Is 1 = 0.999... ? Really?

That might be so but I wasn’t multiplying anything. People who tried that as a proof have that issue, not me.

Do you agree that 1 whole number (1) divided by 3 equal 0.333… ?

Do you agree that 0.333… times 3 equals 0.999… ?

If all that is true, then operators don’t work. At least for base-1.

No I don’t. “0.333…” is not a quantized number, a “quantity”. But 1/3 is a quantity.

I agree that math operators do not work on non-quantity items (anything ending with “…”).

So, obsrvr,

So, This is an interesting theory of numbers!

9/3 = 3
10/3 = 3

I’m not seeing where you are getting that.
Why would 10/3 = 3?

Here’s a proof that (1 = 0).

((1 + 1 + 1 + \cdots) + 1 = 1 + 1 + 1 + \cdots)

Agree?

If the answer is yes, subtract (1 + 1 + 1 + \cdots) from both sides.

What do we get?

(1 = 0)

But if the answer is no, it appears to me that it follows that one of the two sides of the expression is greater than or less than the other – and that means that infinities come in different sizes.

Assuming that I’m wrong, can you help me understand what I’m doing wrong?

Let me see if I understand you.

You have an infinite line and under it you have a dot.

Then you subtract the infinite line away and are left with a dot.

And that confuses you?

And if that confuses you…

When you have 3 parallel lines and subtract 1 parallel line, how many are left?
2

If you then subtract another parallel line, how many are left?
0

2 - 1 = 0

I’m using shorthand before the expansion…

The expansion is .333…

The shorthand works just as well.

9/3 = 3

10/3 = 3

The latter is what Silhouette is arguing

(0.000\dotso1) represents (\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots).

This infinite product never attains (0).

Actually, the only way it can NEVER equal a zero is if it adds 1/10th sequentially. Otherwise, it’s a zero.

What does it mean to add 1/10th sequentially?

1/10th. STOP * 1/10th STOP. * 1/10th STOP etc…

I don’t understand.

It’s means that 1/10th MOVES for EVERY 9 instead of the 1 never occurring (the “the end”)

Wikipedia proofs treat these symbols as representing infinite sums.

(0.\dot9 = 0.9 + 0.09 + 0.009 + \cdots)

Remember that (\infty) has an infinite sum equal to it which is (1 + 1 + 1 + \cdots). This means that if you can’t do arithmetic with infinities (because they are qualities, as you say, and not quantities) then you can’t do arithmetic with infinite sums either. (Which would invalidate Wikipedia proofs.)

Unless, for some strange reason, you don’t think that infinity can be represented as an infinite sum. In that case, I could simply stop talking about infinities and start talking about. . . infinite sums. There would be no difference with regard to my argument.

Do you agree that infinite sums come in different sizes?

Do you agree that ( (1 + 1 + 1 + \cdots) + 1 > 1 + 1 + 1 + \cdots)?

If you do, thank you very much.

But if you don’t, this means that:

((1 + 1 + 1 + \cdots) + 1 = 1 + 1 + 1 + \cdots)

Do you agree that (1 + 1 + 1 + \cdots = 1 + 1 + 1 + \cdots)?

Remember that one of the Wikipedia proofs claims that (0.9 + 0.09 + 0.009 + \cdots = 0.9 + 0.09 + 0.009 + \cdots).

If you don’t agree with this, you also don’t agree with Wikipedia proofs.

If you do agree, let’s subtract (1 + 1 + 1 + \cdots) from the above equation.

What do we get?

We get (1 = 0).

Do you agree with the conclusion?

Are we still doing this?

That’s right.

Except that (0.9 + 0.09 + 0.009 + \cdots) does not equal (\infty). It equals (0.\dot9), which you can easily do arithmetic with.

Your confusing the infinite sum with the result of the infinite sum. 1 + 1 + 1 + … is not just an infinite sum, it also equals (\infty). (0.9 + 0.09 + 0.009 + \cdots), though it is an finite sum, does not.

What I’m saying is that you can’t do arithmetic with (\infty), but you can do as much arithmetic as you want with an infinite number of terms.

I explained all this right after the snippet you quoted above.

Sure, (0.9 + 0.09 + 0.009 + \cdots) < 1 + 1 + 1 + …

I don’t know whether to agree or disagree. I don’t know how to make sense of that.

The problem, Magnus, is that (1 + 1 + 1 + … ) + 1 doesn’t mean anything different than 1 + 1 + 1 + … So yeah, they’re equal.

If you just had a finite sum of 1’s, like this: (1 + 1 + 1) and you added another 1 to it, you’d get this: (1 + 1 + 1) + 1. ← But in that case, why keep the brackets? You can just drop them: 1 + 1 + 1 + 1. ← Now if that was an infinite set of 1’s, you’d write it: 1 + 1 + 1 + … But isn’t that the same as the result above? Just because you have an infinite set of 1s in the brackets doesn’t mean there’s any more reason to keep the brackets. You can drop them for the same reason you can drop them in the finite case (because brackets aren’t needed in sums).

And that, my friend, is why you don’t do arithemtic with infinity.

What?

No, 1 does not then = 0

If they were equal, which they are, because they are both infinite sums of 1, then you have to equally subtract any amount you subtract from one side from the other. You can’t add an extra 1 to the infinite sum, because it is already infinite. You can write a 1 outside of the brackets, but this changes nothing. If you wanted that argument to make sense, you would have to actually write an infinite amount of 1s, and then add one. Which makes no sense, because they are infinite. Math is a notation of quantities, not a representation. Otherwise we would just write 1s until we get the desired amount for any given operation.

The arithmetic holds. 1+(1+1+1+…) = 1+1+1+…

If you take away (1+1+1+…) you get,

0 = 0

Or

1 = 1

Or however many 1s you arbitrarily decide to leave.

Noone said that.

The point is that both (0.\dot9) and (\infty) can be represented as infinite sums.

I am not.

That’s correct. (And yes, you are right, there are infinite sums that evaluate to a finite number. But that’s irrelevant.)

I take this to mean that you agree that infinity can be represented as an infinite sum. In other words, you agree that (\infty = 1 + 1 + 1 + \cdots).

That’s something that has to be proven. One way to prove it is by doing arithmetic with infinite sums. If you can’t do arithmetic with (1 + 1 + 1 + \cdots), because it equals to (\infty) and because (\infty) is not a quantity but a quality, what makes you think you can do arithmetic with (0.9 + 0.09 + 0.009 + \cdots)?

But (\infty) can be represented as an infinite number of terms.

Are you telling me you don’t really know how to do arithmetic with infinite sums?

They had to. One of their proofs is based on the assumption that (0.\dot9 - 0.\dot9 = 0). That’s the same kind of assumption that leads to (1 = 0).

(\infty + 1 = \infty) // subtract (\infty) from both sides
(\infty + 1 - \infty\ = \infty - \infty) // substitute (\infty - \infty) with (0)
(1 = 0)

What’s wrong here is that due to the meaning of the symbol (\infty), it does not necessarily follow that (\infty - \infty) equals (0). It’s indeterminate.

This proof does exactly that. It assumes that the difference between the two underlined infinite sums is equal to zero. Hence the wrong conclusion.

Fine. In that case, you have to accept the conclusion that (1 = 0). But you don’t because you don’t like it – I don’t like it either. We already know that (1 \neq 0), so this is an indication that the proof is fallacious. But where’s the mistake?

Well, in that case, you disagree with this Wikipedia proof because it does arithemtic with infinite sums.

The only way out, it appears to me, is to claim that you can do arithmetic with SOME infinite sums but not ALL of them e.g. you can say you can only do arithmetic with convergent sums (such as (0.\dot9)). But in that case, you’ll have to explain why. And you can’t say it’s because you don’t like the consequences of doing arithmetic with divergent series. That’s not an acceptable answer.

Not really.

If we say that (a = b) this means that (a - b = 0) and that we can substitute every occurrence of (a - b) with (0). If you can’t do that, then (a \neq b). But I guess you want to have it both ways. You want (a) and (b) to be both equal and not equal. Sort of like how Hilbert’s hotel is both full and not full.

The problem is that the way mathematicians define (\infty) implies that (\infty - \infty = 0) is not necessarily true. This means you can’t substitute (\infty - \infty) with (0). It also means you can’t substitute (0.\dot9 - 0.\dot9) with (0).