Is 1 = 0.999... ? Really?

Alright, so let’s look at it a different way.

When does the 1 in 1/10 ever occur? If it always occurs in 1:1 correspondence then that means that .111… equals .0…1.

Problem is, you have the carry from THE LAST DIGIT!!

Slight problem with that, there is no last digit.

Carrying doesn’t work from the first digit

Oh, by the way QED.

I already pm’d prom and told him he can collect his reward.

At this point the same errors are recurring indefinitely, like this new (10^{-\infty}) notation still treating infinity like a finite quantity.

As Ecmandu correctly pointed out, there is no last (10^{-1}) in what that’s supposed to denote, just the same as there is no last (1) in (0.\dot0{1})

Until something comes up that hasn’t already been disproven (\lim_{n\to\infty}10^n) times over, I’m gonna go ahead and leave the nonsense peddlers to it.

Does reading up on last minute math make you an expert as you go along, Sil/Twit? It seems not!

I await both your responses with baited breath! Show me Math! Teach me! I doubt you can, but I know you’ll try.

Actually, I know Magnus well enough that he’s going to take the last two posts very seriously … you’re embarrassing yourself here.

If you don’t like the notation (merely because you don’t understand it, I must say), you may consider its equivalent which is (\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots). That’s a “valid” numerical representation, isn’t it?

I’ve already stated it but you conveniently ignored it opting instead to focus on what you don’t understand.

It seems like you’re one of those rare people who can understand Ecmandu :slight_smile:

There is no last (1) in (0.\dot01). The product (\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots) is infinite. The end is merely in the symbol.

“The end is merely in the symbol”

There is no end!

I’m going to explain this as simply as possible!

You need a last digit to carry.

There’s is no last digit to 0.0…1.

The 1 never gets expressed.

In order for the 1 to get expressed. The added decimal HAS to be 0.111…

The problem with this, is that you have to carry from the first digit!

You can’t carry from the first digit!

Last digit = contradiction
First digit = contradiction

You’re now left with no wiggle room

A couple of questions for Ecmandu.

What does it mean to say that (1) occurs in (\frac{1}{10})?

I am not sure what it means for (1) to always occur in (\frac{1}{10}) in one-to-one correspondence.

What I know is that (0.\dot1) does not equal (0.\dot01) and that it’s very easy to demonstrate this.

Why do you mention carrying?

My post about 1/10 ever being expressed implied your argument of 1/101/101/10… etc…

I thought I could have used short hand, but it doesn’t bother me that we’re trying to be precise here, so, I apologize for the shorthand.

So here’s the deal: “carrying” is when you carry numbers from right to left. This is first grade math.

If I have a number like 99+2… the 2 plus the last 9 is 11. The one gets deposited in the solution column and the 10 gets carried over to the left in the tens column as a 1. 1+9 equals 10, which you put in front of the one that you already have the the solution column, to equal 101.

That’s correct.

Sure. But how is that relevant? That was my question.

Why do you mention carrying?

It’s obvious that you’re talking about the standard algorithm of adding numbers together. The question is: why?

Are you trying to use the standard algorithm to add (0.999\dotso) and (0.000\dotso1) together? If so, why?

Do you think that you can prove that there is no difference between (0.999\dotso) and (1) by proving that the standard algorithm of adding numbers together can’t be used to add (0.999\dotso) and (0.000\dotso1) together?

Ok, fine. Notice I don’t back up to former arguments, I always use new ones.

Let me ask you this from a utilitarian perspective.

Is there anything that 0.9… could be used for that 1 cannot ?

Actually I’ve known all this math since over half my life ago, and I have been regularly keeping up with advanced level maths to this day for fun - I actually really enjoy it.

The problem here though is if others aren’t close enough to the same level, they aren’t going to recognise this and will just assume it’s just another bogus claim from some randomer on the internet because they don’t know any better.

I’m dealing with a self-professed non-mathematician here and it shows. All he has left anymore is “you just don’t understand”, and when I say/prove that I do, he can just respond with incredulity and keep repeating the accusation. That’s all this thread is now - non-mathematicians claiming long-time competent mathematicians don’t understand, whilst repeating the same schoolboy mistakes over and over. Other competent mathematicians here recognise these mistakes too, and keep advising the non-mathematicians that they’re wrong and I’m right. But our old-friend “the backfire effect” is just making them double down harder because admitting you’re wrong is hard for people since it requires emotional maturity and intellectual honesty to overcome the cognitive dissonance. That’s why there’s no longer any point in me trying to help them anymore - they don’t want it.

I only recently learned all the MathJax code to express the math nicely on this forum though - I already thanked Magnus for his only contribution so far of bumping the post from a couple of years ago that mentioned it and linked to its documentation. I also unequivocally accept the utility of exploring what happens when you take something that doesn’t work mathematically, and treat it as something that does work - as I’ve mentioned several times by now. All I’m doing is explaining exactly why it doesn’t work mathematically - I’m not even against doing it anyway because that’s where new math can be born, but there’s still a couple of people left who find even that too much to accept.

See, I’m humble enough to give credit where credit is due, and I’m simply honest about people being wrong when they are.
There’s plenty of topics on this forum that I leave alone because I don’t know enough about them - you won’t find me reading up last minute on those topics and claiming expertise there, and you’ll only find me claiming expertise when I actually have it, such as here. There’s far better mathematicians out there than me, but the dissenters here most definitely aren’t one of them.

Fuck off, Tramp!

If you ask me, you spend a little bit too much time telling us what you think about yourself.

That’s exactly the point of this thread. (0.\dot9\ \neq 1) means the answer to your question is “Yes”.

That’s my side! What the fuck are we arguing about?

Oh! That’s right! You think if you can add something to .9… that it doesn’t need to be added to 1!

So here we are!

Yes, that’s where we agree. Where we disagree is whether or not infinities come in different sizes. You think that they don’t (that’s where you agree with SIlhouette.)

Not sure what that means.

So let’s get this straight!

You, sil and I all agree that 0.9… /= to 1

phew

Glad we got that out of the way!

I’m only speaking for myself, but I think both Sil and I thought you were defending the equality.

Cool! So let’s talk about orders now!

In response to your last question, if 1=0.9…

Then whatever you add (infA) to one, you have to add to the other (which makes the inequalities)

Does that clarify?

I think that Silhouette thinks that (0.\dot9 = 1).

Here’s Silhouette’s “proof” that (9.\dot9 = 10).

I’m afraid we didn’t (:

Not quite.

Am I misunderstanding everyone?!?!

That wouldn’t be the first time.

Let’s let sil answer.

As for the last part:

If 0.999… plus 0.0…1 equals 1, then!

If 0.9… equals 1, then!

1 plus 0.0…1 must equal as well (which it doesn’t)