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☖ Forums → Discussion Forum → Does 0.999... = 1?
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Does 0.999... = 1?
2006-03-09, 11:42 PM #1
Does 0.999... (repeated) equal 1?

It's always interesting to see what people say to this, and the silly debates that follow. So vote.

Edit: Oh, and you have to explain why.
Bassoon, n. A brazen instrument into which a fool blows out his brains.
2006-03-09, 11:47 PM #2
Yes, it does. My Maths teacher once showed me a formula that can prove it.
ORJ / My Level: ORJ Temple Tournament I
2006-03-09, 11:49 PM #3
No. The common arguement is that 1/3 = .33... and so three times that would be 1 = .99...

But that's not true. .33... is not equal to 1/3, it only approaches it. Just like .99... is not equal to 1, it only approaches it.
2006-03-09, 11:50 PM #4
but vin, the limit as x approaches infinity is precisely one.
"it is time to get a credit card to complete my financial independance" — Tibby, Aug. 2009
2006-03-09, 11:50 PM #5
0.999... is always 0.000..1 short of 1.

:p
2006-03-09, 11:52 PM #6
tofu wins.
2006-03-09, 11:52 PM #7
.999 doesn't equal 1.

.999 repeating does.
2006-03-09, 11:53 PM #8
Originally posted by Rob:
.999 doesn't equal 1.

.999 repeating does.

It almost does.
2006-03-09, 11:53 PM #9
Don't make me hit you.
2006-03-09, 11:53 PM #10
Originally posted by tofu:
0.999... is always 0.000..1 short of 1.

:p


That's only true if there's a finite number of 9's. You DO know what the "..." means, don't you? Or am I giving you too much credit?
"it is time to get a credit card to complete my financial independance" — Tibby, Aug. 2009
2006-03-10, 12:00 AM #11
.999... = 1 - 1/inf

?
2006-03-10, 12:02 AM #12
(a decimal point followed by an infinite number of 9's) = 1 - (an infinite number of 0's followed by a 1)

As you can see, there IS no one at the end 'cause there is no end.
"it is time to get a credit card to complete my financial independance" — Tibby, Aug. 2009
2006-03-10, 12:03 AM #13
Originally posted by Freelancer:
That's only true if there's a finite number of 9's. You DO know what the "..." means, don't you? Or am I giving you too much credit?

Maybe you didn't see the smilie, hold on.

[http://i19.photobucket.com/albums/b180/chimaerall/fe7dc720.jpg]

Better?
2006-03-10, 12:03 AM #14
1/inf = 0
Sorry for the lousy German
2006-03-10, 12:05 AM #15
Any real number over infinity = 0. :p
"it is time to get a credit card to complete my financial independance" — Tibby, Aug. 2009
2006-03-10, 12:10 AM #16
If it's truly an infinite series, then it does equal 1:

(one sec might have more stuff)
Attachment: 10930/series1.jpg (8,437 bytes)
一个大西瓜
2006-03-10, 12:11 AM #17
Here's the proof in limit form: If you don't understand it, read a Calculus book until you do.

Code: 
0.9999... =     Sum         9/10^n 
                     (n=1 -> Infinity)

                    =  lim               sum      9/10^n
                     (m -> Infinity) (n=1 -> m)

                    =  lim           .9(1-10^-(m+1))/(1-1/10)
                     (m -> Infinity) 

                    =  lim           .9(1-10^-(m+1))/(9/10)
                     (m -> Infinity) 

                    = .9/(9/10)
                     
                    = 1
"it is time to get a credit card to complete my financial independance" — Tibby, Aug. 2009
2006-03-10, 12:14 AM #18
That's the same proof as I just did except not as cool. You don't even have to know calculus (though the infinite series is calculus-based) to understand mine .. it's just a infinite geo series, of which precalc tells you that the sum is a/(1-r)
一个大西瓜
2006-03-10, 12:17 AM #19
The limit is much clearer imo because it doesn't rely on any other theorems.
"it is time to get a credit card to complete my financial independance" — Tibby, Aug. 2009
2006-03-10, 12:28 AM #20
people who said no are kinda idiots. this proof is elementary and the fact that 0.999... = 1 is something you learn in 1st year high school (maybe even middle school). Limits and calculus aside, here is a proof most of you will understand

PROOF:

Let x = 0.999...

Then:
10x = 9.999...
x = 0.999...

Subtract the bottom eq. from the top equation to get:
10x - x = 9
9x = 9
x = 1

END
2006-03-10, 12:44 AM #21
That smiley hurt my eyes and just made me forget what I was going to say.
"Those ****ing amateurs... You left your dog, you idiots!"
2006-03-10, 12:51 AM #22
Originally posted by ragna:
people who said no are kinda idiots.

No, people who said no don't understand the concept of infinity. Which is fair enough really, it's hardly intuitive.
2006-03-10, 12:52 AM #23
Originally posted by Giraffe:
No, people who said no don't understand the concept of infinity. Which is fair enough really, it's hardly intuitive.


sorry. my bad.
2006-03-10, 1:05 AM #24
If .999(repeatalotalot) = 1, does 1 = .999(repeatalotalot)?

o.0
2006-03-10, 3:12 AM #25
People are perfectionists.

The way I see it, .33333334 = 1/3rd, .66666667 = 2/3rds, .99999999 = 3/3rds, otherwise known as ONE.
2006-03-10, 3:18 AM #26
Originally posted by Delphian:
The way I see it, .33333334 = 1/3rd

Someone doesn't know how to round...
2006-03-10, 3:18 AM #27
Originally posted by Greenboy:
If .999(repeatalotalot) = 1, does 1 = .999(repeatalotalot)?
Yes. And there's no 'if' about it. Laypeople are the only ones who see this matter as open to debate; mathematicians have solved it centuries ago.

Comprehension of 'infinity' is not a prerequisite to understanding this problem. I would bet that all of the people who voted 'no' don't know calculus.

Here are several proofs [Wikipedia.org].
2006-03-10, 3:21 AM #28
.999 repeating can never equal one. It will continue to get .* closer to being 1, but it can never be totally equal to 1.
Think while it's still legal.
2006-03-10, 3:26 AM #29
[QUOTE=Victor Van Dort].999 repeating can never equal one. It will continue to get .* closer to being 1, but it can never be totally equal to 1.[/QUOTE]You don't know calculus.

Edit: I demand a mathematical proof that conclusively demonstrates that 0.999... is not equal to 1.
2006-03-10, 3:31 AM #30
I don't know calculus, but based on the math that I do know, I'm just posting what I think about the situation, even if I am 100% wrong.
Think while it's still legal.
2006-03-10, 3:32 AM #31
[QUOTE=Victor Van Dort]I don't know calculus, but based on the math that I do know, I'm just posting what I think about the situation, even if I am 100% wrong.[/QUOTE]You are 100% wrong.
2006-03-10, 5:28 AM #32
The answer is yes, for reasons already stated.

If you don't know calculus, don't vote in the poll.

Alright, now that this is settled we can move onto the infamous "plane on a conveyor belt" debate (hint: the answer in this case is also "yes"; the plane would take off).
Stuff
2006-03-10, 5:30 AM #33
What's the plane on a conveyor belt question?
2006-03-10, 6:00 AM #34
Originally posted by kyle90:
Alright, now that this is settled we can move onto the infamous "plane on a conveyor belt" debate (hint: the answer in this case is also "yes"; the plane would take off).
No it wouldn't. Planes fly because of airfoil lift: that is, they fly because the air is moving past the wings. A plane on a conveyor belt would have no air moving past the wings. The jet engines provide forward thrust, they don't blow air past the airfoils. This makes even less sense if the airplane is single-prop.

A plane can fly while completely stationary as long as the wind is high enough. The conveyor belt would work, but only if the conveyor belt was also in a wind tunnel.

Edit: Not that the plane would be stationary on the conveyor belt, but I'm still under the impression that this is the assumption of the problem.
2006-03-10, 6:02 AM #35
Of course not, but I can round it up if I want.
Star Wars: TODOA | DXN - Deus Ex: Nihilum
2006-03-10, 6:17 AM #36
Nope. Only comes close.

No matter what crazy formulas you pull out of your hat, it's not true :D
"Jayne, this is something the Captain has to do for himself"

"N-No it's not!"

"Oh."
2006-03-10, 6:36 AM #37
Yes, for reasons already proven on this thread.

it's not so much infinity you need to know but the concept of limits. the proof you need for 0.999... = 1 comes when you get into infinate series though. (about halfway though Calc 2)

Quote:
Let x = 0.999...

Then:
10x = 9.999...
x = 0.999...

Subtract the bottom eq. from the top equation to get:
10x - x = 9
9x = 9
x = 1


this equation is also not a good proof because it involves rounding the numbers if you do it on a calculator and it's impossible to do it by hand thereby defeating the purpose.

the above proof reminded me of this one my calc 2 prof. gave us:
(no fair if you're good at math posting the solution right away)

given: a = b

a^2 = ab

a^2 + a^2 = ab + a^2

2(a^2) = ab + a^2

2(a^2) - 2ab = ab + a^2 - 2ab

2(a^2) - 2ab = a^2 - ab

2(a^2 - ab) = 1(a^2 - ab)

therefore:

2 = 1

END!
“Without education we are in a horrible and deadly danger of taking educated people seriously.” -G.K. Chesterton
2006-03-10, 6:40 AM #38
That's not even a valid proof. The last step divides by 0, which is not possible.
2006-03-10, 6:47 AM #39
I've talked to a good chunk of my math teachers in my school and they've said that it doesn't equal 1. Simple as that. It gets really close, but not quite 1.
"Jayne, this is something the Captain has to do for himself"

"N-No it's not!"

"Oh."
2006-03-10, 6:49 AM #40
[QUOTE=Glyde Bane]I've talked to a good chunk of my math teachers in my school and they've said that it doesn't equal 1. Simple as that. It gets really close, but not quite 1.[/QUOTE]You don't know calculus and neither do they.
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