Now me: If you write 9s for infinite time, have you written infinite 9s? Compared to infinite time, how much time did you spend writing a single 9? The sum total of the time spent writing 9s, over infinite time is what?
If you write nines over an infinite amount of time, with each nine written over a finite amount of time, even a billion years between each 9 you will end with infinite nines. A better way would be to say I write a nine down. Then a minute later another nine. Then 1/2 min later another nine and 1/4 min later another then 1/8 another and 1/16 another etc. After two minutes I will have written an infinite number of nines!
So you agree that you will write infinite 9s. Answer the other two questions, and stop dodging: Over infinite time, what is the amount of time it takes you to write a single 9. Take even the largest amount of time, and compare it to infinite. Is it very small or Zero? Now take the sum of all the time spent writing 9s, individually (that’s the step directly above), is it zero, or infinite? Because according to you, and TennTra, and mathematics, over infinite time, it will take you ZERO time to do any given step, even the longest step. (This is the half distance problem cotton gave you earlier). Thus you must conclude that the sum of the time spent, over infinite time, is zero, correct? And yet, you would probably say it takes infinite time (which it does.) So you have a contradiction, and you’ll avoid it, because dogma.
Let's say that it takes me 1 second to write down a 9 on the piece of infinitely large piece of paper. 1 + 1 + 1 + 1 + 1 + ... Or ∑ 1 (n 1 -> ∞) This is a divergent series and will tend towards infinity. However, the time it takes to write each nine is static: 1 second. Even if you go to infinity, each 9 will take exactly 1 second to perform. So, no, the time it takes to write a 9 does not go to zero. The percentage of the time in total each task takes will go to zero, that is true, but each individual task will take exactly 1 second, no more, no less. Even if we use some astronomical time frame for each 9, say a million years, the time it takes to write each 9 will not change, but you will, over infinite time write down infinite 9s. There is no contradiction.
You are absolutely contradicting yourself. The amount of time between the first second and infinite is infinite. Meaning T sub zero is infinitely far from T sub inf, and T sub 1 is ininfitley far from T sub inf, and T sub 2 is infinitely far from T sub inf. And if you don’t call that zero, since it is infinitely far away, then you cannot call something infinitely close 0, ie: cotton’s half distance problem.
I have no idea why you would expect anyone to call something infinitely far away 'zero' just because you call something infinitely close 'zero', which is not what I am doing anyway, because infinitely close but not the same as zero, it is ε, the infinitesimal. There are no infinitesimals in the Real numbers, but even in the hyperreals and the surreals where ε lives, .999... is not infinitely close to 1, it is 1.
Let us expand 1/3: 3|1.00000000… -> 0.333333333… Is there some remainder even after the performing of infinite computations? I will now argue why that answer is “no.” Doing long division on a piece of paper we can all see how we get that .3333… when we try to convert the fraction 1/3 into a decimal notation. However, it can be argued that 1/3 != .333333…. but is actually equal to .3333333…1, some remainder out there that we would write as a 1. But this is not true, and here is why. We will start the long division and stop after one iteration: `` 0.3 3|1.0 > >0 > ---- > > 10 > > 9 >>---- > > >1 Now, we see we have this stubborn (1) left over at the bottom of the computation, and I think this is where we get hung up because then we want to write it as .333333…1 after infinite iterations. But there is another way to expand 1/3 into decimals without relying on doing long division of 3|1. 1/3 = 3/10 + 1/30 (3/10 = 9/30 so 9/30 + 1/30 = 10/30 = 1/3) 1/3 = 3/10 + 3/100 + 1/300 So, we can see that 1/3 = .3 + .03 + 1/300 Or 1/3 = .33 + 1/300 Now, let’s expand out 1/300: 1/300 = 3/1000 + 1/3000 (3/1000 = 9/3000 so 9/3000 + 1/3000 = 10/3000 = 1/300) 1/300 = 3/1000 + 3/10000 + 1/30000 1/300 = .003 + .0003 + 1/30000 = .0033 + 1/30000 Therefore 1/3 = .33 + .0033 + 1/30000 = .3333 + 1/30000. And what is that remainder of 1/30000? More threes. Do this infinite times, and all you get are 3s. There is no remainder of 1 left over after the infinite threes, because the infinite threes ARE the “remainder”. It is all self-contained. There is no infinitesimal. Therefore, 1/3 is exactly .333333… .
I don't call anything infinitely close, zero. You do. Cotton stated that if he started at X, and moved toward Mexico taking half the distance every time, for infinite, that he'd never get to Mexico (0), because he's constantly taking half. And you said he would get get there, that it would go to zero. I'm saying it gets really, really, really close to zero, but never zero. In other words, it gets infinitely close to zero, but never zero. By saying it is zero, you are stating that infinitely close to zero means zero. That infinitely close to 1 means 1. So back to the 9s, example, you've written 0.9.. It took you 9 seconds to write each 9. T sub inf, you are infintely far away from T sub 0, and T sub 1, and T sub 2, and T sub 3, etc. Ok? So now T sub 0 is infinitely far away from T sub inf. And T sub 1 is infinitely far away from T sub inf. And T sub 2 is infintely far away from T sub inf. Would you not call them equal? And if they are equally infintely far from T sub inf, which is how far away 0 is from T sub inf, would you not call them zero? Meaning they are infinitely close to zero, from the persepective of T sub inf. So how do you differentiate that from what you said to cotton re: Mexico?
Monkey. Long division. He absolutely gets 0.3.. across the top. He absolutely gets infinite remainders, because that's what drives the 0.3.. across the top. If he didn't have a remainder, he wouldn't produce infinite 3s. Period. And because infinite is infinite, and if you do something infinitely get you infinite, he does remainders infinitely, and gets infinte remainders. Just very very small. Thus 1/3 = 0.3.. + [infinitely small remainder]
What is the definition of T(inf)? What is its value? Where is it behind the decimal point? I am afraid you are using infinity as a number but you cannot do that. It is not a number.
1/3 = .3333.... + r where r is [infinitely small remainder] 10/3 = 10 (.33333.... + r) 10/3 = 3.33333.... + 10r Subtract top from bottom 10/3 - 1/3 = 3.33333... - .3333333... + 10r - r 9/3 = 3 + 9r 1/3 = 3/9 + 9/9r 1/3 = 1/3 + r r = 1/3 - 1/3 = 0 The remainder is equal to 0.
T sub 0 is not a number, it is a representation of a concept. T sub inf is not a number, it is a representation of a concept. T sub inf is the representation of “as time approaches infinity” If I’m performing an action for infinite time, I can be said to be performing that action between T sub 0 and T sub Inf. You’ll also see n used, to represent an infinite number of steps. You’d be more familiar calling it i-> sideways8 It’s the same thing. As time approaches infinity, I’ve been writing 9s. And I have written 0.9.. The time since I wrote no 9s is infinite. The time since I wrote the first 9 is infinite. The time since I wrote the second 9 infinite. And on and on. So the time I wrote the first and second 9 are infinitely close to zero, because infinite time is so much larger than that time, such that proportionality, it is so close to zero it’s near indistinguishable from zero. How do you differentiate infinitely close to zero = 0 (as in half distances to Mexico) = 0, and infinitely close to zero when writing 9s <>0? You say it isn’t a contradiction. It is. Both are infinitely close to zero. Thus the same logic must apply. Either both are some very small amount greater than zero, or both are zero.
You are using a flawed set of rules to prove your rule. It is circular. Mathematics doesn’t account for the remainder, so it isn’t included when setting the two things equal. The remainder is there. It is there logically, it is there operationally, and it is there infinitely. You are masking it in your calculation because mathematics is illogical on this subject, owing mostly to a lack of understanding of concepts, due to simple rote regurgitation of “proof” without deeper consideration.
You do realize you are showing that the remainder when subtracting 1/3 - 1/3 is zero, right? Not that 1/3 can’t have a remainder. These are two different things.
Why not leave it as fractions? What ya worried about showing? I don’t mind a discussion; don’t be disingenuous. You have a proof that shows a contradiction. Which means you don’t have a proof.
Treat 0.3.. as a string. Do not manipulate it at all. It is constant at exactly 0.3.. You are allowed no more decimal places than 1. All representations of it will be as 0.3.. You will always arrive at: r = 1/3 - 0.3.. Every time. In other words, the contradiction is that r = 0 and r = 1/3 - 0.3.. It isn’t both. It is 1/3 - 0.3.. which is that infinitely small remainder. And here is the reason: The remainder stems from an operation. That which produces the next decimal place over from step n produces the remainder. If you take even one additional decimal place over, you have cut out the remainder, you lose it. So when Alan does 3.33.. - 0.3.. he’s lost the inherent distance from zero, that would be used to show that the remainder = 1/3 - 0.3.. = very, very small value > 0 And the distance from a fixed point is more important in infinite size than anything else, because infinite is so large. So 0.3.. is different from 0.33.. because 0.3.. has a stated distance from zero of 0.3 whereas 0.33.. has a stated distance from zero of 0.33 And 0.33 <> 0.3 But that one decimal place shift is enough to chop out the remainder. So like I said, treat it as a string, and it will never reduce to less than r = 1/3 - 0.3.. which is very very small, but backs the logic exactly.
