Think about this for a moment.
1/9 = 0.1111111 (the one's continue forever)
2/9 = 0.2222222 (again they repeat forever)
3/9 = 0.3333333 (same thing)
4/9 = 0.4444444 (I hope your getting the point by now)
5/9 = 0.5555556 (rounds up now)
6/9 = 0.6666667
7/9 = 0.7777778
8/9 = 0.8888889
9/9 = 1
Now if you notice the last value is not in fact 1, but a 0.99999 (repeating forever) now this does become 1 when you round any of the 9's in the number.
If this is true, is any number divided by itself truly 1?
Your thoughts.
Why use such lame reasoning? It can be algrebraically proven.
GIVEN: x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1
THEREFORE: 0.999... = 1
Q.E.D.
I've learned this years ago.
10x= 1.11111
- 1x = 0.11111
9x = 1
According to your logic, all numbers >= 5/9 (of the 9 denominator) up to nine has that last repeating digit that rounds. .999999 would round the last digit to .999990 but that extra hundred thousanth has to go somewhere, so you get .99990, .9990, .990, .90, 1.
No, .9999.... does not equal 1.
8/9 = .8888...
But
9/9 is 1. Not .9999...
DTBK:
QUOTE
GIVEN: x = 0.999...
10x = 9.999...
That step is wrong. You can't multiply .9999... by 10 unless you want the end to be ..9990. Otherwise, you would end up with an extra ...0009, which is impossible.
Don't try to use silly math tricks like "move one decimal over" or whatever. Use logic.

QUOTE(DT_Battlekruser @ Mar 1 2006, 01:29 AM)
Why use such lame reasoning? It can be algrebraically proven.
GIVEN: x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1
THEREFORE: 0.999... = 1
Q.E.D.
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you cant just pull numbers/variables out of ur arss, for the 10x - x, you have to minus x on the other side too.....and then it gets all mixed up
if im wrong then plz kill me now for i have not truly understood anything that i have been taught in school
If I recall, I use to have a signature like this...
QUOTE(Blizzard® Entertainment)
lim(m --> 8) sum(n = 1)^m (9)/(10^n) = 1
0.9999... = 1
Thus x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.
1 = 1 & 1 = 1
1 ˜ .9998 & 1 ? .9998
Take that, Blizzard!
http://www.blizzard.com/press/040401.shtmlQUOTE(CheeZe @ Mar 1 2006, 06:06 PM)
No, .9999.... does not equal 1.
8/9 = .8888...
But
9/9 is 1. Not .9999...
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ya. lawl.
1 = 0.999...
The number continues forever. Since there is no fraction for this number, you cannot add/subtract/divide/multiply this number, so instead, you make life easy and round so it's possible to work with. The 0.999... must be rounded, it really isn't an option.
Rounding .9999... makes it 1. But that doens't make .9999... equal to 1.
The thing is, you must round to be able to work with it, understand?
This stuff hurts my head. Wouldn't that mean that I have 1.998 hands?
It's .999 repeating, as in .99999999999999999 etc. Not .999. You're thinking of a simple fraction, we're talking about an irrational number.
It's really weird.
1/3 = .33¯
2/3 = .66¯
3/3 = 1
(Oh right, thirds are ninths too)
0.999... = My way of saying 0.9 repeating.
QUOTE
Proof: 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
[sub]Why are you guys saying 9/9 is .99999? :/[/sub]
the number .9999 will remain the same if untouched, but it will be equal to 1...only way you'd use the .9999 would be that you want to measure something very precisely...but anyway i guess it depends on what you do to it.
Back in my calculus my prof said something about a=b...probably it relates how .9999=1?
QUOTE(SlyShadow @ Mar 2 2006, 01:12 PM)
the number .9999 will remain the same if untouched, but it will be equal to 1...only way you'd use the .9999 would be that you want to measure something very precisely...but anyway i guess it depends on what you do to it.
Back in my calculus my prof said something about a=b...probably it relates how .9999=1?
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Can you be more vague?

QUOTE(HolySin @ Mar 2 2006, 02:23 PM)
Can you be more vague?

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uhh if you don't do anything to the .9999 itll stay the same wouldn't it?but rounding it would make it change to a whole number...
for the second part, the prof taught me a very long process,its about a year ago,but he did it how different variables can equal to each other...the a=b using some arithmetic function, i forgot what he did...but i remember he did prove it somehow...just making it how it related to the .9999=1 question...
Am I the only one who thinks this thread is pointless?
QUOTE(Falcon_A @ Mar 2 2006, 06:01 PM)
Am I the only one who thinks this thread is pointless?
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I will agree with you, that this is a silly thread. However everyone has their right to ask a question, somtimes.
9/9 is 1/1, is 1. Just reduce the fraction. 9/9 * 9 = 1/1 = 1
QUOTE
9/9 is 1/1, is 1. Just reduce the fraction. 9/9 * 9 = 1/1 = 1
That's the way I see it.
QUOTE
I will agree with you, that this is a silly thread. However everyone has their right to ask a question, somtimes.
I must agree with that as well.
QUOTE(CheeZe @ Mar 1 2006, 05:06 PM)
Don't try to use silly math tricks like "move one decimal over" or whatever. Use logic.

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It's not a silly math trick. By definition, 0.999...(10) = 9.999..... It's the foundation of our number system. If you'd like to do it longhand to prove me wrong, I'll see you in infinity years. 0.{9}0 = 0.{9} (where {x} denotes infinite repeating.)
Besdies, LW proved it quite well.
Saying 0.999... ≠ 1 is like saying lim[sub]x→2[/sub] (x[sup]2[/sup]-4)/(x-2) ≠ 4.Just because you put .999999999999999999 into your calculator and get 1 doesn't mean they're equal.
