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☖ Forums → Discussion Forum → Does 0.99999~ equal 1?
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Does 0.99999~ equal 1?
2004-09-26, 10:46 AM #1
Does 0.99999~ equal 1?

What do you think?
Made you look
2004-09-26, 10:49 AM #2
Equal? No. Approximate? Yes.
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2004-09-26, 10:51 AM #3
Yes, it does. You'll learn the proof when you get to Calculus. Sufficed to say, for now, you can think of it this way: 0.3333333.... = 1/3 right? 1/3 + 1/3 + 1/3 = 1 = 0.9999999...
"it is time to get a credit card to complete my financial independance" — Tibby, Aug. 2009
2004-09-26, 10:52 AM #4
0.999 reoccuring ≈ 1.
0.999 reoccuring ≠ 1.

That is all.

Oh, and the whole thirds thing? Since 0.333 reocurring is just that, reocurring, you can only add a number of decimal places, rather than the whole thing. Which is why 0.333 (that's 333 over 1000) plus 0.333 plus 0.333 equals 0.999.
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2004-09-26, 10:56 AM #5
SHUT UP SHUT UP SHUT UP

Edit: Yes.
2004-09-26, 10:56 AM #6
Quote:
Originally posted by jesseg88
Does 0.99999~ equal 1?

Yes, for very large values of 0.99999... ;)
And when the moment is right, I'm gonna fly a kite.
2004-09-26, 10:57 AM #7
0.999... = 9/10 + 9/100 + 9/1000...

Using limits, you can prove that this infinite sum is 1.

The specifics? here, and here.
"it is time to get a credit card to complete my financial independance" — Tibby, Aug. 2009
2004-09-26, 11:08 AM #8
No. The proof is fake. 1 = 1 end of story.
2004-09-26, 11:13 AM #9
No, 0.9 with 9 repeating does not and never will equal 1.

What does equal 1 is the limit of that infinite sum, which is different than the actual number...
2004-09-26, 11:15 AM #10
Code: 
             
    1/3 = 0.3
+   2/3 = 0.6
————————————————
    3/3 = 0.9
       
 .: 0.9 = 1.0
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2004-09-26, 11:27 AM #11
This thread is basically an attempt to understand the concept of infinity.

Good luck with that.
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2004-09-26, 11:59 AM #12
It might as well.
Warhead[97]
2004-09-26, 1:57 PM #13
seems as though it does. :D
I know that you believe you understand what you think I said, but I'm not sure you realize that what you heard is not what I meant.
2004-09-26, 2:02 PM #14
0.999999~ is 1 - 1/infinity.

If 1/infinity = 0, then yes, it does = 1. However, if 1/infinity is just very very small, then it doesn't. To be honest, I don't care, although Freelancer's/Emon's proof sort of has me converted. I guess for all extents and purposes, it does = 1.
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2004-09-26, 2:13 PM #15
Yse it does, you learn that in basic pre-algebra when they teach you to convert decimals to fractions and vice versa. Try converting .99999999999999999999999999 to a fraction. You come out with one.
D E A T H
2004-09-26, 2:16 PM #16
Correct Answer:

No, 0.999999999 repeating does not equal 1. The limit is 1, however.

My Answer:

It's close enough for all practical purposes.
Stuff
2004-09-26, 2:19 PM #17
Quote:
Originally posted by Dj Yoshi
Yse it does, you learn that in basic pre-algebra when they teach you to convert decimals to fractions and vice versa. Try converting .99999999999999999999999999 to a fraction. You come out with one.


No, you come out with 99999999999999999999999999/100000000000000000000000000

:)

Quote:
Originally posted by Freelancer
0.999... = 9/10 + 9/100 + 9/1000...

Using limits, you can prove that this infinite sum is 1.


Very true. But 0.999999~ != 1. The limit of the infinite sum of 9/10^i as i tends towards infinity is one. Not the same thing.
2004-09-26, 2:31 PM #18
0.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
guess how many nines...
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YEAH YEAH YEAH
2004-09-26, 2:31 PM #19
PI IS EXACLTY 3!!!!!!!! [/Prof. Frink]
May the mass times acceleration be with you.
2004-09-26, 2:43 PM #20
*GASP*

Oh yeah.... well Star Trek is better than Star Wars!

*gets killed by Massassians*
Catloaf, meet mouseloaf.
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2004-09-26, 2:48 PM #21
Speaking of infinity, anyone remember the monkey-bible thread?
2004-09-26, 3:28 PM #22
Quote:
The limit of the infinite sum of 9/10^i as i tends towards infinity is one. Not the same thing.

The sum of an infinite series is defined as the limit of the sequence of partial sums. You can say, "1 is the sum of the infinite series," or, "1 is the limit of the sequence of partial sums," but, "1 is the limit of the sum," is complete nonsense. By definition, 1 is the sum.

Anyone debating the equivalence of an infinite series with its sum should recall that the Reimann integral is defined as an infinite series. The last time I checked, the area under a curve is an exact number, not an approximation or "limit of a sum".
2004-09-26, 4:07 PM #23
Damnit guys, now Ive got a headache! :mad: :(
And when the moment is right, I'm gonna fly a kite.
2004-09-26, 4:12 PM #24
:gbk:

And Argath, you're perfectly right. I don't know what the **** I was thinking writing that. I've been going linear algebra all afternoon, calculus is not really what's in my head right now.
2004-09-26, 4:39 PM #25
Quote:
Originally posted by Argath
The sum of an infinite series is defined as the limit of the sequence of partial sums. You can say, "1 is the sum of the infinite series," or, "1 is the limit of the sequence of partial sums," but, "1 is the limit of the sum," is complete nonsense. By definition, 1 is the sum.

Anyone debating the equivalence of an infinite series with its sum should recall that the Reimann integral is defined as an infinite series. The last time I checked, the area under a curve is an exact number, not an approximation or "limit of a sum".
I didn't understand a word of that, so I'm taking it upon myself to memorize it and rattle it off in order to impress my friends.
2004-09-26, 4:44 PM #26
Quote:
Originally posted by Freelancer
Yes, it does. You'll learn the proof when you get to Calculus. Sufficed to say, for now, you can think of it this way: 0.3333333.... = 1/3 right? 1/3 + 1/3 + 1/3 = 1 = 0.9999999...


No. 1/3 = 0.3333333 recurring (the sysmbol is the little line above the three
Pissed Off?
2004-09-26, 4:57 PM #27
The ... means recurring. 0.999... is another notation for .999 with a line over it, which of course means there are an infinite number of nines.
2004-09-26, 5:32 PM #28
Wasn't this like an april fool's joke from blizzard?
2004-09-26, 6:01 PM #29
MUST......DESTROY......MATH.....
2004-09-26, 6:30 PM #30
Argath wins the thread. :) There is no longer any need to reply to it.
"it is time to get a credit card to complete my financial independance" — Tibby, Aug. 2009
2004-09-26, 6:34 PM #31
OMG!?!?! Y h3lo thar!?!
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2004-09-26, 6:55 PM #32
Quote:
Originally posted by Vincent Valentine
Speaking of infinity, anyone remember the monkey-bible thread?
No, but it probably sucked.
Catloaf, meet mouseloaf.
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2004-09-26, 7:05 PM #33
Form for an infinite sequence: an = a1 (r ^ (n-1))

So here's an infinite sequence that will give you .9, .09, .009, and so on:
an = .9 * (.1 ^ (n-1))

Here's the formula for the sum of an infinite sequence: S = a1 / (1-r)

We have r as .1 and a1 as .9. Let's substitute.

S = (.9) / (1 - .1)
S = (.9) / (.9)
S = 1

There you go.
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2004-09-26, 7:16 PM #34
Quote:
Originally posted by gbk
Damnit guys, now Ive got a headache! :mad: :(
Laughing at my spelling herts my feelings. Well laughing is fine actully, but posting about it is not.
2004-09-26, 7:17 PM #35
Quote:
Originally posted by Pagewizard_YKS
MUST......DESTROY......MATH.....


YES!!!!


math > me

:(
2004-09-26, 7:19 PM #36
0.999... = 0.999...
1 = 1
0.999... =/= 1
I fail to see the controversy. You all lose.
2004-09-26, 7:23 PM #37
Quote:
Originally posted by Mikus
0.999... = 0.999...
1 = 1
0.999... =/= 1
I fail to see the controversy. You all lose.


agree'd
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…( `\(o),,_/` ¯ : o : : : o`-, ….
2004-09-26, 7:28 PM #38
.9999~ does not equal 1, just like 1.999~ does not equal 2 etc... It's like that damn curving line on the Y axis, and no matter how long it goes, it wont cross the damn X axis! It just keeps getting closer and closer.
Think while it's still legal.
2004-09-26, 7:33 PM #39
Quote:
Originally posted by SAJN_Master
.9999~ does not equal 1, just like 1.999~ does not equal 2 etc... It's like that damn curving line on the Y axis, and no matter how long it goes, it wont cross the damn X axis! It just keeps getting closer and closer.

Learn calculus.

One distinction I could point out is that we're not watching the equation y = .9(.1 ^ x), we're watching the sum of all of its terms as x approaches infinity.
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2004-09-26, 7:33 PM #40
Quote:
Originally posted by Argath
The sum of an infinite series is defined as the limit of the sequence of partial sums. You can say, "1 is the sum of the infinite series," or, "1 is the limit of the sequence of partial sums," but, "1 is the limit of the sum," is complete nonsense. By definition, 1 is the sum.

Anyone debating the equivalence of an infinite series with its sum should recall that the Reimann integral is defined as an infinite series. The last time I checked, the area under a curve is an exact number, not an approximation or "limit of a sum".
Oh, except that the area under a curve actually is an approximation. :)
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