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

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.

To convince me, however, you metaphorically appeal to the God calculator, and sprinkle in "actual" as adjectives. But that's not convincing for me. I read both the metaphor and the adjective as just reifying.

God didn't give us addition on tablets; we invented it. The "base" definition of addition works recursively down to base cases, so that's fine for finite numbers of terms. But you cannot reduce an infinite recursion down to base cases. So there's no a priori definition of infinite sums.

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

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

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:

From this we conclude that any positive quantity added infinitely sums to 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.

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

I don't insist it doesn't sum to 1. You may well be right. I'm saying we don't know because we can never have an actual infinite sum. An infinity of positive quantities are being summed and any positive quantity summed infinitely, is infinity.

What is the 'last' term in the sequence 1/2, 1/4, 1/8...? No need to answer, it is a rhetorical question. But surely all terms - the whole infinity of them - are positive and > 0. Right? Now sum an infinity of positive quantities...

any positive quantity summed infinitely, is infinity. — EnPassant

That’s where you’re wrong. Just flat wrong. You tried to show that true and I showed it false in just three terms.

The same positive quantity added to itself infinitely many times is infinity, sure. But not every series is like that. No convergent series is like that. A series like that can have no limit. Any series with a limit is unlike that. A series with a limit sums to that limit. That’s what the limit is.

the whole infinity of them - are positive and > 0. Right? — EnPassant

That's insufficient to use your theorem, as I explained in my previous reply.

1/2 + 1/4 + 1/8+....1/x >

1/x + 1/x + 1/x+...+1/x

Now let x go to infinity-

$\sum_{i=1}^{\infty} 1/x = \infty$

1/2 + 1/4 + 1/8+....1/x > 1/x + 1/x + 1/x+...+1/x — 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.

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

x is not static. 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.

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 positive integral x you specify, it will break down at the xth term.

...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

x is a power of 2.

1/2 + 1/4 + 1/8+....+1/1024 >

1/1024 + 1/1024 + 1/1024+....+1/1024 for the same number of terms.

Now go to infinity with x. You are still adding an infinity of positive terms.

1/2 + 1/4 + 1/8+....+1/1024 > 1/1024 + 1/1024 + 1/1024+....+1/1024 — EnPassant

So that's 10 terms. What happens at the 1024th term?

Your left sum is 0.111....11 with 1024 1 bits in binary. Your right sum is 1. Is 0.111...11 with 1024 1-bits greater than 1?

Your left sum is 0.111....11 with 1024 1 bits in binary. Your right sum is 1. Is 0.111...11 with 1024 1-bits greater than 1? — InPitzotl

The sum of terms 1/x is infinity if 1/x > 0.

The sum of terms 1/x is infinity if 1/x > 0. — EnPassant

What are you talking about?

At the 1024th term on the left, we're adding 1024 terms... in binary point, 0.1, 0.01, 0.001, ... , 0.00...001 (with 1 in the 2^-1024th place). That sum is 0.111...11 (with 1024 1's).

At the 1024th term on the right, we're adding 1024 terms, each of which is 1/1024... that is by definition of multiplication equal to 1024*(1/1024), which is 1024/1024=1.

After the 1024th term, we're adding numbers on the left much smaller than 1/1024; in fact, they're smaller than 1/2^1024. And on the right, for each of these, we're just adding 1/1024.

After the 1024th term, we're adding numbers on the left much smaller than 1/1024; in fact, they're smaller than 1/2^1024. And on the right, for each of these, we're just adding 1/1024. — InPitzotl

What is $1/2^c$ added to itself infinitely?

$1/2^c$ added $2^c$ times is $1$

Now add another $2^c$ terms$= 1$

1 + 1 + ... = $\infty$

in fact, they're smaller than 1/2^1024. — InPitzotl

It doesn't matter. An infinite sum of equal infinitesimals must be infinite.

It's just that i think about extremely trivial things, which is not a common trait. — Metaphysician Undercover

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.

What is 1/2c added to itself infinitely? — EnPassant

Your latex is garbled... let me generalize and math this for you. For any positive integral x, no matter how large:
$\displaystyle\sum_{i=1}^{x}{2^i} \lt \sum_{i=1}^{x}{x} = 1$
...intuitively, you can see this by "argument from binary". The left sum is always:
0.111...11 with x 1-bits. That's always less than 1.

...and for each y>x:
$\displaystyle 2^y \lt 1/x$

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. 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...
1. Pick your x.
2. Write 1/x in binary
3. Count the digits; call the number of digits plus one n.
4. 2^n is greater than your 1/x.
5. 2^-n is less than your x.
6. All terms after the nth term are less than 2^-n.
(eta: corrections)

There is no infinite sum of equals on the left side. — InPitzotl

1/2 + 1/4 + 1/8+....+1/1024 > 1/1024 + 1/1024 + 1/1024+....+1/1024

LHS has the same number of terms as RHS

Now let the number of terms run to infinity and the sum on RHS is infinite at infinity.

The next inequality would be-

1/2 + 1/4 + 1/8+....+1/2048 > 1/2048 + 1/2048 + 1/2048+....+1/2048

Now let the number of terms run to infinity and the sum on RHS is infinite. — EnPassant

You keep handwaving through the same argument's flaw.

You have an inequality that's true for term 10:
0.1111111111b > 1024+1024+1024+1024+1024+1024+1024+1024+1024+1024

That's all fine and dandy. But it doesn't hold at term 1024:
0.11111111......111b ≯ 1024+1024+1024+1024+1024+1024+1024....1024
(with 1024 1-bits on the left, and 1024 1024-terms on the right).

In fact, at that term:
0.11111111......111b < 1024+1024+1024+1024+1024+1024+1024....1024
...and after that term, you're adding values less than 2^-1024 on the left, which is << 1/1024... but for each such term, you're adding 1/1024 on the right.

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 passing 1024.

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

I'm not up to speed on binary. I don't think you understand what I'm saying.
Are all of the terms in 1/2, 1/4, 1/8...positive and > 0? Yes.
For all c 1/2^c is positive and > 0.

Let c run to infinity and sum. Now you have an infinite sum of positive quantities > 0 and that's infinite.

There's no point in saying calculus says otherwise because calculus does not deal explicitly in infinite sums. You need the God calculator for that.

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

All we have as evidence for the impossibility od three being divided by nine is your insistence. Further, I am certain that dividing none by three is a reasonably straight forward activity.

Contemplate this; I have three dozen pies. I divide them amongst the nine of us.

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.

You have done this several times in this thread; taking a common way of speaking and arguing that it is incorrect. You did the same in the conversation where you denied that one could calculate the velocity of an object at a given time, and indeed you did much the same thing in the discussion of the Tractatus.

@Pfhorrest, @InPitzotl, @A Seagull: The discussion here is not mathematics, which Meta plainly has misunderstood. I said before that the interest here is in identifying the origin and progress of the crack. It seems to me to lead to Aristotle, and to a curiously inert language, in which "number", "velocity", "fact" and other terms are arbitrarily and unilateral forbidden their usual use. Meta is of interest because of his inability to see that ⅓ is just another number, that velocity is calculable and that facts change over time. The lesson is that one can show someone the solution to their philosophical issues, but they may not see that solution.

Contemplate this; I have three dozen pies. I divide them amongst the nine of us. — Banno

Personally I think mathematics is not really about numbers. Mathematics is more about harmonies and proportion. Numbers are 'markers' in the symphony of proportion and relation. The real music of mathematics is beyond numbers. Just a thought...

I'm not up to speed on binary. I don't think you understand what I'm saying. — EnPassant

I'm going to take a stab at your confusion then. Here is the full form of the inequality:
1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512+1/1024 > 1024+1024+1024+1024+1024+1024+1024+1024+1024+1024

There are ten terms here. The sum on the left as it turns out is 1023/1024. The sum on the right is 10/1024. 1023/1024 > 10/1024, as you said.

I can't write the full form of 1024 terms without basically flooding the channel... 1024 terms itself is large enough, but the value of 2^1024 itself is huge:
179769313486231590772930519078902473361797697894230657273430081157732675805500963132708477322407536021120113879871393357658789768814416622492847430639474124377767893424865485276302219601246094119453082952085005768838150682342462881473913110540827237163350510684586298239947245938479716304835356329624224137216

But we can write the sums in power notation. The sum of 1024 terms of 1/2+1/4+...+1/2^1024 = ((2^1024)-1)/2^1024. That is less than 1 (just barely... by 1 divided by that huge number above, but less is less). But 1024*(1/1024) is equal to 1. So:
1/2+1/4+1/8+...+1/2^1023+1/2^1024 < 1024+1024+...+1024
...because:
((2^1024)-1)/2^1024 < 1

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.

Now you have an infinite sum of positive quantities > 0 and that's infinite. — EnPassant

Dubious. Your argument was based on this theorem:

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

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.

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

Hang on - you can't have it both ways; you can't both say that we know how it works and yet we don't understand it.

If we know the answer, what exactly is missing?

Mathematics consists in explicit patterns. Proportion and relation are elements of these patterns.

Hence mathematics sets out a multiplicity of patterns, of ways of speaking that do not lead to contradiction.

And so it should not be a surprise that some of these patterns, these ways of speaking, can be used for doing things like economics and physics.

1/2+1/4+1/8+...+1/2^1023+1/2^1024 < 1024+1024+...+1024 — InPitzotl

I'm not talking about the 1024th term. I'm saying-

1/2+1/4+1/8+...+ 1/2^c to k terms - whatever the value of k

>1/2^c + 1/2^c + 1/2^c to k terms. There are the same number of terms in each series.

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

I remember all too well, which is why I'm not joining in with the rest of the gang arguing with you. Nice job trolling them all though. Your objection isn't to the theory of convergent infinite series of real numbers. Your objection starts with 2 + 2 = 4. You seem to agree.

I surely didn't ignore the difference between equivalent and identical. On the contrary I wrote many posts carefully explaining the difference; and showing through actual mathematical proof, as well as other methods, that 2 + 2 and 4 designate the same mathematical object. One object, two names.

You do agree that a single object or thing may have more than one name. Earlier you claimed this as some kind of problem. Your position seems absurd on its face.

Indeed, Meta delights in obtuse objections.

>1/2^c + 1/2^c + 1/2^c to k terms. There are the same number of terms in each series. — EnPassant

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.

Or think of it this way. Note that every time you add a term, you change all of the terms. We go from 1/2, to 1/4+1/4, to 1/8+1/8+1/8, and so on. But note also that we can actually sum these partial terms too... 1/2=1/2, 1/4+1/4=1/2, 1/8+1/8+1/8=3/8, and so on.

The general form here is:
$\displaystyle\sum_{i=1}^{k}{\frac{1}{2^k}} = \frac{k}{2^k}$
But when you generalize this, you're talking about what that form becomes, so you really mean:
$\displaystyle\lim_{k\to\infty}{\sum_{i=1}^{k}{\frac{1}{2^k}}} = \lim_{k\to\infty}{\frac{k}{2^k}}$
But your theorem 1 just says:
$\displaystyle\lim_{\to\infty}{\sum_{i=1}^{k}{\frac{1}{x}}}=\infty$ where $0\lt\frac{1}{x}\lt1$
...so doesn't apply.

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

I do not argue against the fact that mathematicians believe that .999..., and 1 refer to the same value. The difference between these two is a difference which does not make a difference, for them, so they say that it is the same value. But that doesn't prevent me from arguing that the claim that there is a difference which doesn't make a difference is a contradictory claim.

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

My argument, if you've read what I posted, is that .999... does not represent a particular quantity. I suggest that you come back when you've got an argument to make. The fact that mathematicians believe that .999... has the same value as 1 is just evidence that they are wrong, it's not an argument that .999... refers to a particular quantity.

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

Sorry, I have no idea of what you're talking about here. I never said .9, or 9/10 is a problem. I said these do not represent any particular quantity, and ought not be considered as numbers. It is the belief that they are numbers which is what I consider to be a problem.

wonder if MU believes in negative numbers either, or just the naturals. Does zero count to him? — Pfhorrest

I believe that zero is a very complex idea with numerous different meanings, some inconsistent with each other, as exemplified by imaginary numbers.

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

I agree that these are all considered by mathematicians, to be numbers. What I haven't seen yet is a definition of "number" which validates this belief. It is possible that a person, or even a whole group of people, believe that a certain thing is such and such a type of thing, but when a clear definition of that type of thing is made, it turns out that the thing is actually not that type of thing. Take Pluto for example. Everyone believed it was a planet, until a clear definition of "planet was made, then the people realized that Pluto actually wasn't a planet. The same might be the case with some of these things which people believe are numbers. Until a clear definition of "number" is produced we will not know if this is the case. According to the definition I proposed, some of these are not numbers.

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 wonder where you get your idea of correct from. That everyone does it, doesn't make it correct, read my example above. You support mob rule?

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

If you knew your example was so bad, why present it? Clearly "three dozen" does not represent a quantity of three, just like "four score" does not represent a quantity of four, and "twenty six" does not represent a quantity of twenty. Sometimes I wonder Banno, how you can go so far as to conceive such bad arguments. It must take strenuous effort to make your arguments so bad.

Meta is of interest because of his inability to see that ⅓ is just another number, — Banno

I've explained very thoroughly why 1/2 is not a number. I have yet to see a counter argument, only your extremely bad example which premises that "three dozen" represents a quantity of three. That false premise disqualifies the argument as unsound.

Earlier you claimed this as some kind of problem. Your position seems absurd on its face. — fishfry

I didn't claim this is a problem, that was Pitzotl''s misinterpretation. I said that if the same thing has two distinct names, there is a reason for that.