that is refuted here. Somehow, this is what String-in-the-butt read.

In all instances, 100% of the apple is equal to 100% of the apple. You can divide it into thirds, but never equally. One portion of the apple must be slightly more, to include the extra remainder. This extra amount may be so small it is insignificant, and we can assume that all three parts of the apple are equal. I can't remember what it is called, Nelson's Law? But there is some long exasperating formula in calculus and trigonometry, which is used to rectify this remainder in critical plotting calculus. I'll admit, I am not some math wizard who knows everything about it, but I am not some dumb hick who doesn't understand what a cotangent or square root is either. It doesn't take a math genius to understand that a remainder is produced whenever you divide one by three. Most 3rd graders understand it, why can't you guys? That's the part I don't get!

Dixie if you divided a 3 leaf clover (where the leaves are equal in size) into 3 would you have any remainder? No.
So imagine if you combined those leaves into a larger leaf and again divided them into 3, why all of a sudden would you have a remainder?
 
Dixie if you divided a 3 leaf clover (where the leaves are equal in size) into 3 would you have any remainder? No.
So imagine if you combined those leaves into a larger leaf and again divided them into 3, why all of a sudden would you have a remainder?

He wonders why there is a breakdown in communication but he poses statements like that. Things can most certainly be divided equally into thirds. You don't just say 'thats 1 apple wide'. Shit has measurements, ya know? All you need to divide it equally into thrids is to take the proper measurement and divide it by 3. For instance, I got a rectangle cake, 1 foot long and 4 inches wide. All I gotta do to divide it equally by a a third, is to take that 1 foot long, and cut every 4 inches. Walla!
 
1 is indeed equal to 1 in base 12, but 1 in base 12 is not equal to 1 in base 10, is it?


Yes it is. All the ones in all the bases are equal to each other.

Nope... the value of "1" changed when you changed base systems. You are therefore, no longer dealing with the same values. In base 12, the value of "1" in base 10, is now changed.

Yes we are. It's just a different way of counting. Instead of counting 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 you count 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, 10. The values in the decimals change similarly. It doesn't change the actual value.

You could even have base 1, where 3 in decimal is 111. 1/3 could actually only be represented by a fraction in this system - 1/111.


Let me blow your little mind some more Waterhead...

If I had an Apple in base 10, and I took it to base 12, it would only be .83333 of an Apple!

"An apple in base 10" is meaningless. If you COUNTED an apple using base 10, it would be ONE apple, and if you COUNTED an apple in base 12, it WOULD BE ONE APPLE. To "convert" between the two systems, you wouldn't even need to do anything. There's no difference until you get to the value represented by 10 in the decimal counting system.

In order to have a whole base 12 Apple,

There's no such thing as a base 10 apple or a base 12 apple. There's just an apple. You can use various systems to count this apple but it's not going to change the amount of apple.

You have absolutely no understanding of set theory or how to convert between bases.

I would have to have my base 10 apple, and .2 of another base 10 apple added to it, then I would have "1" base 12 apple.

:D

1 in base 12 is 1 in base 10. 10 in base 10 is A in base 12. 11 in base 10 is B in base 12. 12 in base 10 is 10 in base 12. Do you get it now? It does not vary how you think it does - the value of ONE does not change., the amount of symbols you must use before you have to add another one is what changes.

A system which changed the value of one would be meaningless and silly.
 
Ron, what do you think about this problem:
"Say that Jane and Joan (who are unrelated) each has two children. We know that at least one of Jane's children is a boy and that Joan's oldest child is a boy. What are the odds that each of the women has 2 boys?"
http://www.justplainpolitics.com/showthread.php?t=16657&highlight=smartest

MottleyDude, me, the smartest woman in the world (where I got this problem from) and countless math professors (who use this problem in class) got the right answer, but Damo, Thorn and Super (and I think a few others) went forever trying to say we were wrong.
I want to know what you think.

It's probably just a problem deliberately designed to not be intuitive.
 
Here is a good example of the breakdown of communication. I don't think I ever argued the point that is refuted here. Somehow, this is what String-in-the-butt read.

In all instances, 100% of the apple is equal to 100% of the apple. You can divide it into thirds, but never equally. One portion of the apple must be slightly more, to include the extra remainder. This extra amount may be so small it is insignificant, and we can assume that all three parts of the apple are equal. I can't remember what it is called, Nelson's Law? But there is some long exasperating formula in calculus and trigonometry, which is used to rectify this remainder in critical plotting calculus. I'll admit, I am not some math wizard who knows everything about it, but I am not some dumb hick who doesn't understand what a cotangent or square root is either. It doesn't take a math genius to understand that a remainder is produced whenever you divide one by three. Most 3rd graders understand it, why can't you guys? That's the part I don't get!

There's a remainder but it's no big fucking deal.
 
It's probably just a problem deliberately designed to not be intuitive.

Sure on first read, in fact it plays on how people think, but if you read it carefully there is no ambiguity and the answer makes sense.
That thread was so ridiculous, I still can't believe that there were some who kept going arguing it even after I revealed that it was a well known logical problem that countless thousands of math professors and students had no problem understanding and did not dispute.
 
???...uh, no.....an object with a weight described as "1" unit in a base 10 system would be equal to a weight of "1.2" units in a base 12 system.....unless of course, you also choose to redefine the meaning of unit.....which might result in saying for example, that "1 ounce" equals "47 tons"......do you begin to understand why this argument is simply about methods of communication rather than physics?......
You have no idea what you are talking about, stick to religion math isn't for everybody. A base twelve system has 12 digits rather than 10. 1 would equal 1 in both systems. The numbers are written differently only in that there are more digits representing the base of the system.

Lets just say that # is the digit that would replace 10, and $ is the digit that replaces 11 in a base 12 system... (because in base 12 it doesn't start again with the 1 until you are representing 12 with "10"...

The numbers would be: 0,1,2,3,4,5,6,7,8,9,#,$

Then it would be 10 through 1$ (which would equal 21), and so forth. 1 is 1 regardless of which system you use. However if you divided 3 into 1 you would get the answer .4 and Dixie could be satisfied because no difficult concepts would have to be encompassed by thought. Well, unless you were dividing 1 by 4...
 
You have no idea what you are talking about, stick to religion math isn't for everybody. A base twelve system has 12 digits rather than 10. 1 would equal 1 in both systems. The numbers are written differently only in that there are more digits representing the base of the system.

Lets just say that # is the digit that would replace 10, and $ is the digit that replaces 11 in a base 12 system...

The numbers would be: 0,1,2,3,4,5,6,7,8,9,#,$

Then it would be 10 through 1$ (which would equal 21), and so forth. 1 is 1 regardless of which system you use. However if you divided 3 into 1 you would get the answer .4 and Dixie could be satisfied because no difficult concepts would have to be encompassed by thought. Well, unless you were dividing 1 by 4...

GREAT... just PERFECT... now we are going to have to read Ditzie ranting about 1/4 in base 12. Thanks Damo...:cool:
 
Imagine, if our culture had been brought up with the base 12 system ingrained in our heads instead of the base 10 system, dixie would be complaining about the impossibility of dividing things into fourths.

It's amazing how people can count this way every day of their life without really understanding the system they're using. (Not that I'm saying we should teach it in elementary school like in the New Math days.)
 
Dixie if you divided a 3 leaf clover (where the leaves are equal in size) into 3 would you have any remainder? No.
So imagine if you combined those leaves into a larger leaf and again divided them into 3, why all of a sudden would you have a remainder?

Dano, I am disappointed in you. The one person I thought would be fully capable of comprehending and understanding the point, and here you are... assuming I've said something outrageous, just like the rest of these pinheads.

I have already stated, very clearly, that we can divide anything into thirds! I have never claimed that we couldn't! We can even assume that all three parts are equal! We do this all the fucking time! Nowhere, have I ever stated, that we can't do this, or don't do this, or that it's not possible to do it. lol

100% divided into thirds, is what??? Give me the three EQUAL parts! If you say 33.33333e%, I will ask you to define 'e' because it is not a number. Eventually, you will realize, the extra part can never be resolved. We can divide it down until we get to a single atom, and when we try to split an atom into three parts, nuclear fission happens and the experiment is over.
 
Here is a good example of the breakdown of communication. I don't think I ever argued the point that is refuted here. Somehow, this is what String-in-the-butt read.

In all instances, 100% of the apple is equal to 100% of the apple. You can divide it into thirds, but never equally. One portion of the apple must be slightly more, to include the extra remainder. This extra amount may be so small it is insignificant, and we can assume that all three parts of the apple are equal. I can't remember what it is called, Nelson's Law? But there is some long exasperating formula in calculus and trigonometry, which is used to rectify this remainder in critical plotting calculus. I'll admit, I am not some math wizard who knows everything about it, but I am not some dumb hick who doesn't understand what a cotangent or square root is either. It doesn't take a math genius to understand that a remainder is produced whenever you divide one by three. Most 3rd graders understand it, why can't you guys? That's the part I don't get!

And here you are again arguing that equal thirds are impossible, which you'll later claim you never said.

Given the right tools you can divide it into equal thirds without any remainder.

The remainder from 1/3 is but a problem of the decimal systems representation of the fraction. It does not effect the reality that the apple can be divided into equal thirds.

Here we could go back to the ruler. But you'll just get lost again.
 
You have no idea what you are talking about, stick to religion math isn't for everybody. A base twelve system has 12 digits rather than 10. 1 would equal 1 in both systems.

If 1 is still equal to 1, then 1 still can't be equally divided by 3, presuming 3 is still equal to 3.
 
"In all instances, 100% of the apple is equal to 100% of the apple. You can divide it into thirds, but never equally. One portion of the apple must be slightly more, to include the extra remainder."

Absolutely and utterly untrue.

Imagine a three pound apple, it could be divided into exactly three equal parts of one pound each. You just would write it as 1/3 and then to Dixie the apple would disappear because it would be impossible to make it equal because you can't write it down decimally without a bar over the three.

The decimal system is just the tool we use to express an idea, it isn't perfect nor is it the only way to express that idea. Hence we came up with the concept of fractions... 1/3 is a way to express that idea without the never ending 3s....
 
If 1 is still equal to 1, then 1 still can't be equally divided by 3, presuming 3 is still equal to 3.
Incorrect because we are no longer using a decimal system, we are using a system based in 12 digits. 1/3 would equal .4 if you were using base 12 math, and here is where your head will explode, even with the numbers being equal because we are expressing the idea differently. It's like a whole new language. Suddenly there is no remainder on the same division problem....

Wow. Magically you can now divide an apple into three equal parts because there is no "remainder" that will get all over you. However, it would be impossible (to you) to divide it into 4 equal parts because that one would produce one of those never ending remainder parties that get stuff everywhere.

If humans were born with 12 digits on their hands rather than 10, we'd live in a world where 1/4 was "impossible"!
 
And here you are again arguing that equal thirds are impossible, which you'll later claim you never said.

Given the right tools you can divide it into equal thirds without any remainder.

The remainder from 1/3 is but a problem of the decimal systems representation of the fraction. It does not effect the reality that the apple can be divided into equal thirds.

Here we could go back to the ruler. But you'll just get lost again.

It's not a problem of the decimal systems representation, it's a problem with your brain not comprehending a relatively simple basic math principle. The reality is, you can never divide 100% of the apple into three equal parts, if you could, you would be able to give me the percentage of each equal part, and you can't. You can have two parts which are 33.33% of the whole, and another part which is 33.34% of the whole, and most retards like Waterhead and Damo, will never notice they have one of the smaller pieces. They will assume, as you do, that the apple has been "evenly" divided into thirds.
 
It's not a problem of the decimal systems representation, it's a problem with your brain not comprehending a relatively simple basic math principle. The reality is, you can never divide 100% of the apple into three equal parts, if you could, you would be able to give me the percentage of each equal part, and you can't. You can have two parts which are 33.33% of the whole, and another part which is 33.34% of the whole, and most retards like Waterhead and Damo, will never notice they have one of the smaller pieces. They will assume, as you do, that the apple has been "evenly" divided into thirds.
Again utterly incorrect. Percentages are not the measure of whether you can equally divide a three pound apple.
 
Incorrect because we are no longer using a decimal system, we are using a system based in 12 digits. 1/3 would equal .4 if you were using base 12 math, and here is where your head will explode, even with the numbers being equal because we are expressing the idea differently. It's like a whole new language. Suddenly there is no remainder on the same division problem....

Wow. Magically you can now divide an apple into three equal parts because there is no "remainder" that will get all over you. However, it would be impossible (to you) to divide it into 4 equal parts because that one would produce one of those never ending remainder parties that get stuff everywhere.

The physical properties of the apple did not change with base systems. In either case, you have the same 100% of 1 Apple. If you divide it into three parts, one part will have to contain the remainder. You can assume that the remainder is insignificant to your application, we often do this. Basically, what you are saying is, you can switch base math systems and recognize the value of something differently, but the value doesn't change. Going back to what Stringy said, the number symbols are only arbitrary representations we assign value to. 100% of the apple still exists, regardless of how you want to look at the apple. You've really not made your case, rather proven my point. You need to use base 12 math to assume the remainder in the division, but this is merely an arbitrarily assigned value you've placed on the problem in order to resolve it. This doesn't change the composition or "value" of the apple. It still remains 100% of an apple!
 
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