Is 1 = 0.999... ? Really?

What is an “atheistic universe”? Everyone there is an atheist?

It’s actually much more elegant than that!!

You can co-exist in the same exact world system and the system is internally and externally consistent for atheists and creationists… this type of precision is mind blowing!!

Bah, humbug.

Atheists think there is no creator and theists think there is a creator. The atheists are wrong if a creator made the universe. Nothing mind blowing.

I’ll add to this (sorry James)

The options are:

1.) one grand creator
2.) more than one grand creator (cocreation) but not everyone
3.) everyone created this together
4.) nobody created this

You can pick any option or any multiple options you want. I pick all 4. There is no wrong answer. I can’t stress that enough

Not that simple. It’s not an excluded middle. You don’t think a creator could create a non created cosmos and a created cosmos at the same time?

I know it sounds absurd at first… one person who believes in a creator, will see all the signs of one eventually… those who don’t will always be able to prove the other ones wrong. It’s really hard to explain unless you’ve seen it… everything constructed by perspective in something so mindowingly elegant … it’s your choice!! You cannot make a wrong choice!

Making a wrong choice is not the end of the world.

And on that, I’ll remain silent.

Perhaps we can return to whether math is reality or a representation of reality, and what this implies in terms of convergence with infinite regress

Edit: my context was the decision about creation stuff. That blanket statement outside of context is something not appropriate to discuss

Unlike some, I try to not think with my balls, thk u.

This is a point of great interest. It illustrates the profound difference between the mathematical real numbers on the one hand, and floating point (or any other) representation of real numbers on a physical computer.

The classic example that shows the distinction is the harmonic series

$$\sum_{n=2}^\infty \frac{1}{n} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots$$

[Some people prefer to start counting at (1), makes no difference].

This series converges on a computer but diverges in the real numbers. What you say does apply to computer math, but it does not apply to the real numbers. Let’s look more closely.

On any physical implementation of floating point arithmetic, there is a natural number (N \in \mathbb N) such that (\frac{1}{n}) is indistinguishable from zero whenever (n > N).

Therefore no matter what the implementation, at some point the entire tail of the series is effectively a string of zeros and adds nothing to the sum. The finitely many nonzero terms add up to the finite sum of the series.

On the other hand, the harmonic series diverges in the real numbers.

$$\underbrace{\frac{1}{2}}{~= ~\frac{1}{2}} + \underbrace{\frac{1}{3} + \frac{1}{4}}{> ~\frac{1}{4} ~+ ~\frac{1}{4} ~= ~\frac{1}{2}} + \underbrace{\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}}{> ~\frac{1}{8} ~+ ~\frac{1}{8} ~+ ~\frac{1}{8} ~+ ~\frac{1}{8} ~= ~\frac{1}{2}} + \underbrace{\frac{1}{9} \dots \frac{1}{16}}{> ~8 ~\times ~\frac{1}{16} ~= ~\frac{1}{2}} + \underbrace{\frac{1}{17} \dots \frac{1}{32}}_{> ~16 ~\times ~\frac{1}{32} ~= ~\frac{1}{2}} +\dots$$

We can always grab the next (2^n) terms of the series to have a finitely long block whose sum is greater than (\frac{1}{2}). No matter how large a number you give me, I can go out far enough in the series and find a partial sum that’s greater than your number. If you challenge me with one million for example, I’ll just grab the next two million (2^n)-sized blocks, each of which sums to more than (\frac{1}{2}). The corresponding partial sum of the series will therefore exceed one million.

So the harmonic series diverges. It does not have any finite sum.

Your idea that the terms get so close to zero that they don’t matter, ONLY APPLIES to the computer version! In the real numbers, all those tiny little crumbs at the end keep adding up and the sum goes higher and higher without any finite limit.

It’s kind of weird to imagine. Every tail diverges. No matter how far you go out, where the numbers are really really tiny, the sum of the terms after that STILL fails to converge.

This example in a nutshell is the difference between the real numbers and computer implementations of floating point arithmetic. It also serves as the standard example of a series that fails to converge despite its terms going to zero.

Conclusion: Computers are inadequate to capture the essential nature of the real numbers. Intuitions about computer arithmetic do not necessarily translate to the reals.

All the more reason to approach these problems from several angles.

Do you get consistent results? Why or why not?

No, I’m saying that when you focus down to the level of each digit that gets derived by the algorithm, it’s incredible easy to understand how that digit is derived. Given that a rational number is just a finite series of such digits (or a repeating pattern of such digits), understanding the derivation of each digit amounts to an understanding of the derivation of the entire series.

What is 25 divided by 4?

Well, I know that you can multiple 4 six times to get 24, and the remainder is 1. So I understand how I derive the first digit: 6.

Then when we divide the remainder 1 by 4, I know that we’ll get 4 quarters. So I understand how we get the next two digits: 6.25.

^ There’s nothing complicated here.

Division never gives us irrational numbers. Irrational numbers are, by definition, not representable as a ratio. So I’m not sure what process you’re talking about that sometimes gives us rational numbers and sometimes gives us irrational numbers. But I jumped into this discussion in response to your comment: “nobody on earth actually knows why some rational numbers numbers do or don’t terminate…” ← If you’re talking about only rational numbers, then I assume you’re talking about the difference between, for example, 25/4=6.25 vs. 10/3=3.333… both of which we can understand why they do or don’t terminate. But why some quantities can only be represented as irrational numbers in our base 10 system, I agree that we (or I) don’t understand.

Here’s a theory though: viewtopic.php?t=162173

What is the argument again?

I don’t think this is as easy to explain.

The argument is that 1/99 = 1 & 0.9… = 31/3 & 3*0.3…

It’s a great argument until you reverse engineer the steps …

10-3 = 7
1.0-0.3= 0.7
1.00-0.33 = 0.77

It’s a complete reverse engineer of the same process that not only gives you the repeating fractions, it also reverse engineers the same logic used to justify that .9… = 1

A calculator will let you get 0.7… step by step, which is the method we use to even abstract this problem, but they are automatically programmed (by assholes),
To answer 0.6… if you ask it 1-1/3

To anticipate the obvious question:

The reason the reverse engineering uses 10 instead of 9 (obviously 9-3=6) is that 1 adds a zero as the expansion occurs in the initial procedure. It’s not a proper reverse engineer, if you start with 0.9…

Actually 1.00-0.33 =0.67

Why you are subtracting 3, 0.3 and 0.33 remains a mystery. :confused:

You’re not making any sense.

I agree that (1/9)*9 = 1.

I agree that 0.9… = 3*(1/3) & 3*0.3…

I agree that 10-3 = 7

I agree that 1.0-0.3 = 0.7

I agree that 1.00-0.33 = 0.77

I don’t see the problem.

But I guess if calculators were programmed by nice guys, we’d get different results.

How the hell did I do that! I started with 10-3 because because the 1 carries over to a ten when you do 3/1… then I quickly typed 1-0.3 which was 0.7, then I extrapolated infinite 7’s, then I typed 1-1/3 and got 6 repeating. Had I typed 1-.33 I’d have seen what you just corrected.

I’m not seeing where that train of thought is going now … I just assumed a factor of 10 for each place would always be 10-3=7 right down the line…


Silly, thx for pointing that out!

It also was occurring to me that 1/9 is 0.1…

As it expands, the ones keep getting smaller and smaller until at this hypothetical convergence it becomes zero …

And if you have 9 of these going at the same time, all nine of them converge at zero (like if you plotted the regress on a graph with a zero axis).

This means that 9/9 = 0

Hmm… strange.

Back to the mistake you pointed out phyllo…

Read above post as well!

That creates a last digit if you reverse engineer it, with an infinity in front of it… which causes pretty serious issues through dimensional flooding (infinite 6’s before a 7 + infinite 3’s before a 4)

I take it that your point is that if you go “in reverse” and you subtract 1- 0.333 all the digits are 6s except the ‘last’ digit which is a 7. But is you “go forward” and divide 2/3 all the digits are 6s. So it appears that you can see the missing 0.000…0001 which is the difference between 0.999… and 1.

It’s remotely possible in the case of 0.9…

That would be intense mental gymnastics which strike me initially as resulting perhaps in gibberish…

The biggest problem I see with reverse engineering is when you add multiple reverse engineered numbers together; which causes a solution greater than 1.

The problem I see as a whole, is that in the easier to understand instance of 1/9, you have a situation where the regress approaches zero… which under convergence hypothesis means that 1/9*9=0 rather than 1!

This topic has been done to death, but ah, what the hell *shrugs.

I would say 0.999> is as close to being 1 as you can possibly be, without being 1.
Why?
Simply because they aren’t exactly the same, only 1 can be 1, only x can be x.
Maybe 0.999> is so close to being 1 that for all intensive purposes it might as well be 1, but strictly speaking it’s not.

Maybe it would help to visualize what 0.999> would even look like in nature?
Could 0.999> even exist in nature?
Can matter and space be infinitely small/big?
Perhaps we will never know.

Assuming they can, what would a domino that’s 0.999> centimeters in height look like, and is it equivalent to a domino that’s 1 cm in height?
Place the two dominos on a perfectly flat, perfectly level table.
You should be able to wave your hand 1 cm above and over the table without touching the 0.999> cm domino and knocking it over, but you shouldn’t be able to wave your hand 1 cm above and over the table without touching the 1 cm domino and knocking it over.

That would be the real world empirical, tangible difference between something that’s 1 cm, versus something that’s anything less, whether it’s 0.1 cm less, or 0.01 cm less, or even 0.000> cm less than 1.
However, if your hand is any closer to the table than 1 cm away from it, even just the teeniest, tiniest bit, you will touch both the 1 cm domino, and the 0.99> cm domino, as you wave your hand over the table, knocking them both over, because 0.999> is as close to being 1 as you possibly can come without being 1, so anything closer than 1 cm away, even if it’s only 0.0000000001 cm closer less is going to be within its range to interact with it.

…Or maybe at that distance your hand will pass over both of them, without touching them, but it should be possible to place your hand at a distance from the table, so the bottom of it occupies the same space as the top of the 1 cm domino, but not the 0.999> domino, so you’ll be able to knock the former over at that distance, but not the latter.