.999... = 1

[owheelj]

Agreed, this isn't a matter of opinion - you'll find it in plenty of text books etc.

[Suspect101]

Agreed, this isn't a matter of opinion - you'll find it in plenty of text books etc.

I would like you to list ANY text book, edition, and page number that claims .99999 = 1.

[Suspect101]

Agreed, this isn't a matter of opinion - you'll find it in plenty of text books etc.

I would like you to list ANY text book, edition, and page number that claims .99999 = 1.

nope, i found one. I am an idiot

[owheelj]

I know you found one, but here is one example;

An Introduction to Algebra, 1st edition, pg 177 - published 1806 (contrary to Prowler's suggestion that this is some kind of recent decision by mathematicians to avoid argument, or whatever he was trying to say). It's by John Bonnycastle. You can read the entire book on google books if you have a look too.

There are, of course, plenty of very modern textbooks that also contain the proofs that 0.999... = 1. I remember one of my textbooks had it. Don't remember the name of the book or anything about it though, except it had an orange cover (like bright orange, which just some white text on the front). I'll see if my parents still have it.

[Suspect101]

Agreed, this isn't a matter of opinion - you'll find it in plenty of text books etc.

I would like you to list ANY text book, edition, and page number that claims .99999 = 1.

nope, i found one. I am an idiot

I still disagree because I found the same problem with this proof. Two numbers are considered equal if their difference is 0.
If you try and take 1 and subtract .999.... you never get an answer. It repeats forever and will never stop, therefore you will never get a difference but you never get 0 either.

You are trying to solve for a solution for an unbounded equation, which gives you a theory as to what it is, not what it actually is. I have found what you are talking about, but they still "assign to the notation "0.999…" is defined to be the real number which is the limit of the convergent sequence (0.9, 0.99, 0.999, 0.9999, …)." The limit of a convergent sequence never reachs its limit. it only gets infinteciamly closer to it.

I still believe the key here is convergence, asymptotes, and limits.

[xelabale]

When I first started ... I ... was a joke -- everybody knows that . Then I actually went to ... a certain group of unnamed individuals ... of the OPINION that it was ... true. I still wasn't convinced.

Then the self-styled ...whiz kid from Australia, the land of individuals descended from English reprobates and criminals so abhorrent to civilized society that they had to be manacled and ostracized to a dry, barren, hellish island thousands of miles away, ... convinced me with some pretty fast talking...

Then I saw Prowler's post... does ... equal ... Dumdum ...

Such a mind is ... unintelligent, whose natural position is a lack of understanding. It is ... a failed attempt at every turn.

Each and every ...digit ... brings you ... reality.

That is why we ... ... laymen... be easily brainwashed into believing it.

My work here....

Hey, this misquoting thing's fun! Thanks KLOBBER, every word your own! I shall call this style kloting, to go with klogic and klobberfacts.

[Suspect101]

A law of mathmatics is: If you take a finite number and subtract an infinte number you get another infinte number.

Like if you take 1 (finite)-.333333...(infinite).... = .6666.....(infinte)

If you take 1 - .999999 you do not get 0, because 0 is finite, you get 0.000000.... which is infinite.

[Timminz]

I still believe the key here is convergence, asymptotes, and limits.

Believing is only necessary when you can't prove. Math is about proofs, not beliefs.

A law of mathmatics is: If you take a finite number and subtract an infinte number you get another infinte number.

Like if you take 1 (finite)-.333333...(infinite).... = .6666.....(infinte)

If you take 1 - .999999 you do not get 0, because 0 is finite, you get 0.000000.... which is infinite.

Okay. Where does 0.000... cease to be zero?

[Suspect101]

I still believe the key here is convergence, asymptotes, and limits.

Believing is only necessary when you can't prove. Math is about proofs, not beliefs.

A law of mathmatics is: If you take a finite number and subtract an infinte number you get another infinte number.

Like if you take 1 (finite)-.333333...(infinite).... = .6666.....(infinte)

If you take 1 - .999999 you do not get 0, because 0 is finite, you get 0.000000.... which is infinite.

Okay. Where does 0.000... cease to be zero?

the same place .99999 ends. when you find it let me know.

[sully800]

I know this is old, but I'm catching up.

Your 0.333recurring is slightly smaller than my 0.333recurring. I know this is hard to grasp.

Does your head hurt?

That first statement made me snort so loud, that yes, my head literally does hurt.

[Snorri1234]

I still believe the key here is convergence, asymptotes, and limits.

Believing is only necessary when you can't prove. Math is about proofs, not beliefs.

A law of mathmatics is: If you take a finite number and subtract an infinte number you get another infinte number.

Like if you take 1 (finite)-.333333...(infinite).... = .6666.....(infinte)

If you take 1 - .999999 you do not get 0, because 0 is finite, you get 0.000000.... which is infinite.

Okay. Where does 0.000... cease to be zero?

the same place .99999 ends. when you find it let me know.

It doesn't end.

I really like to know whether you think 3 times .333.... is 1 though. Do you also believe, like prowler, that 1/3 is .333... but that the same thing doesn't apply to .999...?

[sully800]

http://polymathematics.typepad.com/poly ... y_it_.html

So here's why it's not false:

False assumption: * ".9 repeating doesn't equal 1, it gets closer and closer to 1."

May I remind you that .9 repeating is a number. That means it has it's place on the number line somewhere. Which means that it's not "getting" anywhere. It doesn't move. It either equals 1 or it doesn't (it does of course), but it doesn't "get" closer to 1.

[Neoteny]

Have you people never heard of an asymptote?

[john9blue]

Have you people never heard of an asymptote?

That's 9 letters long, Neo. It might be kind of a stretch.

[sully800]

It's not asymptotic though, it IS 1. Exactly equivalent and interchangeable with the number one.

If you plot the series 9/10 + 9/100 + 9/1000 ... it will asymptotically approach 1.0, that is true.

But 0.999recurring, by definition, is a number with infinite 9's following the decimal point. The number doesn't change as you add more levels of accuracy, and it doesn't approach anything. It simply is a number, and it is equivalent to 1.

[Suspect101]

Equation:

1/(1-x)
if x = 1
1/(1-1) = DNE

If x = (1-1/∞)
1/(1-.999...) = ∞

[Snorri1234]

Equation:

1/(1-x)
if x = 1
1/(1-1) = DNE

If x = (1-1/∞)
1/(1-.999...) = ∞

Yeah...

Wut?

[Suspect101]

Equation:

1/(1-x)
if x = 1
1/(1-1) = DNE

If x = (1-1/∞)
1/(1-.999...) = ∞

Yeah...

Wut?

DNE is not the same as ∞. Therefor 1 is not the same as .99...

[Timminz]

Equation:

1/(1-x)
if x = 1
1/(1-1) = DNE

If x = (1-1/∞)
1/(1-.999...) = ∞

You have already been shown absolute proof that 0.999... = 1, so 1/(1-0.999...) =/= ∞, 1/(1-0.999...) = 1/(1-1) = 1/0 = DNE, DNE = DNE, therefore your proof is flawed.

[Suspect101]

Equation:

1/(1-x)
if x = 1
1/(1-1) = DNE

If x = (1-1/∞)
1/(1-.999...) = ∞

You have already been shown absolute proof that 0.999... = 1, so 1/(1-0.999...) =/= ∞, 1/(1-0.999...) = 1/(1-1) = 1/0 = DNE, DNE = DNE, therefore your proof is flawed.

It is not an absolute proof. that is the whole point. you are using numbers with an infinite ammount of decimal places, therefor it is impossible to calculate. The theory that .99.. = 1 is based off of assumptions that are not provable, therefor the theory is not provable.

The only reason you say my proof is flawed is because you assume that 1 =.99999 if you take the limit, you see that as x -> 1 y-> infinity, which does exist. when x = 1 y DNE. Two different things.

1/3 does not = .3333.... this is beacuse it is an interpretation of a number with infinite decimal places.

.3 +.3+.4 = 1
.33+.33+.34 = 1
.333+.333+.334 = 1
you can put as many 3 inbetween as you like, but there have to have the remainder added in to = 1 other wise it is not = 1.

1/3 + 1/3 +1/3 = 1 becuase it is assumed that the remainder is added in.

Just because some people print something does not make it true and does not mean we should blindly follow it, otherwise the world would still be flat.

[Timminz]

9.999... - 0.999... = 9.

Show me where that is incorrect.

[a.sub]

9.999... - 0.999... = 9.

Show me where that is incorrect.

timmy, ur arguing logically
in other words
you are trying to fill a well with a spoon sized shovel

[AAFitz]

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

sure you can: infinity - (1-.999recuring)
you dont get much closer than that.

The post was a joke. However, practically speaking, if .999999 recuring = 1, then 1 is infinity for all intensive purposes. If it hasnt, the .999s keep going, and cant =1 fully.

But again, for those that missed my sarcastic tone, it was and is a joke.

The entire debate is about 1- .99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
recurring.

If you cant see that most of the people arguing it are taking the piss as the brits call it... youre missing the whole point.

[e_i_pi]

Just like the difference between 1 and .9999recurring (ie. 1 minus .9999recurring) approaches 0.

Yeah sorry owheelj, I agree with Prowly here (rare that that happens really).

As you're dealing with a number that is essentially a series 10^-x, or a limit, then you're not going to be able to make the two equal no matter how hard you try. q.v. Asymptotes / Limits / etc etc etc

0.9 = 1 - 10^-1
0.99 = 1 - 10^-2
0.999 = 1 - 10^-3
0.999... = 1 - 10^-∞

Let f(x) = 10^-x

Now, lim(x->∞) f(x) = 0

But, the definition of a limit is the point it approaches - it never reaches that point. So, if

10^-x -> 0

then

1 - 10^-x -> 1 - 0 -> 1

It approaches 1, it never equals 1.

[Timminz]

But, as Sully has already pointed out, 0.999... is not a series. It is a single number. It can not be said to be approaching anything.