Equal Thirds of One

Damocles said:
Negative numbers are represented in Binary by putting the - sign in front of them.

You use a "sign bit"...

Here is a nice chart for you:

0 1 1 1 1 1 1 1 = 127
0 0 0 0 0 0 1 0 = 2
0 0 0 0 0 0 0 1 = 1
0 0 0 0 0 0 0 0 = 0
1 1 1 1 1 1 1 1 = −1
1 1 1 1 1 1 1 0 = −2
1 0 0 0 0 0 0 1 = −127
1 0 0 0 0 0 0 0 = −128

The first digit of each of these is the "sign bit". In binary it is possible to express odd things like -0 and other such.
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Yep I was not sure if that was universal or not though.
It is dependent on computer hardware.
 
AssHatZombie said:
Who knows how to make a number negative in binary? 10 bonus points.
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Don't confuse binary with necessarily being computer methods.

You can do it with a negative sign (-) or you can use the leftmost bit. Just like 1 / 3 it is just a symbol and neither is necessarily more correct than the other.
 
Damocles said:
Negative numbers are represented in Binary by putting the - sign in front of them.

You use a "sign bit"...

Here is a nice chart for you:

0 1 1 1 1 1 1 1 = 127
0 0 0 0 0 0 1 0 = 2
0 0 0 0 0 0 0 1 = 1
0 0 0 0 0 0 0 0 = 0
1 1 1 1 1 1 1 1 = −1
1 1 1 1 1 1 1 0 = −2
1 0 0 0 0 0 0 1 = −127
1 0 0 0 0 0 0 0 = −128

The first digit of each of these is the "sign bit". In binary it is possible to express odd things like -0 and other such.
Click to expand...

Flip the bits and add 1. Yes. that is correct.
 
Ok board.

I am going to the mathematics department sometime this week. I swear to god, I will get a professor to look over dixie's comments. I will report back to you with the findings.
 
Grind said:
Ok board.

I am going to the mathematics department sometime this week. I swear to god, I will get a professor to look over dixie's comments. I will report back to you with the findings.
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Grind, why don’t you take his comments to the psychiatric department instead? Do you have a neurology department, they might diagnose him with a brain tumor?
 
Grind said:
Ok board.

I am going to the mathematics department sometime this week. I swear to god, I will get a professor to look over dixie's comments. I will report back to you with the findings.
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LOL. I'm holding you to this.
 
Entertaining thread.....you can't read it and not laugh out loud...

Damo and others have explained it time and time again....

Dix is just stuck in the decimal base 10 number system and can't think outside of it....
He doesn't understand its limitations and gives those those limitations he finds more power and meaning and they warrant...
give it up...
 
Stay in Base 10, within the topic of the discussion guys. Nowhere have I stated that thirds can't exist or be defined in other base number systems. It's just a really simple elementary math problem, and while I am not surprised at the pinheads wanting argue about it, I am quite surprised at Damo and Grind, I thought you two had some common sense.
 
This is where Dixie tries again to weasel out from what he said. But he will be back if you poke him a little.

Dixie said:
Stay in Base 10, within the topic of the discussion guys. Nowhere have I stated that thirds can't exist or be defined in other base number systems. It's just a really simple elementary math problem, and while I am not surprised at the pinheads wanting argue about it, I am quite surprised at Damo and Grind, I thought you two had some common sense.
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You have argued that equal thirds cannot exist.

Your equation assumes the remainder. As I said, we assume the remainder and presume the problem as resolved, we have no other choice. Your assumption is not incorrect, it just isn't definitively equal. 1/3=.3333E and when you add that together 3 times, it comes out to .9999E, which is not 1, but as close as we can get to 1 without being 1. The remainder becomes insignificant at some point, depending on the application, but the remainder still exists. Maybe it is defined as "equal" to the closest tenth, hundredth, thousandth, millionth, billionth, or trillionth... depends on the application, but unless you know of some way to resolve the remainder, it can't be defined and must be assumed at some point down the line.
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WTF does this mean other than you think base 10 has some impact on the application of dividing 1 by 3 in the concrete?

Nowhere have I stated that thirds can't exist or be defined in other base number systems.
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Using base 10 we can represent one third with .333333e. In base 10 .333333e = 1/3 and we do not need to assume any remainder anywhere in the equation when multiplying it by 3 to get 1. And that is simple elementary math.
 
You have argued that equal thirds cannot exist.
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Nope... I never said that... pay attention, one can't be divided evenly by three because a remainder is produced. Since this remainder can't be resolved, it is assumed. We assume 'equal' thirds, which means they exist.

WTF does this mean other than you think base 10 has some impact on the application of dividing 1 by 3 in the concrete?
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Because other base math systems are not applicable to my argument. It would be like saying there are 27 letters in the alphabet, therefore I am wrong. I have already stated that things can be divided by three in other base math systems, things can even be divided by three in base 10, and a perception or assumption of equality made. So these idiotic arguments designed to change my argument into something I never said, are not admissible....Sorry.

Using base 10 we can represent one third with .333333e. In base 10 .333333e = 1/3 and we do not need to assume any remainder anywhere in the equation when multiplying it by 3 to get 1. And that is simple elementary math.
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Well if you don't need to assume the remainder, what happened to it? It clearly exists in your equation, so it has to be rectified or you have to assume it. There is no other option but to assume it, we can't calculate infinity. One divided by three equals .3 with a remainder of .1, and you can keep continuing to divide that remainder forever, you will not resolve it. If it can't ever be resolved, it has to be assumed, there is no other alternative. Eventually, one of the three parts of the thirds will have to include this extra portion, or remainder, there is no other way to resolve the problem. And YES, it can be resolved in different base math, and YES it can be assumed or perceived as resolved in base 10 math, that is not my argument. Never has been.
 
Dixie said:
Nope... I never said that... pay attention, one can't be divided evenly by three because a remainder is produced. Since this remainder can't be resolved, it is assumed. We assume 'equal' thirds, which means they exist.
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Do you know what the word assume means? In this context, it means we suppose it for the sake of argument. But we have no need to do that whatsoever. 1 / 3 * 3 = 1, fucking end of story retard. It does not equal .9, .99, .999, .9999, etc. It equals 1.
 
RStringfield said:
Do you know what the word assume means? In this context, it means we suppose it for the sake of argument. But we have no need to do that whatsoever. 1 / 3 * 3 = 1, fucking end of story retard. It does not equal .9, .99, .999, .9999, etc. It equals 1.
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The expression "1/3" is a division problem, not a numeric value.
 
That is not really relevant to what I said (that 1 / 3 * 3 = 1) but again you are wrong. BTW, I put the spaces to represent it as division rather than a fraction though there is no real difference, because 1 / 3 = 1/3.

1/3 is a numeric value. It is a representation of one third just as "one third," .3333e in base 10, .1 in base 3, etc. are representations of one third and it is in numeric form.

http://www.merriam-webster.com/dictionary/fraction

Main Entry:
frac·tion Listen to the pronunciation of fraction
Pronunciation:
\ˈfrak-shən\
Function:
noun
Etymology:
Middle English fraccioun, from Late Latin fraction-, fractio act of breaking, from Latin frangere to break — more at break
Date:
14th century

1 a: a numerical representation (as 3⁄4, 5⁄8, or 3.234) indicating the quotient of two numbers b (1): a piece broken off : fragment (2): a discrete unit : portion2: one of several portions (as of a distillate) separable by fractionation3: bit , little <a fraction closer>
 
Case in point... If I tell you I have 5 liters of gas in my tank, you can convert the value of 5 liters into gallons and tell me how many gallons of gas are in my tank... However, if I tell you I have 1/3 tank of gas, can you tell me how many gallons I have? Nope! 1/3 is a division calculation. When you work the calculation, it is unresolved because of the remainder, therefore, it has to be assumed. 1/3=.333E (the E means 'eternally') If you add two of these suckers together, it equals .666E and three of them equals .999E, we assume the remainder and call it 1, because we can't resolve an infinite value.
 
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