>>2
That's a horrid proof, but yes, as everyone else said, it does.
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Anonymous2005-11-16 10:48
IT DOESN'T NEED PROOF. IT'S LIKE THAT BY DEFINITION YOU FUCKING IDIOTS.
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Anonymous2005-11-16 12:40
if it's like that by definition then it doesn't have a proof
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Anonymous2005-11-16 15:20
>>8
Nothing in mathematics is like something by definition. Everything in mathematics can be proven (or disproven).
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Anonymous2005-11-16 16:11
>>10
no. the + function is defined. sqrt, *, / ... the list goes on. you have to define some things to contruct other things and prove other things assuming the definitions.
key word is assume here. ex: if we assume the derivative of ln(x) = 1/x, then we can prove that the derivative of e^x is e^x.
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Anonymous2005-11-16 17:35
Actually, that would be "if we assume the derivative of f(x), f'(x), is the slope of f(x) at every value x, then the derivative of e^x is e^x."
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Anonymous2005-11-16 18:58
Saying that .9~=1 is like saying that 9=10. You're ignoring basics of math. You can't distort equations to prove your ignorance.
Different values cannot be equal! .9~ and 1 are different. As the linked forum topic says: Infinity DOES NOT EXIST- it's just an abstract word.
Therefore, if infinity does not exist, .9~ actually means .9..9 And that isn't one. That's .9...9. It's a matter of believing and infinity and not. You're probably just a bunch of Christian fundamentalist engineers with too much time on your hands who want to distort your own knowledge to prove God.
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Anonymous2005-11-16 19:42
Infinity does exist; it's just a concept. Similarly, thoughts exist, but not physically.
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Anonymous2005-11-16 19:44
.9(...)9 is equal to 1 in a "real life" sense, even though .9(...)9 is actually infintesimally smaller than 1.
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Anonymous2005-11-16 19:49
For all purposed .999...9 might as well be 1. So it should be defined to be 1. Unless it's inconvenient lol.
if you believe that every subset of the natural numbers has a least element, then you are forced to discard the concept of an infinintesimally small number.
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Anonymous2005-11-16 22:11
maybe you could think of 1-.9999... as 1/infinity or lim(x->inf, 1/x)
You've made us all stupider. Read the massive quantities of explanation linked to by >>6 before sharing any further idiocy. .9 repeating is not, and never will be, .9..9.
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Anonymous2005-11-16 23:38
the thread in >>6 is unreadable. what's with all the [smiley.gif] crap? someone should link to a nice readable proof or some math fag saying it's defined or something.
Your second assertion, that .9 repeating and 1 are different simply because they look different, is patently silly. A single number has an infinite number of representations. 2/6 is the same number as 1/3, even though the look different. .9 repeating is simply another name for 1. The numbers exist independent of our representation of them.
the notation is part of a mental effort of >>15 to cling to the concept of the infinintesimal. What is implied is that you could add 0.0000....00001 (infinitely many zeroes) to .9(...)9 in order to get one. What is further implied, or desired, is that performing this particular sum is somehow distinct from adding 0 to 1, when in fact it is not.
The interesting thing is that such a number as .9(...)9, if it even exists, doesn't belong to the set of reals. While .9... (where .9... is .9 repeating) most certainly does. >>15,>>25 should read >>6's link more thoroughly. This is discussed there in great detail.
>>15,>>25
There is no "infinitesimal" difference between .9... and 1. Check the linked post for Icarus' Misconception 8 (That "There is a least number greater than or greatest number less than a given real number.") Your view seems to be centered on this misconception. 0.9... is not, and cannot be, the "greatest number that's less than" 1.
it's real. why wouldn't it be real? just as .9, .99, .999 are improving approximations of .9..., .99, .999, .9999, are improving approximations of .9(...)9. In fact, you could write .9(...)169328472169894823 where ... represents another infinite string of nines, and IT ALL EQUALS ONE.
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dv2005-11-17 9:22
OMG YOU SHITS
1) REAL NUMBERS ARE THE CLOSURE (SEE BASIC TOPOLOGY) OF RATIONAL NUMBERS.
2) ALL CAUCHY SEQUENCES IN THE REAL NUMBERS CONVERGE TO A _UNIQUE_ REAL NUMBER
3) THE SEQUENCES 1,0,0,...,0 and 0,9,9,...,9 CONVERGE TO THE SAME FUCKING REAL NUMBER (LIM(N->INF, 1) = 1, AND LIM(N->INF, 1-(1/10^N)) = 1)
HENCE THEY ARE THE SAME FUCKING NUMBER
QUOD ERAT DEMONSTRANDUM
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dv2005-11-17 9:27
LOL OOPS. I MEANT COMPLETION NOT CLOSURE. I SUCK COCK
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Anonymous2005-11-17 9:32
ERRATA #2 (whoa dude, this is turning out like a Rudin book): I MEANT THE SEQUENCES OF APPROXIMATIONS 1.0, 1.00, 1.000, ... AND 0.9, 0.99, 0.999, ... CONVERGE TO THE SAME REAL NUMBER
doh'
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Anonymous2005-11-17 10:52
>>30
Yes there is, .9 repeating is infintesimally smaller than 1. However, for every day use it can equal 1 because, well, no one really uses infintesimals in their day-to-day lives outside of work.
Shut up until you understand decimal representations of real numbers, because currently you clearly don't. Refer to Misconceptions 1 ("Infinite Decimals are Approximations"), 7 ("There are numbers without decimal representations"), and 8 ("There is a least number greater than or greatest number less than a given real number") in >>6's link because you seem to be falling afoul of them with your statement of decimals being "inaccurate".
And consider yourself bitchslapped per >>36's request.
READ THIS AND BE PERSUADED OF IT FOR IT IS THE GOSPEL TRUTH OF MATHEMATICS, AS FAR AS THE REAL NUMBERS ARE CONCERNED.
You wish to describe an infinitesimal number. Call it 1/infinity, or 0.000...0001, or whatever else. You wish to maintain that it is less than any other strictly positive real number, but more importantly, you wish to maintain that it is nonzero. call it x. Then the infinitesimal x is greater than 0. You might like to say that it is a real number, or something else. At any rate, if we show that it's merely equal to zero, we discard your claim. Suppose I write 1/(2*infinity). you would rightly claim that this is equal to your 1/infinity, while attempting to maintain that neither is equal to zero. But what is (1/(2*infinity))? It is just (1/2)(1/infinity). And there's only one real value I know of that's half of itself, and that's zero. This is not rigorous, but it is appropriate to your level of contemplation of the problem. It is also legitimate, since we obeyed regular arithmetic throughout.
THERE IS NO SUCH THING AS A REAL NUMBER WHICH IS LESS THAN EVERY POSITIVE REAL, YET GREATER THAN ZERO.
IN EXACTLY THE SAME SENSE THAT THERE IS NO INFINITELY LARGE POSITIVE REAL NUMBER, SO TOO IS THERE NO INFINITELY SMALL POSITIVE REAL NUMBER.
Let's beat this into your head some more, with respect to the original question.
x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.
While not RIGOROUS, this proof works, since it's not like the fallacious 1=2 proof where one divides by zero (zing). All that it does is to show that the thing we mean by .999... and by 1 have the same numerical value. 1/3 + 1/3 + 1/3 = .333... +.333... +.333... = .999... =1. Precisely 1. Nothing other than 1.
INFINITESIMALS ARE RIGOROUSLY BANISHED FROM THE REAL NUMBER SYSTEM.
Some history! Newton and Leibniz used infinitesimal quantities and came up with appropriate results. The rigorous difficulty arose in the fact that they divided by a quantity which was pretended to be nonzero, and later discarded like it was zero. It took the foundations of limits and functions much later to give calculus a firm(er) base. Have a mathworld link while we're at it: http://mathworld.wolfram.com/Infinitesimal.html
Infinitesimal: "A quantity which yields 0 after some limiting process"
'Infinitesimals are legitimate quantities in the non-standard analysis of Abraham Robinson.'-Wikipedia. Moreover, they were necessary imaginitive tools in the initial development of the calculus. The point of which you will now be convinced is this: .999... = 1 and your infinitesimal turns out to be zero when you try to do anything with it as a number in and of itself. If you want to comment on an non-archimedian field, you are free to read up on it and do so. But your object has no place in the real number system. .999... is in every possible sense, 1, and that's the truth. The object you are referring to has no reality in the only number system on which you are prepared to comment on it.
You have no idea what you're talking about. You're asserting with neither proof nor understanding. And you're being willfully ignorant out of pure stubborness. Mathematics at this basic, fundamental level never take into account "realistically" versus "theoretically". Realistically and theoretically, .9... is identically equal to 1. Just as .3... is not an approximation, but rather is exactly 1/3, .9... is not an approximation, but rather exactly 3/3.
First of all, 1/3 isn't "modelled" as 0.3..., 1/3 *is* 0.3...: they're two different names for exactly the same thing. I assume you understand that, though, and simply made a poor choice of words.
I have no idea what you're talking about wrt "what if the number [what number?] is not concerning 1/3?"
And what does any of this have to do with irrational numbers?
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Anonymous2005-11-19 13:00
1/3 is not 0.3...
There are real numbers and rational numbers. Rational numbers are embedded in real numbers.
1/3 is rational, meaning "the ratio of 1 to 3."
It is represented as 0.3... when in decimal form. Decimals are also called decimal fractions. Decimals are base 10, and are a ratio based on a power of 10.
1/3 is not based on a power of ten.
0.3... can only be infinitely "close" to 1/3.
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Styrofoam!DWDMFPPpRw2005-11-19 13:59
GOD DAMN YOU ALL ARE FUCKING STUPID
WHY DOES THIS THREAD REAPPEAR EVERY FEW DAYS
.999~ = 1 EXACTLY
PROOF:
let x = .999~ (where ~ indicates repeating forever)
multiply each side by 10
10x = 9.999~
subtract x from each side
9x = 9 (because x = .999~, so we subtracted .999~ from 9.999~)
divide both sides by 9
x = 1
and we started by saying x = .999~, so...
.999~ = 1
NOW SHUT THE FUCK UP ABOUT IT
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Anonymous2005-11-19 17:52
How would subtracting .999~ from 10 equal 9 and not 9.111~ ?
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Anonymous2005-11-19 18:49
>>48
Math.
.9 repeating ~= .999 (because you and >>47 obviously like threes)
10 - .999 = 9.001
9.001 != 9.111
1 = definately 0.999...
when you're in a store and you order 0.999...pound of meat and the sailsman says "might it be a little bit more" you say "NO IT HAS TO BE 0.9999... OR I'LL CUT OFF YOUR BALLS" because you are an obsessive little bastard and you gotta have 0.999... AND NOT 1.
now does this story sound rite to you? i didnt think so...
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Anonymous2005-11-20 13:36
HEY FUCKASS THAT'S NOT A VERY GOOD ANALOGY: YOU CAN'T SAY "HEY I WANT PI POUNDS OF MEAT" OR "I WANT THE SQUARE ROOT OF TEN POUNDS OF MEAT" EITHER
You are all wrong. Read the god damn proofs written earlier in the thread. It's a mathematical proof using only algebra. Infinities don't enter in to it.
You really shouldn't talk about infinity when you don't understand its most fundamental concept. There is no "last" 9 in an infinite string of digits which will result in the magical transfinite ...01.
1 - 0.999~ = 0.000~ = 0
Also...
Infinite strings of digits, such as 0.333~, are not approximations of the rational notation, such as 1/3. They are exactly equivalent and describe exactly the same number. 0.333~ * 3 does equal 1, and does equal 0.999~, because 1 = 0.999~.
test
[tex]\prod_{n}^{1}\cdot cx^{n-d}[/tex]
\prod_{n}^{1}\cdot cx^{n-d}
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Anonymous2010-11-12 14:22
go away.
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Anonymous2010-11-13 6:45
I think you can all agree that:
0.999... = 0.9 + 0.09 + 0.009 + ... = the sum from i=1 to n of (9*10^(-i))
Let me propose that: sum from i=1 to n of (9*10^(-i)) = 1 - 10^(-n) for all natural n
Show for n=1:
sum from i=1 to 1 of (9*10^(-i)) = 9*10^(-1) = 9/10 =
1 - 10^(-1) = 1 - 1/10 = 9/10
Assume true for n=k, show for k+1 etc:
sum from i=1 to (k+1) of (9*10^(-i))
= ( sum from i=1 to k of (9*10^(-i)) ) + 9*10^(-(k+1))
= 1-10^(-k) + 9*10^(-(k+1))
= 1-10^(-k) + (10-1)*10^(-(k+1))
= 1-10^(-k) + 10^(-k) - 10^(-(k+1))
= 1-10^(-(k+1))
Proved that sum from i=1 to n of (9*10^(-i)) = 1-10^(-n)
Now consider this when n -> infinity
lim (1-10^(-n)) as n -> infinity
= lim (1-1/10^n) as n -> infinity
= 1 - lim(1/10^n) as n -> infinity
= 1 - 0
= fucking 1
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Anonymous2010-11-13 16:43
You can split a circle into 3 perfectly even parts but you can't divide the number 1, 10, or 100 by 3. Interesting.