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.99999 repeating = 1

Name: Anonymous 2007-02-18 15:23

x= .999 repeating
10x = 9.999 repeating
10x-x = 9.999 repeating - .999 repeating
9x= 9
x=1

Name: Anonymous 2007-02-18 15:42

No.

.9999 repeating = 1 - .0000infinityinhere1

Name: Anonymous 2007-02-18 15:43

>>2
No. You are wrong.

.9 repeating = 1, and there are at least three separate proofs for it.

Name: Anonymous 2007-02-18 15:58

>>3
Those proofs all use expoits in mathematics to trick idiots.

Name: Anonymous 2007-02-18 16:01

op here, i just found a fault in my proof, .9999 x 10 does not equal 9.999999 repeating.

Name: Anonymous 2007-02-18 16:31

>>5
good, let's never speak of this again

Name: Anonymous 2007-02-18 17:01

This thread delivers

Name: Anonymous 2007-02-18 17:06

.9999 ... = 1
there seems to be general consensus about this.

However i don't know if that proof in OP is valid.

Name: Anonymous 2007-02-18 17:39

So does this mean .7 repeating = .infinite 7s and then one 8?

Name: Anonymous 2007-02-18 18:04

Rather than saying "giving infinity a value," it's perhaps a bit
clearer to say, "giving the concept of a limit of an infinite sequence
of numbers a value."

.9 is not 1; neither is .999, nor .9999999999. In fact if you stop the
expansion of 9s at any finite point, the fraction you have (like .9999
= 9999/10000) is never equal to 1. But each time you add a 9, the
error is less. In fact, with each 9, the error is ten times smaller.

You can show (using calculus or other methods) that with a large
enough number of 9s in the expansion, you can get arbitrarily close to
1, and here's the key:

THERE IS NO OTHER NUMBER THAT THE SEQUENCE GETS ARBITRARILY CLOSE TO.

Thus, if you are going to assign a value to .9999... (going on
forever), the only sensible value is 1.

There is nothing special about .999...  The idea that 1/3 = .3333...
is the same. None of .3, .33, .333333, etc. is exactly equal to 1/3,
but with each 3 added, the fraction is closer than the previous
approximation. In addition, 1/3 is the ONLY number that the series
gets arbitrarily close to.

And it doesn't limit itself to single repeated decimals. When we say:

1/7 = .142857142857142857...

none of the finite parts of the decimal is equal to 1/7; it's just
that the more you add, the closer you get to 1/7, and in addition, 1/7
is the UNIQUE number that they all get closer to.

Finally, you can show for all such examples that doing the arithmetic
on the series produces "reasonable" results:

Since:

1/3 = .333333...
2/3 = .666666...

1/3 + 2/3 = .999999... = 1.

By the way, there is nothing special about 1 as being a non-unique
decimal expansion. Here are a couple of others:

2 = 1.9999...
3.71 = 3.709999999...
2.778 = 2.77799999999999...

...and the student who says you're trying to show that 1 = 1/infinity
is wrong.

Name: Anonymous 2007-02-18 18:04

>>9
no, it's just .7 repeating = 7/9. You get the 8 at the end when you round it at which it becomes not repeating forever.

Name: Anonymous 2007-02-18 18:37

>>4
http://qntm.org/pointnine
http://descmath.com/diag/nines.html
http://www.cut-the-knot.org/arithmetic/999999.shtml
http://en.wikipedia.org/wiki/0.999... (and before you cry foul, note that nearly every paragraph has a footnote, of which there are more than sixty. It's not unsourced and unverifiable.)

So, given multiple corroborating proofs from multiple sources, the burden of proof is on you. Where's the mathematical trickery?

Name: Anonymous 2007-02-18 23:14

Where's the trickery?  It was pencilled in the margins while the guy held it with his anus.

Name: Anonymous 2007-02-18 23:38

>>1
Here's a more eloquent form of the same proof.

http://www.anonib.com/mathchan/images/6/0.99999.jpeg

The mathchan is very inactive, so it makes a good image repository.

Name: Anonymous 2007-02-19 1:59

No one like that stupid number anyway, it can fuck off

Name: Anonymous 2007-02-19 6:02

>>2 - >>15
yhbt.

Name: Anonymous 2007-02-19 10:51

yhbt.

Iz dat simon lol

Name: Anonymous 2007-02-19 13:22

>>12
Theyre tricking you, dude. Its a math joke.

Name: Anonymous 2007-02-20 11:06

>>12
This is wrong. All those pages make the same fallacy: they're using infinity in regular mathematics. It doesn't work like that.

I suggest you read "Hotel Infinity": http://scidiv.bcc.ctc.edu/Math/InfiniteHotel.html
It may clear things up a bit.

Name: Anonymous 2007-02-20 12:30

>>19
OH jesus fuck you people are so god damn stupid

Name: Anonymous 2007-02-20 12:55

>>19
This is wrong.  You have no concept of infinity, please go back to school and learn Calculus.  It will clear things up a bit.

Name: Anonymous 2007-02-20 16:37 ID:mQs8UbXn

>>20
YHBT ;)

Name: Anonymous 2007-02-20 22:05 ID:Heaven

>>10

This is _the_ classic math troll.  Just let the children play.

Name: Anonymous 2007-02-20 22:12 ID:SPnDISQ2

>>19
Please explain how the concept of an infinite geometric series is misusing infinity.

And you can take your shitty hotel metaphors with you when you GTFO.

Name: Anonymous 2007-02-20 23:38 ID:TfbhBCm1

Infinity is not a real number.

Name: Anonymous 2007-02-21 3:24 ID:D4Pcez+l

.99999 repeating is one there are proof's of it and even a wiki

my proof 1 is the same as 3/3 correct?
and 1/3 = .33333333333
and if we ad another 1/3 to get 2/3 that =.6666666 correct
so if we ad 1/3 or .3333repeating  to 2/3 or .6666repeating  we get both 1 and .99999 repeating

Name: Anonymous 2007-02-21 3:29 ID:D4Pcez+l

you think your so smart don't you, you dirty fucks with all you "fancy" words like "Calculus" "Maths" and "2"

3/3 = 1

that is all the proof needed

Name: Anonymous 2007-02-21 4:51 ID:tl5dVrYu

3/3 = 1
1/3 = 0.333...∞
2/3 = 0.666...∞

1/3 + 2/3 = 0.333...∞ + 0.666...∞
0.333...∞ + 0.666...∞ = 0.999...∞ + ...∞
0.999...∞ + ...∞ = 1

0.999...∞ ≠ 1

QED. You guys fail to see that 0.999...∞ still misses the infinitely tiny fraction ...∞ to make it 1.

More evidence:

I have 0.999...∞, let me take off an infinitely small fraction ...∞

Now I have 0.999...∞ - ...∞ = 0.999...∞

As you can see, I have taken a number with an infinitely small fraction less than 1, took off an infinitely small fraction and still have the same number with an infinitely small fraction smaller than 1! Amazing, isn't it!

Like Hotel Infinity should make clear to you, you can't use infinity in regular arithmetic. 0.999...∞ repeating is a CONCEPT and the only thing you can derive from it is that it is smaller than 1. An infinitely small fraction smaller than 1, which is why we can only use it as a concept, and not in mathematical formulas.

Name: Anonymous 2007-02-21 7:31 ID:JV3T/sTk

>>28
You are wrong.

Name: Anonymous 2007-02-21 7:51 ID:Heaven

>>29
What an amazing argument you got there.

Name: Anonymous 2007-02-21 14:04 ID:IOInmI9A

>>30
Why should >>29 bother posting an argument? People have already presented dozens of arguments (which, unlike yours, are flawless and actually written by someone who has finished junior high) and you ignored all of them.

Name: Anonymous 2007-02-21 14:34 ID:tl5dVrYu

>>31
Flawless how? I pointed out exactly what flaw was made and provided proof for this reasoning. You disprove it, if you can. If this simply is not enough to explain things to you, I don't think you are capable of understanding the concept matter of infinity.

The problem is you are trying to put a finite point in your mind to an infinitely tiny fraction, which makes it not infinitely tiny anymore. As I explained, infinity is a concept, and if you don't understand that concept well you will make mistakes like "0.999... = 1"

Name: Anonymous 2007-02-21 15:30 ID:RNtSRkWM

Let us never speak of this horrible number again

Name: Anonymous 2007-02-21 15:50 ID:Heaven

>>32
You didn't point out a flaw, you pointed out that you have ABSOLUTELY NO FORMAL TRAINING IN MATH.

Name: Anonymous 2007-02-21 15:56 ID:yXM2Rzjs

>>No sir, it is you who does not understand the concept of infinity.

>Now I have 0.999...∞ - ...∞ = 0.999...∞
That is a flawed argument. You can't take sigma(9/10^n) from {n = 0 to ∞} and subtract 9/10^∞. hell, 9/10^∞ isn't even a number. What you can do though is take the limit as n approaches infinity of 9/10^n and subtract that.

sigma(9/10^n) from {n = 0 to ∞} - limit(9/10^n) as {n approaches ∞}

Unfortunately for you the limit as n approaches infinity of 9/10^n is ZERO, ZEROZEROZEROZEROZEROZEROZEROZEROZEROZERO!
In Effect:

sigma(9/10^n) from {n = 0 to ∞} - 0 = sigma(9/10^n) from {n = 0 to ∞}

Nice try dumbshit.

Name: Anonymous 2007-02-21 17:10 ID:C+qsu6k9

people should seriously start sdudying maths, calculus and analisys before demonstrating shitty things

Name: Anonymous 2007-02-21 17:40 ID:5QdPT5h7

okay
it's like with the graph of a limit
it touches one, but it isn't one. it's so close it doesn't matter and it's not worth arguing about. If you physically produced something with that accuracy, chances are it would get quantum particles touching or something, and theyd jump around and switch places as particles will do, and it would be the same fucking thing anyhow.

Name: Anonymous 2007-02-21 18:10 ID:Heaven

>>37
It's not "close to one", it IS 1. End of story. If you don't like it, go post in a fucking philosophy board you useless asshole. The real numbers are defined as the set of limits of cauchy sequences of rational numbers, and the cauchy sequence of rational numbers {0.9, 0.99, 0.999, ... } converges both to 0.999... and 1. Hence, they are equal.

Name: Anonymous 2007-02-21 18:42 ID:xo6mgDQP

>>28
Your math is so fucking broken retard.

1/3 + 2/3 = 0.333...∞ + 0.666...∞
0.333...∞ + 0.666...∞ = 0.999...∞ + ...∞

thats just plain wrong.. it clearly is 0.999.....
3+6 = 9
this repeats infinitely.. its not even hard to understand. Youre talking as if 0.333.. repeating infinitely is 0.333... + some fraction.
You have totally misunderstood infinity and math, even though you claim that youre the one who claims to understand it. Just gtfo.


1 - 0.99... = 0.000000000000...
1 = 0.99...

Name: Anonymous 2007-02-21 19:25 ID:xvBnJwcw

Does 0.000... = 0? 1 - 0.999... = 0.000..., and since 0.999... ≠ 1, 0.000... must be infinitesimally greater than 0. Probably represented as 0.000...1.

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