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In it's current state it's false because it's not .999999 recurring. You can put as many 9s as you want, it doesn't make it recurring. And most NGers learned this when we were 12 so this is nothing special.

0.999999... is as high as a number can go without equalling or being greater than one, so there for it does not equal one. Simple as that.

lim(x>1)x=1

x=1

False. It's a number starting with 0.9 & therefore can never be above or equal to a whole number.

At 9/6/09 12:07 PM, UberCream wrote: 0.999999... is as high as a number can go without equalling or being greater than one, so there for it does not equal one. Simple as that.

No, actually, if you say that it's as high as a number can go without equalling or being greater than one, that means it IS equal to 1. because a simple axiom of the real numbers is that they are not countable, if you have two numbers that are not equal, there is always an infinite number of real numbers between those two, if there is nothing between, it means they are equal.

At 9/6/09 12:11 PM, Michaelas10 wrote: lim(x>1)x=1

x=1

???

First off, you generally take the limit of an expression, not an equation. Plus this really has nothing to do with proving 0.9... = 1

0.999 recurring isn't 1. It's pretty close, but it isn't.

At 9/6/09 12:21 PM, RedHatCore wrote: 0.999 recurring isn't 1. It's pretty close, but it isn't.

Wowo, I like that logic. Mind providing any sort of proof to that?

At 9/4/09 07:39 AM, BlammerReviewer wrote:

At 9/4/09 07:36 AM, gregaaron89 wrote:

At 9/4/09 07:23 AM, TheMaster wrote: It does, and the proof is easy.

1/3 = 0.3333...
0.9999... / 3 = 0.3333...

So 1 = 0.9999...

Wow I've never thought about it that way. You're totally right.

Damn it why did someone quote on that post. Yes that is the working out and the answer is true. The thread ends now.

Uh no, that's wrong, 1/3 is 0.333 RECURRING where as 0.9/3 is .3 flat. So no, they are not the same.

That's like saying because 100/20 is 5 and 55/11 is 5, 100 and 55 are the same number.

People keep quoting a third as a good argument.

Are you forgetting that recurring numbers are irrational and cannot fully add up properly because of this?

1/3 = 0.333 RECURRING
2/3 = 0.666 RECURRING but if it had an end it would force itself up to 0.666'7
3/3 = 1. A whole. But 1/3 and 2/3 don't add together properly so this would continue forever until the universe ended.

In lamens terms, yes. In a scientific mindset, no.

Simple. Finished. Done.

Thar you go people.

At 9/4/09 07:23 AM, TheMaster wrote: It does, and the proof is easy.

1/3 = 0.3333...
0.9999... / 3 = 0.3333...

So 1 = 0.9999...

It's a paradox! Don't listen or your head will explode!!!

At 9/4/09 08:14 AM, BananaBreadMuffin wrote: I don't know why you're saying it that way round, since decimals are pretty shitty whereas fractions are pretty absolute.

But anyway, yeah, 1 = 0.999..., and there are a whole variety of proofs.

My favourite is this one

x = 0.999...

10x = 9.99....

10x-x = 9.999...-0.999...

9x = 9

x = 1

0.999... = 1.

Actually no, that doesn't make sense. Infinity cannot be multiplied, therefore 0.999... cannot be multiplied.

This makes perfect sense, but it's still not true.

0.99999999999999999999999999999999999999 9999999999999999999999999999999999999999 9999999999999999999999999999999999999999 9999999999999999999999999999999999999999 9999999999999999999999999999999999999999 9
Goes on forever, but 1 will always be greater than any decimal that starts with a 0, such as 0.56778885 or 0.38475, maybe even 0.1234567890

At 9/4/09 08:40 AM, Zerostar wrote: It's basically a limit problem, because .9 recurring is infinitely approaching 1.

.9 recurring can be written as 1-10^(-x) for all positive integer values of x. As x approaches infinity, 10^(-x) approaches 0, and therefore .9 recurring is equivalent to 1.

and zerostar posts the only actual proof that is in this thread. this is extremely basic stuff that you learn on day one of any real analysis course. the only reason this is even a "debate" is because kids haven't learned enough maths and don't understand the meaning of infinity.

But 3/3 doesn't represent 1. It represents a whole. Which isn't neccesarly 1 in all cases?

Um. Yeh, I suck at math okay.

Alright, whatever, I'll get my post in on this thread.

It is false, it is less then 1. Ok then.

At 9/4/09 08:40 AM, Zerostar wrote: It's basically a limit problem, because .9 recurring is infinitely approaching 1.

.9 recurring can be written as 1-10^(-x) for all positive integer values of x. As x approaches infinity, 10^(-x) approaches 0, and therefore .9 recurring is equivalent to 1.

Ooo, I like this proof very much. It's alot more straightforward than other similar proofs.

At 9/4/09 07:18 AM, BlammerReviewer wrote: think you can work it out? True or false?

1=1 duh?

0.9 repeating is one hundred percent equal to one. Blows my mind in some ways, but it's true.

if you accept that 1/3 = 0,333... then you have to accept that 0,999... = 1.
The thing is 0,333... is just not a good way to describe 1/3.

I put 0.999999999999999999999999999999 into a calculator and it said, 0.9-=1.0

At 9/6/09 08:28 PM, JKMonkey wrote: I put 0.999999999999999999999999999999 into a calculator and it said, 0.9-=1.0

I just got out my TI-83. You're right.

Depends on your significant figures

here's something to think about for those that are saying false.

there is a property of the real numbers that states that if two numbers, x and y, are different, then there exists another real number, z, such that x<z<y. this is intuitively obvious, as you just take the average of x and y.

now if you're saying that 0.999~ is less than 1, you're saying that there's a number between the two of them. someone tell me what it is and I'll paypal you $0.999~

At 9/7/09 06:11 AM, jonthomson wrote: here's something to think about for those that are saying false.

there is a property of the real numbers that states that if two numbers, x and y, are different, then there exists another real number, z, such that x<z<y. this is intuitively obvious, as you just take the average of x and y.

Most of you people are forgetting that 'recurring decimals' are just a convention for writing fractions that don't have a finite decimal equivalent.

I'm going to pose to you that 0.9~ =/= 1 because 0.9~ doesn't actually exist rationally in the real number system, since there is no rational function that will equal that decimal.

1 = 1
Simple.

At 9/4/09 07:23 AM, TheMaster wrote: It does, and the proof is easy.

1/3 = 0.3333...
0.9999... / 3 = 0.3333...

So 1 = 0.9999...

...You hurt my brain.