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Topic: 0.999. (Read 105935 times) 

mentor
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0.999.
« on: Jul 26^{th}, 2002, 10:27am » 

There's a method to remove reccurence form a decimal: FIrst, take the number and time by 10^(no of places before reccurence): 0.999. x 10 = 9.99. Now take the original number away from the answer: 9 If we repeat the same process with a variable x instead of the decimal, we get: 9x We now equate the two and solve for x, which gives us: 1 so, 0.999. = 1


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mentor
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Re: 0.999.
« Reply #1 on: Jul 26^{th}, 2002, 11:40am » 

An Addendum: it's a short step (left as an excercise to the reader) to show that 1/(ifinity) is 0.


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Frothingslosh
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Re: 0.999.
« Reply #2 on: Jul 26^{th}, 2002, 12:52pm » 

Since .9999... = 1, there are any number of ways to prove this. When I find someone who can't accept some of the classic proofs, I resolve the matter once and for all with the following: 1/3 = .333333...... 2/3 = .666666...... ____ ___________ 3/3 = .999999....... Just add each side together and it becomes obvious!


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Phezik
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Re: 0.999.
« Reply #3 on: Jul 26^{th}, 2002, 4:10pm » 

Another even easier way to think of this for those who won't accept the classic proof is 1  .9999... = 0.0000... If you take the difference between two numbers, and the difference is 0, then they have to be the same number.


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Kozo Morimoto
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Re: 0.999.
« Reply #4 on: Jul 26^{th}, 2002, 5:16pm » 

First off 1/3 is NOT 0.3333... that's just the decimal approximation. I don't understand the first solution. And 0.0000... is NOT 0, very close to zero, but not zero, so that equal sign in the middle is wrong. 1 is not equal to 0.999... I suppose you have to define what EQUAL is... I've heard that something with a probability of less than 1 out of 10^40 is mathematically classified as impossible, although it isn't 0.


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Rhaokarr
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Re: 0.999.
« Reply #5 on: Jul 26^{th}, 2002, 9:10pm » 

I want to see who would take up the argument that 0.999' > 1


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Davers
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Re: 0.999.
« Reply #6 on: Jul 27^{th}, 2002, 3:32am » 

A nonmath solution: What number is between .9999... and 1? See? There isn't one. Hence: .9999... = 1


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Kozo Morimoto
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Re: 0.999.
« Reply #7 on: Jul 27^{th}, 2002, 6:32am » 

"A nonmath solution: What number is between .9999... and 1? See? There isn't one. Hence: .9999... = 1" Of course there is, we just can't express it with the Arabic numeral that we use. Lets create a new digit # which is bigger than 9 but less than 10. Then .999999... < .####..... < 1 so there is a number between .999... and 1, but we just can't express it with the digits that we NORMALLY use. Just because we can't express it doesn't mean that it isn't there.


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NickH
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Re: 0.999.
« Reply #8 on: Jul 27^{th}, 2002, 8:01am » 

Suppose, for the sake of argument, that # = 9.5. Then, .## = .95 + .095, which is already greater than 1! In fact, for # = 9.5, .#####... = 19/18. Further, .#####... > 1, for any # > 9. Your argument that there exists a # such that 9 < # < 10 does not imply the existence of a number @ such that .999... < @ < 1. There is no such @.


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cnmne
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Re: 0.999.
« Reply #9 on: Jul 27^{th}, 2002, 10:58am » 

> First off 1/3 is NOT 0.3333... that's just the decimal approximation. that is kind of the point. 0.9999... is the decimal approximation of 3/3 = 1


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obtuse
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Re: 0.999.
« Reply #10 on: Jul 27^{th}, 2002, 1:56pm » 

I recall reading an article defining a counting system in which you could easily write infinitely large and infinitely small numbers. It incorporates ideas from Zeno's Paradox. I hope my explanation makes some sense. ^ = up arrow V = down arrow The first arrow signifies a change of 1. Every arrow in the same direction as the first counts as another increment of 1. ^^ = 2 VVV = 3 The first arrow to change direction meant a 'decrement' of 0.5, and every arrow thereafter was a 'decrement' of half the previous value. ^V = 1  0.5 = 0.5 ^VV^ = 1  0.5  0.25 + 0.125 = 0.375 Expressing a repeating pattern was done by adding a bar over the top, which I will exchange for an underline. Multiple bars were allowed, to allow embedded patterns and multiple infinities. ^ = infinity ^^ = infinity plus 1 (there IS a difference here!) ^VV^ = 1/3 (if I'm doing my math right) It defined an iota as ^V, meaning the smallest possible number above zero. Thus: 0.999... + 0.000... = 1 or ^^V^ + ^V = ^ I assume this isn't what the problem was looking for, but interesting regardless.


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domn
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Re: 0.999.
« Reply #11 on: Jul 27^{th}, 2002, 4:35pm » 

10 x .9999... = 9.9999... 9.9999...  .9999... 9 9/(101) = 1


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Frothingslosh
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Re: 0.999.
« Reply #12 on: Jul 27^{th}, 2002, 8:32pm » 

>> First off 1/3 is NOT 0.3333... that's just the decimal approximation. > that is kind of the point. > 0.9999... is the decimal approximation of 3/3 = 1 You guys are dead wrong. 1/3 = .333....... exactly. The .... is an indication that the 3's continue on to infinity. This is not an aproximation, it is the exact value, represented by the dots to save time from writing an infinite number of 3's. But it is the exact answer, if you didn't learn that much math in grade school there isn't much point in further discussion. I suggest you both do a little research and try to come up with anything to support your position.


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Jeremiah Smith
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Re: 0.999.
« Reply #13 on: Jul 28^{th}, 2002, 12:46am » 

Actually, .99999... is a geometric series... 9/10 + 9/100 + 9/1000, each term is 1/10 of the last. I remember a formula from my calculus class dealing with the sum of geometric series, and although it's much too late (or early...) for me to remember, I do remember that it showed that .9999... = 1. Maybe I'll go find that formula tomorrow.


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Kozo Morimoto
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Re: 0.999.
« Reply #14 on: Jul 28^{th}, 2002, 6:37am » 

"You guys are dead wrong. 1/3 = .333....... exactly. The .... is an indication that the 3's continue on to infinity." You are wrong. .333... is an aproximation of 1/3. That's why 1/3 is a rational number and thus can not be represented by decimal notation. So .333... is an approximation. Otherwise it wouldn't be a rational number. Rationals are a subset of reals and a superset of integers. "Suppose, for the sake of argument, that # = 9.5. Then, .## = .95 + .095," No, .###... = .###.... and can not be represented by normal decimal notation in my example. If I could I would need to use the new symbol/digit "#". It's still a number between 0.999... and 1 by definition.


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NickH
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Re: 0.999.
« Reply #15 on: Jul 28^{th}, 2002, 8:44am » 

"You are wrong. .333... is an aproximation of 1/3. That's why 1/3 is a rational number and thus can not be represented by decimal notation. So .333... is an approximation. Otherwise it wouldn't be a rational number. Rationals are a subset of reals and a superset of integers." I think you are confused by the concepts of "rational" and "representable in decimal notation." A rational number is a real number that can be represented as a/b, where a and b are integers, b != 0. Some rational numbers have a decimal expansion that terminates; example: .25. All other rationals have a decimal expansion that repeats indefinitely from some point on; example 1/11 = .09090909..., where '09' is the repeating group. Conversely, any number whose decimal expansion either terminates or repeats from some point on is rational. As jeremiahsmith says above of .999..., .333... is mathematical shorthand for the infinite geometric series: 3/10 + 3/100 + 3/1000 + ... The sum of such a series is given by a/(1r), where a is the first term and r is the common ratio, if r < 1. In this case we have a = 3/10, r = 1/10, yielding a sum of 1/3. This is the exact answer. It is what we get when we sum to infinity. Let me ask you this: if the sum to infinity of 3/10 + 3/100 + 3/1000 + ... is not equal to 1/3, what is it equal to? "No, .###... = .###.... and can not be represented by normal decimal notation in my example. If I could I would need to use the new symbol/digit "#". It's still a number between 0.999... and 1 by definition." I think you need to explain more clearly exactly what "#" is, what ".#" means, and what ".###..." means. Clearly # is not an integer, for there are no integers > 9 and < 10. Is # a real number? If so, then what is .#? What number base are you working in? Once you've explained the concepts you are using, attempt a proof that .999... < .###... < 1.


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S. Owen
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Re: 0.999.
« Reply #16 on: Jul 28^{th}, 2002, 9:13am » 

NickH is 99.9999...% right. By which I mean 100%. I agree that the problem here is that people are throwing around ".333..." like any other decimal. Indeed, it's a way of expressing the value 1/3 as the sum of an infinite geometric series. ".333.." just looks like an ordinary decimal, and that's the trick of this question. But it's an "infinitely long decimal" which is quite an imaginary thing to begin with. So .999... = 1. It's just a roundabout way of expressing "1". And no, there is no value between .999... and 1, because they are in fact equal! This business about a digit "#" equalling 9.5 is misguided.


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Gareth Pearce
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Re: 0.999.
« Reply #17 on: Jul 28^{th}, 2002, 6:50pm » 

0.9999999... = 1 is simple but can you prove 0.9999... < 1 ? That is, can bother answers 1 and 2 be true at the same time... I dont think so, but i havent proven it.


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Kozo Morimoto
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Re: 0.999.
« Reply #18 on: Jul 28^{th}, 2002, 9:49pm » 

In my example, if # is half way between 9 and 10, then 0.# would be halfway between 0.9 and 1. So 0.## would be half way between 0.99 and 1 and so on. So 0.###... would be half way between 0.999... and 1


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jmlyle
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Re: 0.999.
« Reply #19 on: Jul 28^{th}, 2002, 11:41pm » 

Kozo, it's time to step away from this problem. Quote:if # is half way between 9 and 10, then 0.# would be halfway between 0.9 and 1. So 0.## would be half way between 0.99 and 1 and so on. So 0.###... would be half way between 0.999... and 1 
 Imagine that there is a number that is half way between 8 and 10. Let's call this number "9." A little substitution: Quote:If 9 is half way between 8 and 10, then, 0.9 would be halfway between 0.8 and 1. So 0.99 would be half way between 0.88 and 1 and so on. So 0.999... would be half way between 0.888... and 1. 
 And maybe we should also imagine that math works completely differently than it does. jmlyle


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NickH
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Re: 0.999.
« Reply #20 on: Jul 29^{th}, 2002, 12:25am » 

"In my example, if # is half way between 9 and 10, then 0.# would be halfway between 0.9 and 1. So 0.## would be half way between 0.99 and 1 and so on. So 0.###... would be half way between 0.999... and 1" Ah, I think I see what you mean by .##. You mean .995. Similarly, .### = .9995. Now, it's true that for any finite number of #'s, .9999...9 < .###...# < 1. However, when we sum to infinity, both .999... and .###... are equal to 1.


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anshil
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Re: 0.999.
« Reply #21 on: Jul 29^{th}, 2002, 12:52am » 

<i>if 9 is half way between 8 and 10, then, 0.9 would be halfway between 0.8 and 1. So 0.99 would be half way between 0.88 and 1 and so on. So 0.999... would be half way between 0.888... and 1.</i> I don't want to disappoint you, but take out the calculator and find out what's halfway between 0.88 and 1. Well it's (1  0.8 / 2 + 0.88 right? Well that's 0.94, not 0.99 I guess that is the pretty end of the # theory Don't forget what the decimal system is actually about. It's just a shortcut for writing like 0.99 is 0 * 10 ^ 0 + 9 * 1 * (10 ^ 1) + 9 * (10 ^ 2) You can't introduce a new decimal but leaving the base equal, of course you can create a new numbering system, with the base of 11, so we need a new digit in this system to represent 10, let it be #. So 10 decimal would #. 11 would be "10". 0.## would be written in the new system: 0 * 10 ^ 0 + # * 10 ^ 1 + # * 10 ^  2. Translated to the decimal system: 0 * (11 ^ 0) + 10 * (11 ^  1) + 10 * (11 ^  2) And thats equal to: 0.991736 Well we only have generated a new calculation system, but that still didn't create a new number per se. So in our new system we still can say: 0.########......... = 1 It's importand to understand that both systems are valid, and well equal, but you can't mix them Like using the number representation # in the decimal system. There are systems with a base other than ten which have their very own sex. Like of course the binary system (base 2), used in every computer, beside the decimal (base 10) the most important system today. The tertinary (base 3) almost not used, but having it's own very interesting behaviours. The hexadecimal system (base 16) used for easy representation of binary numbers. And of course a very old system, the system with base 12, forgot a name for it. Guess it's even older than the decimal, since 12 is devidable by 3, 4 and 6. While ten is only devideable by 2 and 5. Still our years are devided in 12 months, our days in 12 * 2 hours and so on. But I'm babbeling already Actually math works on Paper. Sometimes you have to believe in what your calculations proof and show.


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Kozo Morimoto
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Re: 0.999.
« Reply #22 on: Jul 29^{th}, 2002, 6:40am » 

OK, nobody seems to be getting the '#' besides me so I'll drop it... How about this. 0.9 is pretty close to 1, an approximation of 1. You add 0.09 to make it 0.99 and its even closer to 1 than 0.9 but still not 1. Repeat to infinity. You get ever closer and closer to 1 from the low side, but you never get there? Using the ideas given above, does it mean that 1.000 ... 001 with infinite zeroes between the two 1s make it equal to 1? So Does it mean that 0.999... = 1.000...001 ? How about 0.999...998 = 0.999... = 1.000...001?


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anshil
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Re: 0.999.
« Reply #23 on: Jul 29^{th}, 2002, 7:15am » 

"OK, nobody seems to be getting the '#' besides me so I'll drop it... " Well thats a very arrogant argument. Maybe all the people but you do "get" it, what you mean, but it just makes mathematically not much sense. " 0.9 is pretty close to 1, an approximation of 1. You add 0.09 to make it 0.99 and its even closer to 1 than 0.9 but still not 1. Repeat to infinity. You get ever closer and closer to 1 from the low side, but you never get there? " There you are, infinity and "near to infinity". Yes you get closer ever step you do, and as you are approaching infinity you are approaching 1 that is true. However when if you hit infinity you hit 1 also. Using the ideas given above, does it mean that 1.000 ... 001 with infinite zeroes between the two 1s make it equal to 1? No not, it's inifnitly small larger than one. So Does it mean that 0.999... = 1.000...001 ? No. 0.999999... .... 9998 would be smaller than 1, okay? But 0.9999....9999 _is_ 1. Like 1.00000......00000 is also 1.


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S. Owen
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Re: 0.999.
« Reply #24 on: Jul 29^{th}, 2002, 11:33am » 

on Jul 29^{th}, 2002, 6:40am, Kozo Morimoto wrote: 0.9 is pretty close to 1, an approximation of 1. You add 0.09 to make it 0.99 and its even closer to 1 than 0.9 but still not 1. Repeat to infinity. You get ever closer and closer to 1 from the low side, but you never get there? 
 Well, you "never get there" in the sense that there is no finite number of steps of this loop that gest you to 1. But "0.999..." is probably best viewed here as an expression of where you are headed to: 1. on Jul 29^{th}, 2002, 6:40am, Kozo Morimoto wrote: Using the ideas given above, does it mean that 1.000 ... 001 with infinite zeroes between the two 1s make it equal to 1? So Does it mean that 0.999... = 1.000...001 ? How about 0.999...998 = 0.999... = 1.000...001? 
 What does "1.000...0001" mean? It doesn't make sense to talk about a terminating decimal with a 1 after an infinite number of zeroes. Likewise for "0.999...998".


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