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Topic: 0.999. (Read 125649 times) 

towr
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Re: 0.999.
« Reply #150 on: Dec 29^{th}, 2002, 5:21am » 

but you forget that 0.99999 isn't 0.99999 infinity +1 doesn't equal infinity :p (infinity +1  infinity = 1, infinity  infinity = 0, as long as the infinities are equal)


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lukes new shoes
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Re: 0.999.
« Reply #151 on: Dec 29^{th}, 2002, 6:21pm » 

i havent read the whole way thru this thread, but the main argument ive seen for 0.999 being equal to 1 is: n=0.999 10n=9.999 10nn=9.999  0.999 9n=9 n=1 but according to towr 0.9999 is different to 0.9999, so the infinite amount of nines in step one is going to be different to the infinite amount of nines in step two, and therefore when they are subtracted will not equal zero.


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Robbie Man
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Re: 0.999.
« Reply #152 on: Dec 29^{th}, 2002, 10:30pm » 

.99999... is merely a number that represents the closest possible number to 1. The fact that .99999 is a decimal (and 1 is not) is enough proof for me. I further think it is a mistake whenever we try to use .99999 is a mathimatical equation. You see, 1.9999 = 0.0001, so 1.9999.... must = 0.0000 with a very very small .0001 somewhere at the end of infinite.


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towr
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Re: 0.999.
« Reply #153 on: Dec 30^{th}, 2002, 1:04am » 

on Dec 29^{th}, 2002, 6:21pm, lukes new shoes wrote:i havent read the whole way thru this thread, but the main argument ive seen for 0.999 being equal to 1 is: n=0.999 10n=9.999 10nn=9.999  0.999 9n=9 n=1 
 What's happening here is really just a geometric series.. you have 0.999 which is: 9*10^{1} + 9*10^{2} + 9*10^{3} + ... + 9*10^{n} where n goes to infinity to solve it you subtract from 10*9*10^{1} + 10*9*10^{2} + 10*9*10^{3} + ... + 10*9*10^{n} (ten times the original value) which is 9*10^{0} + 9*10^{1} + 9*10^{2} + ... + 9*10^{(n1)} After subtraction you're left with 9*10^{0}  9*10^{n}, so this is 9 times what you started with. All terms except the first and last have canceled each other out. Now divide by 9 to get the value for the original series 10^{0}  10^{n} = 1  10^{n} The limit for n to infinity of 10^{n} is 0, so you're left with 1. The discussion seems to hinge on the question "is the limit for n to infinity of 10^{n} really 0 ?" You could f.i. say it's really the smallest number just above 0. But then, in the real number domain there are always others numbers between any two numbers (between any different x and y there lies f.i. (x+y)/2 ). So any two numbers that aren't seperated by more numbers have to be the same (different > seperated => ~seperated > ~different).

« Last Edit: Dec 30^{th}, 2002, 1:08am by towr » 
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S. Owen
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Re: 0.999.
« Reply #154 on: Dec 30^{th}, 2002, 6:29am » 

on Dec 29^{th}, 2002, 10:30pm, Robbie Man wrote:.99999... is merely a number that represents the closest possible number to 1. 
 There is no "closest possible number to 1"... well, say that this number is X. Then what would you call (X+1)/2? It seems like you can view 0.999... as a number just barely less than 1  as barely less as possible, but that ultimately doesn't make sense. If it's less than 1, then there is some string of 9s which is closer, which is a contradiction.


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Cyrus
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Re: 0.999.
« Reply #155 on: Dec 30^{th}, 2002, 12:42pm » 

Howdy all, I'm kind of a moron. But I read ALL the posts on ALL the pages and kind of understood most of them. And all the arguments I understand seem to point that .9999..... = 1. This may be completely unrelated but saying that .9999... = 1 isn't that the same as saying that the graph of y=1/x would EVENTUALLY meet the y (or x) axis if you carried it out to infinity. And if it NEVER meets the axis then the axis and the line must be running parrallel because only parrallel lines will never cross. I don't really know what I'm talkin about I took a calculus class when I was in high school but I forget all that stuff now, so I'm just hoping that someone can point out my mistake and explain this to a very NON mathmatics degree person. I don't think I'm an idiot, just ignorant.


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S. Owen
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Re: 0.999.
« Reply #156 on: Dec 30^{th}, 2002, 1:23pm » 

on Dec 30^{th}, 2002, 12:42pm, Cyrus wrote:This may be completely unrelated but saying that .9999... = 1 isn't that the same as saying that the graph of y=1/x would EVENTUALLY meet the y (or x) axis if you carried it out to infinity. 
 Hardly unrelated or moronic  I think you've got the point exactly. I believe that some people are effectively trying to argue something like "the graph of y = 1(1/x) never intersects y = 1 for any x > 0." That's right, but it does not show that 0.999... < 1. Assuming that those two things are basically the same thing is the easy mistake to make, and the crux of this riddle. To continue the analogy, saying that 0.999... = 1 is more like saying "the graph of y = 1(1/x) approaches (but does not reach) y = 1 as x approaches infinity." on Dec 30^{th}, 2002, 12:42pm, Cyrus wrote: And if it NEVER meets the axis then the axis and the line must be running parrallel because only parrallel lines will never cross. 
 The graph itself isn't a straight line, so it's not parallel or unparallel to anything. The tangent lines to the graph are lines, but there is no value of x such that the tangent at x is a horizontal line, and thus parallel to the xaxis. So the graph of 1/x is always decreasing, but that does not mean it gets to 0, because the rate at which it decreases also decreases.


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lukes new shoes
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Re: 0.999.
« Reply #157 on: Dec 30^{th}, 2002, 4:29pm » 

towr wrote: The limit for n to infinity of 10^{n} is 0, so you're left with 1. the limit is 0, which means it approaches 0 but never actually reaches it, so you're not left with 1. (getting off track, if i have an endless amount of apples and some gives me another apple how many apples do i have?)

« Last Edit: Dec 30^{th}, 2002, 4:39pm by lukes new shoes » 
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redPEPPER
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Re: 0.999.
« Reply #158 on: Dec 30^{th}, 2002, 5:24pm » 

on Dec 29^{th}, 2002, 5:21am, towr wrote:but you forget that 0.99999 isn't 0.99999 
 0.99999 isn't a real number at all. There's no way you can set a specific value for the last decimal of a number with infinite decimals. Such a number doesn't have a final digit. This is why lukes new shoes' demonstration is not "mathematically legal", as he suspected. Quote:infinity +1 doesn't equal infinity :p 
 It doesn't? Could you explain? Quote:(infinity +1  infinity = 1, infinity  infinity = 0, as long as the infinities are equal) 
 What are such things as "equal infinities"? You seem to be treating infinity like a regular number here. x+1 not = x, x+1x=1 ... but that's not correct. Infinity  infinity might take a number of values, including infinity. Example: there's an infinity of natural numbers. Among these, there's an infinity of even numbers. If you remove all these from the natural number (infinity  infinity) you're left with an infinity of even numbers, and zero (so infinity  infinity = infinity + 1).


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redPEPPER
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Re: 0.999.
« Reply #159 on: Dec 30^{th}, 2002, 5:36pm » 

on Dec 29^{th}, 2002, 10:30pm, Robbie Man wrote:.99999... is merely a number that represents the closest possible number to 1. The fact that .99999 is a decimal (and 1 is not) is enough proof for me. I further think it is a mistake whenever we try to use .99999 is a mathimatical equation. You see, 1.9999 = 0.0001, so 1.9999.... must = 0.0000 with a very very small .0001 somewhere at the end of infinite. 
 There's no end to the infinite. You started well with "the closest possible number to 1". Let's accept that 0.999... is that number. But there's a basic property of real numbers that is, two distinct real numbers have another real number between them. Actually they have an infinity of numbers between them, but one is enough. As you cannot possibly write a number that would be between 0.999... and 1, they must be the same number. It's a little like in geometry. There's no such thing in geometry as "adjacent" points. Between two distinct points you can add more points, you can measure a distance. If you can't, then the points are not distinct.


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lukes new shoes
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Re: 0.999.
« Reply #160 on: Dec 30^{th}, 2002, 6:06pm » 

i've decieded that 0.999 is greater that one. equation 1 infinity + infinity = infinity equation 2 infinity  infinity = 0 add equation 1 to equation 2 infinity + infinity + infinity  infinity = infinity + 0 3*infinity  infinity = infinity infinity  infinity = infinity 0 = infinity add 0.999 to both sides 0 + 0.999 = infinity + 0.999 0.999 = infinity therfore 0.999 > 1


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redPEPPER
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Re: 0.999.
« Reply #161 on: Dec 30^{th}, 2002, 6:07pm » 

on Dec 30^{th}, 2002, 4:29pm, lukes new shoes wrote:towr wrote: the limit is 0, which means it approaches 0 but never actually reaches it, so you're not left with 1. 
 Achilles shoots an arrow at a target. The distance to the target is one mile. The arrow first crosses half of the distance (1/2 mile). Then the arrow crosses half of the remaining distance (1/4 mile). Then half of what's left (1/8 mile) and so on... The limit_{x > inf} of that sum_{i=1 > x} 1/2^{i} is obviously 1 mile. Does that mean the arrow approaches the target but "never actually reaches it"? Would you volunteer as the target to verify this?


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Brandon
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Re: 0.999.
« Reply #162 on: Dec 30^{th}, 2002, 11:48pm » 

I have a 1 gallon jug (j). It is empty (j=0). I fill 9/10ths of the empty space (j+=0.9). I fill 9/10ths of the empty space (j+=0.09). If I am able to do this forever, I will never fill the entire jug (j<1). Please discuss.


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towr
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Re: 0.999.
« Reply #163 on: Dec 31^{st}, 2002, 2:39am » 

ok.. let's look at infinity(s) a bit closer.. can be let's call this x but is also , let's call this one y x/x=y/y=1, x/y = 0, y/x = They're different.. Maybe a better way to look at the different order of magnitudes of infinity is to look at number sets, for example rational and real numbers. Both these sets are infinite, yet the real set is infinitely larger. On the other hand you might be inclined to say that the set of rational numbers is larger than the set of natural numbers, yet these are the same size, since there is a one to one mapping from one set to the other.


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Cyrus
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Re: 0.999.
« Reply #164 on: Dec 31^{st}, 2002, 11:14am » 

Ok, I THINK I got it now. Thanks S Owen and the billions of other explanations I've read. So looking at Bradon's example of the jug, or the crossing of the football field example I remember reading earlier on someone's post. Is the explanation that ... you never actually FILL the jug, or REACH the other line. Just like the line y = 1(1/x) never actually intersects the axis. But you are so INFINITESIMALLY (no idea how you spell that word) close that you are in fact there. So you've filled the jug so full that it's INCOMPREHENDABLY full, but it's not actually full. Am I understanding or getting more confused. THAT I don't know. And maybe THAT is the TRUE riddle.


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S. Owen
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Re: 0.999.
« Reply #165 on: Dec 31^{st}, 2002, 11:32am » 

on Dec 31^{st}, 2002, 11:14am, Cyrus wrote:But you are so INFINITESIMALLY (no idea how you spell that word) close that you are in fact there. 
 The argument's not so much that 0.999... is infinitesimally close to 1  but that it is 1. 0.999... = 1 is just saying that 0.9, 0.99, 0.999, ... gets ever closer to something, and that something is 1. It's making a statement about the limit of the sequence, not any element in the sequence. Or, saying 0.999... = 1 is like saying the level of the water in the jug will approaches 100% full.


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redPEPPER
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Re: 0.999.
« Reply #166 on: Dec 31^{st}, 2002, 6:08pm » 

on Dec 30^{th}, 2002, 11:48pm, Brandon wrote:I have a 1 gallon jug (j). It is empty (j=0). I fill 9/10ths of the empty space (j+=0.9). I fill 9/10ths of the empty space (j+=0.09). If I am able to do this forever, I will never fill the entire jug (j<1). Please discuss. 
 Actually, if you're able to do this forever, you WILL fill the entire jug, just like the arrow will reach the target in my previous example. Imagine you fill 9/10ths of the empty space in one second initially, and then in half the time it took to fill the previous iteration after that. In one second you will fill 9/10th of the jug. In 1.5 seconds, 99/100 of the jug. In 1.75 seconds, 999/1000 of the jug... After two seconds, the jug is full. Not almost full: completely full. As sure as the arrow will reach the target. If you go for a finite number of iterations, you approach the limit. But if you go infinitely, you DO reach the limit.


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SPAZ
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Re: 0.999.
« Reply #167 on: Jan 1^{st}, 2003, 12:57am » 

to clarify the initial math.... given: let x = .999... (going on for ever) multiply each side by 10: 10x = 9.999... subtract x from each side: 10x  x = 9.999...  x simplify: 9x = 9.999... .999... 9x = 9 devide each side by 9 9x = 9 9 9 simplify: x = 9 9 9 = 1 9 x = 1 thus: .999... = 1


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Cyrus
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Re: 0.999.
« Reply #168 on: Jan 2^{nd}, 2003, 8:57am » 

on Dec 30^{th}, 2002, 5:36pm, redPEPPER wrote: There's a basic property of real numbers that is, two distinct real numbers have another real number between them . . . As you cannot possibly write a number that would be between 0.999... and 1, they must be the same number. It's a little like in geometry. There's no such thing in geometry as "adjacent" points. Between two distinct points you can add more points, you can measure a distance. If you can't, then the points are not distinct. 
 I really like that real number/geometry example. Easy for a simpleton like myself to understand and agree with. I like it when people use different forms of math to prove their answers, like using geometry to make a point about real numbers. Good teaching skill! But again . . . I go back to the graph of y=1(1/x). Does that mean that the line DOES intersect 1 if you follow x to infinity? Just like the jug does eventually get completely full. Because if it does, then if you look at (x = infinity + 1) wouldn't y be greater than 1?? Or corespondingly the jug would overflow. Maybe I don't understand this concept of infinity + 1. Does it exist?? Jeeze, just when I think I understand, I get confused again. Somebody tell me to shut up if I'm being stupid.


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S. Owen
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Re: 0.999.
« Reply #169 on: Jan 2^{nd}, 2003, 10:46am » 

on Jan 2^{nd}, 2003, 8:57am, Cyrus wrote:But again . . . I go back to the graph of y=1(1/x). Does that mean that the line DOES intersect 1 if you follow x to infinity? Just like the jug does eventually get completely full. 
 I think it all hinges on what one means by "following x to infinity". You can't evaluate the value of y = 1(1/x) graph at x = infinity. But, quite naturally, you can evaluate the limit of y as x approaches infinity, and that is 1. If anything, you can say that this is the value at x = infinity, but I feel that is misleading, because then you naturally want to ask what is the value at infinity+1, and it would seem that it must be > 1. I'd like to go back to the jug example... I think there needs to be a little clarification there. I think the original post meant something like this: I add 9/10 of a gallon in 1 second. Then, I add another 9/100 of a gallon in 1 second. Then, I add 9/1000 of a gallon in 1 second. Does the jug have 1 gallon after any finite number of steps? No. Does the jug have 1 gallon after any finite number of seconds? No. However here is the second situation that was discussed: I add 9/10 of a gallon in 1 second. Then, I add another 9/100 of a gallon in 1/2 second. Then, I add 9/1000 of a gallon in 1/4 second. Does the jug have 1 gallon after any finite number of steps? No. Does the jug have 1 gallon after any finite number of seconds? Yes  2 seconds. Anyway, my point is that 0.999... = 1 is analgous to the first situation if anything, in my opinion, not this second one. In the second one (which is a riddle unto itself!) you do reach 100% full in a finite amount of time, because the time for each step decreases fast enough, and it might be misleading to relate that to 0.999... = 1 to say that this is why the are equal.

« Last Edit: Jan 2^{nd}, 2003, 10:48am by S. Owen » 
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towr
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Re: 0.999.
« Reply #170 on: Jan 2^{nd}, 2003, 11:28am » 

on Jan 2^{nd}, 2003, 10:46am, S. Owen wrote: Anyway, my point is that 0.999... = 1 is analgous to the first situation if anything, in my opinion, not this second one. 
 It depends on what part you are looking at. Lets look at the time steps in the second situation, in binary: 1 second + 0.1 second + 0.01 second etc = 1.111... seconds = 10 seconds Which is really the same problem, only in a different base.

« Last Edit: Jan 2^{nd}, 2003, 11:30am by towr » 
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redPEPPER
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Re: 0.999.
« Reply #171 on: Jan 2^{nd}, 2003, 11:36am » 

on Jan 2^{nd}, 2003, 10:46am, S. Owen wrote:I think it all hinges on what one means by "following x to infinity". You can't evaluate the value of y = 1(1/x) graph at x = infinity. But, quite naturally, you can evaluate the limit of y as x approaches infinity, and that is 1. If anything, you can say that this is the value at x = infinity, but I feel that is misleading, because then you naturally want to ask what is the value at infinity+1, and it would seem that it must be > 1. 
 Yeah, people tend to treat infinity like a number. See some of the demonstrations above, with things such as infinity + or  1. But infinity is more a concept in itself than a number. One of its definitions could be: it's the end of a line that doesn't have an end. Saying the graph above never reaches 1, or saying that it reaches it at infinity is not very different. But you can't assume from the latter that you can focus on the place where it reaches 1 and look what happens farther right on the graph. Because it doesn't reach 1 at any x coordinate that's a real number. It reaches 1 at the very edge of the graph... that doesn't have an edge, as it's measured by real numbers. Ah it's not easy to explain. I'm not sure I understand it myself Quote:I'd like to go back to the jug example... I think there needs to be a little clarification there. I think the original post meant something like this: I add 9/10 of a gallon in 1 second. Then, I add another 9/100 of a gallon in 1 second. Then, I add 9/1000 of a gallon in 1 second. Does the jug have 1 gallon after any finite number of steps? No. Does the jug have 1 gallon after any finite number of seconds? No. However here is the second situation that was discussed: I add 9/10 of a gallon in 1 second. Then, I add another 9/100 of a gallon in 1/2 second. Then, I add 9/1000 of a gallon in 1/4 second. Does the jug have 1 gallon after any finite number of steps? No. Does the jug have 1 gallon after any finite number of seconds? Yes  2 seconds. Anyway, my point is that 0.999... = 1 is analgous to the first situation if anything, in my opinion, not this second one. In the second one (which is a riddle unto itself!) you do reach 100% full in a finite amount of time, because the time for each step decreases fast enough, and it might be misleading to relate that to 0.999... = 1 to say that this is why the are equal. 
 My goal was to show that something that's infinite can be "contained" in something that's finite. People confuse the notion of infinity with something like an arbitrarily high number. 0.999... is described as adding 9s to 0. and then adding more 9s and then more, and then more... If you add one figure per time unit, you'll never be done. The number you're writing will never equal 1. It will come closer and closer, but will never reach it, or will reach it after an infinite time, which is not convenient if you want to use that number afterwards. People will say that, no matter how far you are, you can always write a bigger number that's smaller than 1. But if the time needed to add a digit decreases with each iteration, it's possible to be done in a finite amount of time. After two seconds, the number is complete! It has an infinite number of digits, not an arbitrarily high finite number. I figured this could help understanding. This reminds me of a riddle that I'll post here. Look for four beetles and a square.


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S. Owen
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Re: 0.999.
« Reply #172 on: Jan 2^{nd}, 2003, 11:37am » 

Oh sure, agreed, all of this is the same question in the sense that it's a question of infinite sums, series, whatever. I was just trying to say that in that second situation, we're talking about a process that happens to finish in a finite amount of time... you "do get to 1" in the second case... but that does not directly correspond to why 0.999... = 1. The argument is not that 0.999... = 1 because 0.999... "gets to 1" somehow. Just wanting to be clear on exactly how these problems relate.


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lukes new shoes
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Re: 0.999.
« Reply #173 on: Jan 5^{th}, 2003, 1:06am » 

another way to show that 0.999...=1 0.999... divided by 9 equals 0.111... 1 divided by 9 equals 0.111... so 0.999... equals 1


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Kozo Morimoto
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Re: 0.999.
« Reply #174 on: Jan 6^{th}, 2003, 4:14am » 

on Jan 5^{th}, 2003, 1:06am, lukes new shoes wrote:another way to show that 0.999...=1 0.999... divided by 9 equals 0.111... 1 divided by 9 equals 0.111... so 0.999... equals 1 
 That's assuming that 0.999... can actually be divided by 9 to give you 0.111... The whole point of the 'riddle' was that you can't make such assumptions unless you can prove it.


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