.999... = 1

[SultanOfSurreal]

Let's say that the smallest possible difference between two numbers is 0.0... 1; zero followed by an infinite amount of zeros and then a one.

Your 0.333recurring is slightly smaller than my 0.333recurring.

this thread is a goddamned lolacaust

good job, me

[TheProwler]

To what degree are those two numbers different? How many digits to I have to go through those numbers before they diverge, and which number changes from a 3 and what does it change to?

Or are you saying that at particular point along those infinite lines of 3s, the 3 on one line will be worth less than the three on the other?

First, you must open your mind. Only then will I teach you the great secrets that only few men understand.

Ahhh heck, who am I kidding? I don't have the time.

I don't care, honestly, if you believe or understand that 1<>0.999recurring.

It is clear that you will blindly follow someone who will tell you that writing a number this way 0.333recurring is as accurate as writing the number this way 1/3. It is not. But it would take me an infinite amount of time to explain it to you. That's my bad. I take full responsibility.

I've been approaching my time limit...hey, here's some wiki for you: http://en.wikipedia.org/wiki/Limit_(mathematics)

So I extend my hand to you for the customary "Let's shake and curse about each GD stupid MF other behind their back" sign of respect.

Bye!

[Snorri1234]

Let's say that the smallest possible difference between two numbers is 0.0... 1; zero followed by an infinite amount of zeros and then a one.

Your 0.333recurring is slightly smaller than my 0.333recurring.

this thread is a goddamned lolacaust

good job, me

Seriously, I can't even express the joy I have over this thread in real numbers. Let me try to approach it.


0.9348983984938493849839999999999944999999999recurring


that much

[Snorri1234]

To what degree are those two numbers different? How many digits to I have to go through those numbers before they diverge, and which number changes from a 3 and what does it change to?

Or are you saying that at particular point along those infinite lines of 3s, the 3 on one line will be worth less than the three on the other?

First, you must open your mind. Only then will I teach you the great secrets that only few men understand.

shit, we need to make a thread with posts that made you laugh uncontrolabbly for 10 minutes. Seriously, this is comedy gold.

[owheelj]

I understand limits thanks.

You still haven't answered my question - the last digit is a 2 or a 4? or is one of the 3s worth less than the corresponding 3?

Those the options - 2 or a 4, or a 3 worth less than another 3. Which is it?

I'm not going to shake hands, I'm going to keep arguing until you give up or until you admit that I'm right. This has nothing to do with me believing other people, it's to do with me being able to calculate 3x3 as well as 1/3 and 9/3. Merely with those abilities I can work out for myself that 1=0.999...

It's funny that you state that 1/3 does not equal 0.333... What does it equal then? Work it out on a bit of paper. 1/3 - first digit 0. Decimal point. Next digit 3. Next digit 3. Next digit 3 etc. At what point do you realise that it's going to be 0 followed by a decimal point followed by an infinite number of threes. Which is obviously expressed as 0.3 recurring. If we go along that line of infinite threes and multiple each individual digit by 3 what do we get? A 9 every time - all the way down the line. At no point do we have a reason to stop writing down nines. Therefore we *know* that 0.333... x 3 = 0.999... We also know that 0.333... = 1/3. and we know what we get when we multiply 1/3 with 3. If any of these equations are incorrect, don't tell me what my mistake was, tell me what the actual correct answer is.

[Snorri1234]

I understand limits thanks.

You still haven't answered my question - the last digit is a 2 or a 4? or is one of the 3s worth less than the corresponding 3?

Those the options - 2 or a 4, or a 3 worth less than another 3. Which is it?

I'm not going to shake hands, I'm going to keep arguing until you give up or until you admit that I'm right. This has nothing to do with me believing other people, it's to do with me being able to calculate 3x3 as well as 1/3 and 9/3. Merely with those abilities I can work out for myself that 1=0.999...

It's funny that you state that 1/3 does not equal 0.333... What does it equal then? Work it out on a bit of paper. 1/3 - first digit 0. Decimal point. Next digit 3. Next digit 3. Next digit 3 etc. At what point do you realise that it's going to be 0 followed by a decimal point followed by an infinite number of threes. Which is obviously expressed as 0.3 recurring. If we go along that line of infinite threes and multiple each individual digit by 3 what do we get? A 9 every time - all the way down the line. At no point do we have a reason to stop writing down nines. Therefore we *know* that 0.333... x 3 = 0.999... We also know that 0.333... = 1/3. and we know what we get when we multiply 1/3 with 3. If any of these equations are incorrect, don't tell me what my mistake was, tell me what the actual correct answer is.

You don't understand! Prowler once went to highschool!

[pimpdave]

WHY ARE YOU GOING TO SPACE MONKEY?

[Night Strike]

Wow, I didn't realize this thread had gone to 9 pages long.

NERD ALERT!!!

[InkL0sed]

Wow, I didn't realize this thread had gone to 9 pages long.

NERD ALERT!!!

More like...

[boxxy]TROOOLLLLLLLLLL!!1111[/boxxy]

[john9blue]

Nonbelievers: do you agree that .333... = 1/3?

If so then:

.333... = 1/3
(1/3)*3 = 3/3 = 1
Therefore, .333...*3 = 1

.333... = sum of all 3/(10^n) as n goes from 1 to infinity.
(3/(10^n))*3 = 9/(10^n)
sum of all 9/(10^n) as n goes from 1 to infinity = .999...
Therefore, .333...*3 = .999...

If .333... = 1 and .333... = .999..., then .999... = 1. QED,MF.

I'll ignore the mistake...I know what you meant.

Nah, we'll never know until you write it out in long form.

Again, you've only exposed a limitation of common methods when dealing with infinitely long numbers.

What mistake? You're saying that .333... = 1/3, but it's a different .333... than what you're talking about.

Law of math: a value can be expressed in multiple ways, but an expression cannot have multiple values (assuming we are working with simple arithmetic). Don't go making stuff up, just admit you were wrong or leave the thread or both.

[owheelj]

The mistake is your bottom line; 0.333...=/=1. it is in fact a third of 1. Nor does 0.333...=0.999... That's like saying 3=9. Obviously merely a typo.

[SultanOfSurreal]

holy shit you guys

i think i broke klobber

[maasman]

I really enjoy this thread, we just started talking about limits in precalc, so its fun to be able to understand this and I do believe that .999recurring=1, though it is moderately difficult to grasp.

[InkL0sed]

Look at this:



See that? I made programmed that. It involved lots of math.

Therefore, I am right.

QED

[Suspect101]

.9999..... does not equal 1.

Think of an asymptote:

y = (x )/(x +1)

This gives you two asymptotes, a vertical at x=-1 a Horozontal asymptote at y=1.

Lets look at the horozontal (y=1):

The bigger X gets(the same as the more 9's you place in the sequence), the closer y gets to 1, but it never reaches 1. This is the same as .9999.... it never actually reaches 1 other wise it would be 1.

[InkL0sed]

.9999..... does not equal 1.

Think of an asymptote:

y = (x )/(x +1)

This gives you two asymptotes, a vertical at x=-1 a Horozontal asymptote at y=1.

Lets look at the horozontal (y=1):

The bigger X gets(the same as the more 9's you place in the sequence), the closer y gets to 1, but it never reaches 1. This is the same as .9999.... it never actually reaches 1 other wise it would be 1.

Just like .999... never reaches the end.

Or in other words, it reaches the asymptote at infinity, which is never.

[a.sub]

i found his mistake
[quote="the prowler]But don't believe it. Because 9.9999recurring is approaching 10 just a wee bit slower than 0.9999recurring is approaching 1.[/quote]
WRONG
i would like to point out that .999999 and 9.9999999 arent racing, they digits dont run otu like a computer spittingout numbers
ALL INFINITE 9'S EXIST SIMULTANEOUSLY
they have always existed and dont race each other
that is the concept you fail to understand prowler

[xelabale]

Wow, Prowler and Klobber, I'm impressed. 10 pages on a dead topic, you guys are genius. I smell a secret alliance...

[jonesthecurl]

I understand limits thanks.

You still haven't answered my question - the last digit is a 2 or a 4? or is one of the 3s worth less than the corresponding 3?

Those the options - 2 or a 4, or a 3 worth less than another 3. Which is it?

I'm not going to shake hands, I'm going to keep arguing until you give up or until you admit that I'm right. This has nothing to do with me believing other people, it's to do with me being able to calculate 3x3 as well as 1/3 and 9/3. Merely with those abilities I can work out for myself that 1=0.999...

It's funny that you state that 1/3 does not equal 0.333... What does it equal then? Work it out on a bit of paper. 1/3 - first digit 0. Decimal point. Next digit 3. Next digit 3. Next digit 3 etc. At what point do you realise that it's going to be 0 followed by a decimal point followed by an infinite number of threes. Which is obviously expressed as 0.3 recurring. If we go along that line of infinite threes and multiple each individual digit by 3 what do we get? A 9 every time - all the way down the line. At no point do we have a reason to stop writing down nines. Therefore we *know* that 0.333... x 3 = 0.999... We also know that 0.333... = 1/3. and we know what we get when we multiply 1/3 with 3. If any of these equations are incorrect, don't tell me what my mistake was, tell me what the actual correct answer is.

You don't understand! Prowler once went to highschool!

Only to visit.

[AAFitz]

I studied this sometime in 1st or 2nd or 3rd or 4th year university (it was so long ago!). I was in Applied Math with Computer Science (an honours mathematics program) at the University of Waterloo - one of the top mathematics universities in the world.

I was once ranked in the top 99.97% of maths students in Australia (in my year group). I was doing university maths when I was in 13. I don't know about my maths teacher, but I think I could have got in to your little university.

I enjoyed that.

Now, I have some university math, but I dont enjoy it. However, it seems to me that arguing that .999recuring not equaling one, is as useful as arguing that 1 does not equal 1. It is true on a basic level at times, but not extremely practical.

However, in 10,000 years, if we have matter transportation capability, and my starship is 500000000000000000000000000.999recurring miles away, and you have to beam me onto it, I do hope you dont just round up, and fuse me into the hull. (though at some point you have to stop the computation, otherwise the number will never be displayed, so it cant be used in the equation, since 9's will continue to be added to it forever.)

It's not round up, .999 recurring is 1.

Well, as prowler stated, thats a matter of opinion...

but yes, of course it is, until you get near infinity, and it isnt.

Infinity is only a theory after all.

[jonesthecurl]

you can't get "near" infinity...

[xelabale]

Infinity is only a theory after all.

So is 1.

[KLOBBER]

...I smell....

[Timminz]

...I smell....

Excellent point. You've won me over.

[Snorri1234]

Well, as prowler stated, thats a matter of opinion...

Math isn't a matter of opinion.