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0.999... = 1

Name: Anonymous 2007-11-17 12:16

Oh snap!

Name: Anonymous 2007-11-17 13:25

troll

Name: Anonymous 2007-11-17 13:40

>>2
Aren't we all trolls on this forum?

Name: Anonymous 2007-11-17 14:06

fapfapfapfapfap

come on baby, another 100 posts!

Name: Anonymous 2007-11-17 14:08

>>4
Why sage?

Name: Anonymous 2007-11-17 14:21

x= 0.9999
10x = 9.9999
10x - x = 9.9999 - 0.9999
9x = 9
x=1
way old.

Name: Anonymous 2007-11-17 14:27

>>6
This guy tells the truth.

Name: Anonymous 2007-11-17 15:51

That's enough of that.

Name: Anonymous 2007-11-17 23:33

>>1
you cunt

Name: Anonymous 2007-11-17 23:35

>>6
circular reasoning, see the 6 previous threads on the topic

Name: Anonymous 2007-11-18 0:33

>>5
fine, so sage this time
fapfapfapfapfap

Name: sage 2007-11-18 0:56

sequence can be shown to converge to 1
this has been beaten to death a dozen times already

Name: Anonymous 2007-11-18 10:39

>>10
Nothing circular about it. The fact that you don't understand 0.(9) * 10 = 9.(9) doesn't mean it's not true.

Name: Anonymous 2007-11-18 15:16

>>10
Actually, what >>6 said is pure, airtight proof.  It establishes a definition of x being the same as .999... and going from there with only valid arithmetic and algebraic operations.

Name: Anonymous 2007-11-18 18:24

>>13
>>14
Incorrect morons.  To accept the procedure, you must first accept that the numbers are the respective implied limits in the first place.  If you don't take that as an assumption, you aren't able to ever complete the arithmetic operations necessary to solve the equation.  Ergo, you are attempting to prove that the expression 0.9999... is the limit 1 by first assuming that 0.9999... is the limit 1.  Hence, circular reasoning.

Name: Anonymous 2007-11-18 18:34

>>12
what does that mean to you, "converge"

Name: Anonymous 2007-11-18 18:36

>>15
Your post is nonsense, I hope you realise this.

Name: Anonymous 2007-11-18 18:38

>>17
Learn to read, jackass

Name: Anonymous 2007-11-18 18:58

>>18
Learn math, jackass.

Name: RedCream 2007-11-18 20:57

0.999... expressed as a limit, converges.  The result of the convergence is 1.

All else is jackoffery.

Name: Anonymous 2007-11-18 21:04

>>20
All numbers are limits. You're embarrassing yourself more than usual.

Name: Anonymous 2007-11-18 22:18

>>15
i don't understand

Name: Anonymous 2007-11-18 22:41

>>16
The basic definition of convergence of a sequence over a metric space (in this case, R with d(x,y)=|x-y| is given by:

{x_n} is said to converge to x iff for every epsilon > 0, there exists an natural number N such that for all n > N, d(x, x_n) < epsilon.

Basically you take x_1 = 0.9, x_n = x_(n-1) + 9/10^n as your sequence, which clearly becomes 0.999... as n approaches infinity and show that it converges to 1.

Name: RedCream 2007-11-18 22:50

>>21
Only the former is true.  0.999... expressed as the usual limit is 1.  You made no point whatsoever except to affirm the premise by which I made my conclusion.

What's the embarrassment in that?  In fact, the embarrassment here should be YOURS, since you failed epicly in pointing out the flaw in this chain of reasoning:

1. 0.999... has a limit.
2. That limit converges at infinity.
3. That convergence value is 1.

Name: Anonymous 2007-11-18 23:16

>>19
Stop trolling, you can't read can you
>>24
he unfortunately isn't claiming that 0.999... has a limit, he is claiming that 0.999... is it's own limit, ie circular reasoning, therefore no need to prove that 0.999... is 1 because he simply defined that it is 1 in the first place.
>>23
you never once said what the word "converge" means in all of that.

Name: Anonymous 2007-11-18 23:35

>>25
Although I gave you the definition of convergence of a sequence, I guess you want the more intuitive explanation of what convergence means. You beautiful layman, you.

Essentially, as n->inf the amount that x_n may differ from x approaches 0. You could think of this as the diameter of the set of points x_n from some n going to infinity monotonically decreasing as n increases. If the diameter of a set is 0, then it is a set of one element (in this case, the value the sequence converges to).

Name: Anonymous 2007-11-19 1:43

>>25
Oh, wow. Mathtards in my /sci/. 0.999... doesn't have a limit because it's not a function or a sequence. It's a goddamn number. A number cannot be used to express a sequence of values. It's a value. One value.

Name: Anonymous 2007-11-19 1:52

>>24
Numbers don't HAVE limits, numbers ARE limits. Of Cauchy sequences of rational numbers.

>>25
It's not circular reasoning, it is the DEFINITION OF THE REAL NUMBERS.

Name: Anonymous 2007-11-19 11:36

>>27

"0.999..." is not a number.  It is a numeral, a written expression that represents a number.

The "..." notation in this numeral means nothing more nor less than that you are taking a limit.  It is an abbreviation for a more complicated expression involving an explicit limit.

You fail.

Utterly.


Name: Anonymous 2007-11-19 13:28

>>29
The "..." notation in this numeral means nothing more nor less than that you are taking a limit.
No, it's not. It's a shorthand notation for an infinite number of 9s, just like 1 is a shorthand notation for "1." followed by an infinite number of 0s.
Since you're the first to mention the difference between numerals and numbers in this thread, it's pretty ironic that you'd get it that wrong.

Name: Anonymous 2007-11-19 14:16

>>29
You can't just make up your own system of mathematics then pretend like it's the one we've been using all along. Get with the program or get out.

Name: Anonymous 2007-11-19 15:06

>>14 here.

>>15
The initial assumption in >>6 isn't that .999...=1.  It's that x = .999 (which we are allowed to define as true).  From there, we use operations that keep the original equation intact, until we come to 9x = 9.  We divide both sides by 9, and viola!  .999...=x=9/9=1.  lrn2math, fool.

Name: Anonymous 2007-11-19 17:43

>>6 here

guys i really didn't want to mess things up. I just revised it again for uni short time ago so i was able to prove it.

As far as i know that proof is mathematically right and logically too.

Just think if it's wrong, by contradiction you should prove that it exists a number between .(9) and 1. Do it, then I'll say i'm wrong. No problem.

in b4 (.(9) + 1) / 2  :)

Name: Anonymous 2007-11-19 18:05

In fact if they are different then you can write a terminating decimal that lies between them.

>>30
And what is the meaning of the written symbol 0.9999999 (with an infinite number of 9s)?  Oh, that's right, a limit!

Name: Anonymous 2007-11-19 18:43

>>33
How about you prove there's no number between .(9) and 1 instead of shifting the burden of proof.

Name: Anonymous 2007-11-19 18:57

The fact is you need to define arithmetic on recurring decimals, before you can use elementary operations on them, and show that it is a consistent model.

The way you do that is in the construction of the real numbers, and thus it's pointless to use that sort of reasoning, because if you bother to spend the time proving that your proof is valid, you essentially prove the equality of 0.999.... and 1 along the way

Name: Anonymous 2007-11-19 19:23

>>34
Right, and "1" is shorthand for "1." followed by infinitely many 0's, which is a limit. EVERY real number is a limit, because that is how they are DEFINED.

>>35
It's been proven countless times. Or rather, it has been proven that 0.(9) = 1, which in turn implies that there is no number strictly between 0.(9) and 1.

Name: Anonymous 2007-11-19 19:59

FAPFAPFAPFAPFAPFAPFAP


>>35
by the way, you fail furiously

Name: Anonymous 2007-11-19 21:00

>>37
If that how you want to define your "limits", then they have no bearing whatsoever on this discussion. GTFO.

Name: Anonymous 2007-11-19 23:07

>>39
No, moron, that is not how I define my limits. That is how I define the real numbers. This has been explained before in the thread.
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