.999... = 1

[xelabale]

You say 1 is not in the set but 0.999... is in the set. If you take this as your initial assumption it is easy to prove they are different. However if you say they are equal your theory doesn't stand up. You are not proving anything, you are just getting different results depending on your starting assumptions.

[Suspect101]

You say 1 is not in the set but 0.999... is in the set. If you take this as your initial assumption it is easy to prove they are different. However if you say they are equal your theory doesn't stand up. You are not proving anything, you are just getting different results depending on your starting assumptions.

you can't just say they are equal.

[Suspect101]

every proof here that people are claiming is corret depends on the assumption that 1/infintiy =0.

[SultanOfSurreal]

you can't just say they are equal.

yes you can

this is because it happens to be true

[xelabale]

You say 1 is not in the set but 0.999... is in the set. If you take this as your initial assumption it is easy to prove they are different. However if you say they are equal your theory doesn't stand up. You are not proving anything, you are just getting different results depending on your starting assumptions.

you can't just say they are equal.

I agree, there are other proofs for that. But this maths is not a proof because using it you can either prove 0.999...=1 or 0.999...=/=1 depending on your starting assumptions. This makes it invalid.
Amazing, we don't even need the trolls to continue this...

[Snorri1234]

You say 1 is not in the set but 0.999... is in the set. If you take this as your initial assumption it is easy to prove they are different. However if you say they are equal your theory doesn't stand up. You are not proving anything, you are just getting different results depending on your starting assumptions.

you can't just say they are equal.

You can't just say they're not equal either. Your proof hinges on the assumption that 0.999... isn't the same as 1 already.

[1,0] is a closed set. In this set all real numbers between 1 and 0 are included.

(1,0] is an open set. In this set 1 is not included, but .999.... is.

This is your premise. It already says 1 is not the same as .999..., you have defined it as so. If 0.999 was the same as 1 it would not be included in the set.

Your proof is therefore absurd.

[sully800]

every proof here that people are claiming is corret depends on the assumption that 1/infintiy =0.

That is true. The proofs depend on the mathematical definitions of infinity. 2/infinity is not greater than 1/infinity and such.

[Hologram]

Wow, I see a lot of practical jokers using a lot of bunk mathematics to prove an untrue and unimportant point. Goodbye.

[Suspect101]

Even though I am saying you are right......

You are all just using proofs done by other people and not using any of your original thought on the matter to disprove or reprove in a new and exciting way, anything at all. You are just repeating things you have read in a text book.

The statement .9..=1 does have limitations and assumptions, but in the Real number set (which is probably the only set that matters) this is true. The only reason which this is true is because of the Archimedean property, which I did not even think about. This states that no number in the Real number set can be infinitely small or infinitely large. Which means that even subtracting 1-.99.., produces no Real number, because it is infinitely small, which by definition makes it not a Real number and does not exist in the set. This makes 1-.99.. = 0. If there is no difference between two numbers, then they are the same.

Thank-you for the problem, it really got me thinking about Math I have not done in years and got me thinking of a few more theories I would like to retry. It was fun. You are right, .99..=1

[Snorri1234]

Even though I am saying you are right......

You are all just using proofs done by other people and not using any of your original thought on the matter to disprove or reprove in a new and exciting way, anything at all. You are just repeating things you have read in a text book.

The statement .9..=1 does have limitations and assumptions, but in the Real number set (which is probably the only set that matters) this is true. The only reason which this is true is because of the Archimedean property, which I did not even think about. This states that no number in the Real number set can be infinitely small or infinitely large. Which means that even subtracting 1-.99.., produces no Real number, because it is infinitely small, which by definition makes it not a Real number and does not exist in the set. This makes 1-.99.. = 0. If there is no difference between two numbers, then they are the same.

Thank-you for the problem, it really got me thinking about Math I have not done in years and got me thinking of a few more theories I would like to retry. It was fun. You are right, .99..=1

We were more repeating things we read in textbooks to get the idea accross.

We have to first agree on the issue before we can talk about what it means. The issue of .999... being 1 does bring up a lot of interesting mathematical and philosphical questions, it's just that we first have to agree that they are the same before discussing those things.

[TheProwler]

every proof here that people are claiming is corret depends on the assumption that 1/infintiy =0.

That is true. The proofs depend on the mathematical definitions of infinity. 2/infinity is not greater than 1/infinity and such.

2/infinity approaches zero less quickly that 1/infinity approaches zero. Actually, at half the rate.

[Timminz]

0.999... does not approach anything. It is a single value which does not change.

[TheProwler]

0.999... does not approach anything. It is a single value which does not change.

I'm still kinda hoping someone can come up with an equation that will result in an answer of 0.999recurring without using recursion or a recursive number or some declared value that is "infinitely small". If you can't, then I would like someone to explain how 0.999recurring can even exist. Obviously, if something can't exist, it can't equal 1.

[xelabale]

0.999... does not approach anything. It is a single value which does not change.

I'm still kinda hoping someone can come up with an equation that will result in an answer of 0.999recurring without using recursion or a recursive number or some declared value that is "infinitely small". If you can't, then I would like someone to explain how 0.999recurring can even exist. Obviously, if something can't exist, it can't equal 1.

So you're asking if infinity can exist? In theory it can, whether it actually does is a philosophical debate in itself.

[SultanOfSurreal]

every proof here that people are claiming is corret depends on the assumption that 1/infintiy =0.

That is true. The proofs depend on the mathematical definitions of infinity. 2/infinity is not greater than 1/infinity and such.

2/infinity approaches zero less quickly that 1/infinity approaches zero. Actually, at half the rate.

1/x and 2/x approach zero at different rates as x increases but if it were possible to divide by infinity (it isn't) then 1/infinity and 2/infinity would be the same, ie 0

prove me wrong

[sully800]

Even though I am saying you are right......

You are all just using proofs done by other people and not using any of your original thought on the matter to disprove or reprove in a new and exciting way, anything at all. You are just repeating things you have read in a text book.

The statement .9..=1 does have limitations and assumptions, but in the Real number set (which is probably the only set that matters) this is true. The only reason which this is true is because of the Archimedean property, which I did not even think about. This states that no number in the Real number set can be infinitely small or infinitely large. Which means that even subtracting 1-.99.., produces no Real number, because it is infinitely small, which by definition makes it not a Real number and does not exist in the set. This makes 1-.99.. = 0. If there is no difference between two numbers, then they are the same.

Thank-you for the problem, it really got me thinking about Math I have not done in years and got me thinking of a few more theories I would like to retry. It was fun. You are right, .99..=1

I am satisfied with that explanation as well. In the real number set, there is no difference between 0 and 0.0 and 0.00 and 0.0repeating infinitely.

[e_i_pi]

Because 1 and 0.999... are equivalent. There is no difference between the two numbers, not even ε. You are defining 0.999... to be smaller than 1 here, and then saying that it is proof that its smaller than 1 because they are not in the same set. But if they are equivalent, then your statements are false and the are indeed equal. Defining the numbers by sets like that will not work, since your definition determines the outcome.

[owheelj]

every proof here that people are claiming is corret depends on the assumption that 1/infintiy =0.

That is true. The proofs depend on the mathematical definitions of infinity. 2/infinity is not greater than 1/infinity and such.

2/infinity approaches zero less quickly that 1/infinity approaches zero. Actually, at half the rate.

Sorry what?

Replace infinity with 100.

Does 2/100 approach anything faster than 1/100? Of course not. These are equations with specific answers. The same applies if you replace 100 with infinity. 1/infinity and 2/infinity are not limits and they do not approach anything. They would be limits if we said f(x)=1/x and then we could say that as x approaches infinity f(x) approaches 0. However that's obviously not what we're talking about!

[e_i_pi]

Even though I am saying you are right......

You are all just using proofs done by other people and not using any of your original thought on the matter to disprove or reprove in a new and exciting way, anything at all. You are just repeating things you have read in a text book.

The statement .9..=1 does have limitations and assumptions, but in the Real number set (which is probably the only set that matters) this is true. The only reason which this is true is because of the Archimedean property, which I did not even think about. This states that no number in the Real number set can be infinitely small or infinitely large. Which means that even subtracting 1-.99.., produces no Real number, because it is infinitely small, which by definition makes it not a Real number and does not exist in the set. This makes 1-.99.. = 0. If there is no difference between two numbers, then they are the same.

Thank-you for the problem, it really got me thinking about Math I have not done in years and got me thinking of a few more theories I would like to retry. It was fun. You are right, .99..=1

I'm not at all married to the Archimedean property.

Firstly, the definition of "infinitesimals" is dodgy at best. It basically says if you can't measure something really small, it doesn't exist. So I guess atoms only came into existence with the electromicroscope At the moment, we have electron microscopes which can magnify about 2 million times I believe, but there is a measurement in particular that I'd like people to consider - the yoctometre. 10^35 yoctometre's = 1 metre.

Now, according to physics, the Planck length is about 1.6 x 10^-35 metres, or 1.6 yoctometres. Also according to physics, one Planck length is the smallest distance or size about which anything can be known. Let's have a look at this in relation to the Archimedean property.

1 Planck length exists as a finite and recognised length. 1/2 a Planck length does not exist as a finite and recognised length. So, we have one very small length that exists, one even smaller length that doesn't exist. Let's look at what the Archimedean property says:

Denote by Z the set consisting of all positive infinitesimals. This set is bounded above by 1. Now assume by contradiction that Z is nonempty. Then it has a least upper bound c, which is also positive, so c/2 < c < 2c. Since c is an upper bound of Z and 2c is strictly larger than c, 2c must be strictly larger than every positive infinitesimal. In particular, 2c cannot itself be an infinitesimal, for then 2c would have to be greater than itself. Moreover since c is the least upper bound of Z, c/2 must be infinitesimal. But 2c and c/2 cannot have different types by the above result, so there is a contradiction. The conclusion follows that Z is empty after all: there are no positive, infinitesimal real numbers.

In the case of the Planck constant, if we take c to be half a Planck constant, and therefore infintesimal (by definition of not being measurable), then we have c/2 being infinitesimal, and 2c being finite. So, not only is the Archimedean property proved by contradiction, it can be "un"-proven by contradicting it, as it contradicts modern science. Not to mention that the Archimedean property was first coined in about 300BC, 2000 years before the term "infinitesemil" was defined.

Secondly, the Archimedean property defines a number x as infinite if it satisfies the conditions |x|>1, |x|>1+1, |x|>1+1+1, ..., and infinitesimal if x≠0 and the same set of conditions holds for 1/x. A number system is said to be Archimedean if it contains no infinite or infinitesimal members.

Let's have a look at the first part of that definition: "a number x [is] infinite if it satisfies the conditions |x|>1, |x|>1+1, |x|>1+1+1, ...".

The Natural numbers have a countably infinite number of elements. I'll restate that for the slow ones - countably infinite number of elements. So, by definition, the Natural numbers have an infinite number of elements. But no element is infinite in value. So, Archimedes, or whoever is holding his baton, please explain how the Natural numbers can be countable, infinite, and not have an infinite value in them. Try counting out loud from 1 upwards - where is the Archimedean brick wall you hit where no number exists for you to voice any more?

[sully800]

But the difference between 1 and 0.99recurring is not a really small number. It is zero with an infinite number of zeros after the decimal place. Which is identical to zero with infinite degrees of accuracy.

It's not to small to be measured. The difference, by definition of the recurring 9, is non existent.

[sully800]

Because 1 and 0.999... are equivalent. There is no difference between the two numbers, not even ε. You are defining 0.999... to be smaller than 1 here, and then saying that it is proof that its smaller than 1 because they are not in the same set. But if they are equivalent, then your statements are false and the are indeed equal. Defining the numbers by sets like that will not work, since your definition determines the outcome.

I didn't say they are equivalent and then write that it was a proof that they are equivalent like the poster who tried to disprove the relationship. I said they are equivalent which is why he was wrong. Plenty of mathematical proofs have been supplied in this thread, as well as numerous definitions of what constitutes a real number and why there is no real number in between 1 and 0.99recurring.

[AAFitz]

Try counting out loud from 1 upwards - where is the Archimedean brick wall you hit where no number exists for you to voice any more?

its where .999=1

[e_i_pi]

But the difference between 1 and 0.99recurring is not a really small number. It is zero with an infinite number of zeros after the decimal place. Which is identical to zero with infinite degrees of accuracy.

If it is zero, then you wouldn't be writing 9's after the decimal place, and 1 before it. Besides, that wasn't the point - proof by ignoring the other persons argument is not, I assure you, a recognised logical method.

Archimedes said that infinitesimals do not exist due to the Archimedean property. But the Archimedean property does not hold. Therefore infinitesimals can exist, meaning that an immeasurable distance between 0.999... and 1 can be defined, given the series (1-x) where x = 10^-n, for n an element of the natural numbers.

[AAFitz]

0.999... does not approach anything. It is a single value which does not change.

I'm still kinda hoping someone can come up with an equation that will result in an answer of 0.999recurring without using recursion or a recursive number or some declared value that is "infinitely small". If you can't, then I would like someone to explain how 0.999recurring can even exist. Obviously, if something can't exist, it can't equal 1.

Actually, it is the exact reason that it cant "exist" that it does equal one

1 essentially is .999 because .999 is infinite. If it wasnt infinite, then there would be a difference, but because it is infinite, it = 1

1 is the point where .999 recurs infinitely. You dont have to believe it exists, and certainly the entire idea of infinity makes it impossible to prove. Further, as you imply infinity cant exist, you actually help make the point that it =1. to insinuate that .999 cant go on forever, and that it has to stop, is the exact reason it =1. It does stop. .999 recurs, and =1.

Every other amount of .999 recurring that stop, have a point that does not =1, albeit small. But at an infinite amount of recurring , .999 simply =1. The second you call it .999 recurring it =1, because it simply cannot not equal one.

The only thing more infinite or pointless about this conversation, really is its uselessness.

Right or wrong, the difference if there was one between 1 and .99... would be infinitely small. As an infintely small number, it would be = to 0.

(.999=1) = (.999...does not =1) the diffence between the two answers is infinitely small, (its not really small, or extremely small, its infinitely small and therefore must =0) Because of this, both answers are correct, and since .999... in every practical and useful meaningful way, will always and always has =1, then .999...=1. The only time it does not, is if .999 is not infinitely recurring and is not infinite.

[sully800]

Besides, that wasn't the point - proof by ignoring the other persons argument is not, I assure you, a recognised logical method

His "proof" that the two numbers were unequal relied on his initial assertion/definition that they are unequal. My statements were merely to show that it is no proof at all to say the numbers are in different domains, because you cannot say that before determining whether or not they are equal. I did not ignore his argument, I brushed it away as rubbish

This is all very silly though, its clear that neither side will convince the other, because we have been restating the same arguments for many pages. I will stick with the commonly accepted mathematical philosophy, if you continue to disagree with the mathematicians that is fine.