Here's what I read is going on. You want to talk about an "actual sum" in a meaningful sense, outside of the provided definition. You intuit that it means something, but I'm not convinced it actually does.What I'm saying is very simple. Suppose you had a kind of God calculator that would print out the actual addition of 9/10 + 9/100... what would that be 1 or infinity? That's what I mean by the actual sum. — EnPassant
To show how quirky infinite sums are consider the following (this is not meant to answer anything, it is just to illustrate how strange things become at infinity) — EnPassant
It could be infinity; but it doesn't have to be infinity. You have to define what you mean by infinite sums first before you even get to say this sum is infinity. But let's grant that theorem; it works at least for one definition:Theorem 1
Define 1/x such that 0 < 1/x < 1. If 1/x is summed to itself infinitely often, the sum is infinity. — EnPassant
...then this still does not follow. There are infinite sequences of terms in the range (0,1) such that for any such term x, there's only a finite number of terms greater than or equal to that x. In fact, 9/10, 9/100, 9/1000, ... is such a sequence. So even if you're going to try to apply some variant of a squeeze theorem to prove that this sum is infinite, you just can't do it... because there is no term in this series for which you're adding it or any larger number an infinite number of times.From this we conclude that any positive quantity added infinitely sums to infinity — EnPassant
The point of a limit is that the sum never exceeds it. No matter how many terms you add to that convergent series, it will never exceed 1. Why then would you think it could ever add to to infinity? If it could, that would make it a divergent series, one with no limit, by definition. — Pfhorrest
any positive quantity summed infinitely, is infinity. — EnPassant
dubious. Take x=10^9. I happen to know off the top of my head the left hand side goes below 1/10^9 at term 30 (because I work with computers). So in binary, the sum on the left is 0.111...11 with 30 1's. Take that sum and divide it by 1/10^9, you get a finite number... call that number's ceiling y. At term y, the sum on the right equals the sum of the left before term 30, and you're just adding smaller and smaller terms on the left. In fact, by the time you reach term 10^9 on the right, the right sum becomes 1; and the left sum by that term is simply 0.111...11 with 10^9 1's in binary, which is less than 1. After that, every term you add is going to be less than 10^9 on the left, and equal to 10^9 on the right.1/2 + 1/4 + 1/8+....1/x > 1/x + 1/x + 1/x+...+1/x — EnPassant
This is false soon as the number of terms is greater than or equal to x, after which point the bottom sum is greater than 1 and the top sum is still less than 1. — Pfhorrest
...and you'll find the inequality always breaks down for some number of terms, and all terms after that. In fact, you can cheat... whatever positive integral x you specify, it will break down at the xth term.I'm saying if there are the same number of terms in each. Now increase x indefinitely with the same number of terms top and bottom. — EnPassant
...and you'll find the inequality always breaks down for some number of terms, and all terms after that. In fact, you can cheat... whatever integral x you specify, it will break down at the xth term. — InPitzotl
What are you talking about?The sum of terms 1/x is infinity if 1/x > 0. — EnPassant
It's just that i think about extremely trivial things, which is not a common trait. — Metaphysician Undercover
Your latex is garbled... let me generalize and math this for you. For any positive integral x, no matter how large:What is 1/2c added to itself infinitely? — EnPassant
There is no infinite sum of equals on the left side. For any positive x, no matter how small, there are only a finite number of terms greater than x in that infinite sum. Quick proof...It doesn't matter. An infinite sum of equal infinitesimals must be infinite. — EnPassant
There is no infinite sum of equals on the left side. — InPitzotl
You keep handwaving through the same argument's flaw.Now let the number of terms run to infinity and the sum on RHS is infinite. — EnPassant
So that it works at term 10 is irrelevant, because the inequality fails at term 1024 and for all terms after it. You can't go from 10 into infinity without going passing 1024. — InPitzotl
Now you've struck the heart of the problem. Some quantities cannot be divided in certain ways. It is impossible. Three cannot be divided by nine, it is impossible. Nevertheless, mathemagicians are an odd sort, very crafty, wily like the fox, devising new illusions all the time. They like to demonstrate that they can do the impossible. Some people even believe that they actually do what is impossible. That is a problem. — Metaphysician Undercover
Contemplate this; I have three dozen pies. I divide them amongst the nine of us. — Banno
I'm going to take a stab at your confusion then. Here is the full form of the inequality:I'm not up to speed on binary. I don't think you understand what I'm saying. — EnPassant
Of course you don't, because you keep replying to me. But that's not what the problem is. The problem is that you don't understand what you're saying.I don't think you understand what I'm saying. — EnPassant
Dubious. Your argument was based on this theorem:Now you have an infinite sum of positive quantities > 0 and that's infinite. — EnPassant
There is no such term 1/x that is added to itself infinitely often in 1/2+1/4+1/8+...; nor is there a "squeeze term" such that in that sequence there are terms >=1/x added to themselves infinitely often. For this reason you cannot apply Theorem 1. If you disagree, name the number; but I already gave you a generic refutation... for any number you name, I can tell you how many finite terms there are in the sequence >=1/x, and you cannot name a positive number such that there are an infinite number. Theorem 1 requires something that's not there... therefore, you cannot apply it.Theorem 1
Define 1/x such that 0 < 1/x < 1. If 1/x is summed to itself infinitely often, the sum is infinity. — EnPassant
Yes, we know the answer, we know how it works.
The sorry fact is, that we cannot either describe or simply cannot understand infinity as clearly as we would want. Or infinitesimal and it's relation to numbers. — ssu
Seems you have a short memory. What we previously determined is that I do not believe that 2+2 is the same thing as 4. Remember? You argued that 2+2 is identical to 4, ignoring the difference between equivalent and identical. — Metaphysician Undercover
Okay, I think I got it (incidentally, c=k here, right?), but the same objection applies. You still can't apply theorem 1, because you still can't name an x for which you have an infinite number of terms of the value 1/x such that 0<1/x<1. Every x you name is finite; therefore, every term in your sequence is finite. You don't have an infinite number of 1/x for any 0<1/x<1, so you can't apply Theorem 1.>1/2^c + 1/2^c + 1/2^c to k terms. There are the same number of terms in each series. — EnPassant
You're all over the place here. You have a definition of number that refers to a value (read the newer version of OED; cf to definition 1b of your revision). 1 and .999... being equivalent means they refer to the same value. And don't think I didn't catch that suddenly "refer to" changed to "are"; nevertheless, it's common language to use forms of "to be" to represent equivalence under equality.. — InPitzotl
If .999... represents the same "particular quantity" that 1 does, they refer to the same value, which is what it means to say that they are the same thing. — InPitzotl
What conversation pray tell are you even talking about? How can .999... have a second meaning if .9 means 9/10 and 9/10 is allegedly a problem? And how come you can't be honest about what you're inviting me to do? The problem isn't that you're missing that conversation about why there are numbers that have two representations in the decimal system... the problem is that you don't believe decimals are possible because you have a quixotic quest against fractions, and yet you present to claim that you believe .999... has a meaning at all. I'm not the problem here, MU; I can easily have that conversation with someone who isn't so wrapped up in your fictional world of fraction-denial. I just can't have this conversation with you because you can't face the fact that there's a thing to discuss. — InPitzotl
wonder if MU believes in negative numbers either, or just the naturals. Does zero count to him? — Pfhorrest
I think your problem lies with the distinction between pure and applied maths rather than a distinction between 1 and 1/9. 1, 1/9, -1/9 , 0.9, 0.999i are all numbers in the realm of pure maths. — A Seagull
Everyone here does that. No, what I'm curious about is the apparent absence of humility. Given that others have thought about these issues - many others, over centuries - and given that your way of thinking is so at odds with the way these others have approached the topic, I wonder at the absence of self-correction. — Banno
I guess your response will be to the effect that there are 36 pies, not three dozens, and hence that this is not an example of 3 divided by 9, but of 36 divided by 9. — Banno
Meta is of interest because of his inability to see that ⅓ is just another number, — Banno
Earlier you claimed this as some kind of problem. Your position seems absurd on its face. — fishfry
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