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Topic closed

2006-11-29 15:45:45
Onionpaste
Member
+6|6656
Dunno if this has already been covered, if so, sorry for the spam, not intended.

So:

Let's call .99999~ (that's a decimal followed by an infinite number of 9's) a variable x.

So that means that x = .9999~

So by that logic, 10x = 9.9999~

If you subtract x from 10x, you get 9.00000~, as in the equation 10x - x = 9.00000~

This means 9.0000~, or just 9 (I put the 0's there to show that all the 9's after decimal went away), equals 9x.
9x = 9

Divide each side by 9, and you get x = 1.

That means that x is equal to .99999~ and 1 at the same time.  Therefore, .99999~ = 1!

If any math majors have found some sort of error in this reasoning, do not hesitate to reply so I can debate it with you

-Onion
2006-11-29 15:49:27
maffiaw
ph33r me 傻逼
+40|6694|Melbourne, AUS
algebra seems ok. of course .99 recurring would be so close to 1 that it would be indiscernible
2006-11-29 15:51:00
Kmar
Truth is my Bitch
+5,695|6874|132 and Bush

The number "0.9999..." can be "expanded" as:

      0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ..
Xbone Stormsurgezz
2006-11-29 15:52:04
BigmacK
Back from the Dead.
+628|7024|Chicago.
I see no major flaws in your reasoning. But personally, I prefer rounding.
2006-11-29 15:52:56
Bertster7
Confused Pothead
+1,101|6854|SE London

Isn't it to do with the fact that you can't represent the number small enough to be the difference between 0.9999999999 recuring and 1?

I'm no maths expert, I did study it at uni for a while as part of my engineering course though.
2006-11-29 15:53:14
RedTwizzler
I do it for the lulz.
+124|6810|Chicago
That is the error in the decimal system (Base 10). In Base 8, something similar happens with 3.9999~ and 4, I think.

Either way, everyone has A) Seen it before, and not cared, or B) Never seen it before, and doesn't care.
2006-11-29 15:53:23
Yaocelotl
:D
+221|6923|Keyboard
No, when you substract in one side you must perform the same in the other side as well. Leaving you:

10x - x = 9.99999~ - x

Although it throws the same result you have to do the things correctly. This probes why the decimal point values from < .0000001 are null in everyday problems. But this only affects in matters of precision calculations.

Last edited by Yaocelotl (2006-11-29 16:01:00)

2006-11-29 15:54:08
Kmar
Truth is my Bitch
+5,695|6874|132 and Bush

Bertster7 wrote:

Isn't it to do with the fact that you can't represent the number small enough to be the difference between 0.9999999999 recuring and 1?

.
Close

A common objection is that, while 0.999... "gets arbitrarily close" to 1, it is never actually equal to 1. But what is meant by "gets arbitrarily close"? It's not like the number is moving at all; it is what it is, and it just sits there, looking at you. It doesn't "come" or "go" or "move" or "get close" to anything.

Last edited by Kmarion (2006-11-29 15:55:34)

Xbone Stormsurgezz
2006-11-29 15:54:20
Gen. Payne
Member
+50|6980|USA
Now that's interesting!

Yaocelotl wrote:

No, when you substract in one side you must perform the same in the other side as well. Leaving you:

10x - x = 9.00000~ - x

Although it throws the same result you have to do the things correctly. This probes why the decimal point values from < .0000001 are null in everyday problems. But this only affects in matters of precision calculations.


--------------------------------------------------------------------------------
No, it's 10x - x = 9.99999~ - x

Last edited by Gen. Payne (2006-11-29 15:57:26)

2006-11-29 15:55:53
Bertster7
Confused Pothead
+1,101|6854|SE London

Yaocelotl wrote:

No, when you substract in one side you must perform the same in the other side as well. Leaving you:

10x - x = 9.00000~ - x

Although it throws the same result you have to do the things correctly. This probes why the decimal point values from < .0000001 are null in everyday problems. But this only affects in matters of precision calculations.
But you never get (*edit* practical) calculations using infinite length figures (except pi, which is rounded).

Last edited by Bertster7 (2006-11-29 15:56:19)

2006-11-29 15:57:42
Kmar
Truth is my Bitch
+5,695|6874|132 and Bush

http://www.purplemath.com/modules/howcan1.htm
Xbone Stormsurgezz
2006-11-29 15:59:23
Gen. Payne
Member
+50|6980|USA
I remember there was something out there that could prove 1=2. I gotta try to find that.

Found it (it is, in fact, false; try to find the error, you may get karma)

                 a = b
               ab = b²
          - (ab) = - (b²)
          a²-ab = a² - b²
         a(a-b) = (a+b) (a-b)
                a = a+b
                a = 2a
                1 = 2

Last edited by Gen. Payne (2006-11-29 16:16:02)

2006-11-29 16:00:32
deadawakeing
Ummmmmmmmmmmmm
+145|6755

Onionpaste wrote:

That means that x is equal to .99999~ and 1 at the same time.  Therefore, .99999~ = 1!
I think 10x - x= 1

Last edited by deadawakeing (2006-11-29 16:01:27)

2006-11-29 16:00:55
Onionpaste
Member
+6|6656

Yaocelotl wrote:

No, when you substract in one side you must perform the same in the other side as well. Leaving you:

10x - x = 9.00000~ - x

Although it throws the same result you have to do the things correctly. This probes why the decimal point values from < .0000001 are null in everyday problems. But this only affects in matters of precision calculations.
Ok, so in that case, if I subtract x from both sides, that's 10x - x = 9.9999~ - x.  9x = 9.000, because we substitute the right-side x for it's value, or .999~.  Go from there.
2006-11-29 16:02:48
Bertster7
Confused Pothead
+1,101|6854|SE London

deadawakeing wrote:

Onionpaste wrote:

That means that x is equal to .99999~ and 1 at the same time.  Therefore, .99999~ = 1!
I think 10x - x= 1
You think wrong then.

That would equal 9x.

Last edited by Bertster7 (2006-11-29 16:03:53)

2006-11-29 16:05:40
jonsimon
Member
+224|6768
1-.9~ equals an infinitesimally small difference.
2006-11-29 16:06:36
Yaocelotl
:D
+221|6923|Keyboard

Onionpaste wrote:

Yaocelotl wrote:

No, when you substract in one side you must perform the same in the other side as well. Leaving you:

10x - x = 9.00000~ - x

Although it throws the same result you have to do the things correctly. This probes why the decimal point values from < .0000001 are null in everyday problems. But this only affects in matters of precision calculations.
Ok, so in that case, if I subtract x from both sides, that's 10x - x = 9.9999~ - x.  9x = 9.000, because we substitute the right-side x for it's value, or .999~.  Go from there.
Ok, in this case you have (not by my proposed solution to it) that:

9x = 9.99999~

We can see that x = 9.99999~ / 9

Throwing x = 1.11111~
2006-11-29 16:07:36
blakrobe
Member
+3|6803
Just because you get the same results in the same equations with two different numbers doesn't make them equal to each other. Look at it this way...

1 - x = 0.0000~0001
1 - 1 = 0
x - 1 = -0.000000~0001

Now lets say:

a = 1 - 1
b = 1 - x

a x infinity = 0
b x infinity = infinity

Massive difference in the numbers now, seemed so insignificant before though.
2006-11-29 16:07:43
Yaocelotl
:D
+221|6923|Keyboard

Bertster7 wrote:

deadawakeing wrote:

Onionpaste wrote:

That means that x is equal to .99999~ and 1 at the same time.  Therefore, .99999~ = 1!
I think 10x - x= 1
You think wrong then.

That would equal 9x.
Furthermore, x = 1/9
2006-11-29 16:08:26
Des.Kmal
Member
+917|6891|Atlanta, Georgia, USA
no, 1 = 1

i dont care what you say. .99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999~ = .99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999~ NOT 1

ur a dumbfuck if u think thats true. i mean cmon, even BASIC math skills say that every number is a number, their own number, nothing else.

1 = 1

.999999~ = .999999~
Add me on Origin for Battlefield 4 fun: DesKmal
2006-11-29 16:10:22
Fenris_GreyClaw
Real Хорошо
+826|6792|Adelaide, South Australia

Des.Kmal wrote:

no, 1 = 1

i dont care what you say. .99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999~ = .99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999~ NOT 1

ur a dumbfuck if u think thats true. i mean cmon, even BASIC math skills say that every number is a number, their own number, nothing else.

1 = 1

.999999~ = .999999~
/win.

thread done.
2006-11-29 16:10:24
Yaocelotl
:D
+221|6923|Keyboard

Des.Kmal wrote:

no, 1 = 1

i dont care what you say. .99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999~ = .99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999~ NOT 1

ur a dumbfuck if u think thats true. i mean cmon, even BASIC math skills say that every number is a number, their own number, nothing else.

1 = 1

.999999~ = .999999~
But the thing here is that 1 != 0.99999~
2006-11-29 16:12:24
{BMF}*Frank_The_Tank
U.S. > Iran
+497|6851|Florida

Onionpaste wrote:

If any math majors have found some sort of error in this reasoning, do not hesitate to reply so I can debate it with you
HA HA

https://i71.photobucket.com/albums/i143/christurner85/lookwhaticando.jpg
2006-11-29 16:12:40
-=raska=-
Canada's French Frog
+123|6899|Quebec city, Canada
kmal come on

there are lot of proofs that state that 0.99~ = 1. There is a big article about that on wiki and lots of proofs

0,99999 = 1 because the 0.0001 is infinitely small, therefore its like :

    lim            0.0   ...       1 = 0
n -> infinity      |n times|

also 1/3 = 0.33...
1/3 x 3 = 1
0.33... x 3 = 0.999... = 1

Last edited by -=raska=- (2006-11-29 16:19:52)

2006-11-29 16:15:23
-=raska=-
Canada's French Frog
+123|6899|Quebec city, Canada

Fenris_GreyClaw wrote:

thread done.
no thread is not done, thread is right.

if not, prove that the statements (verified by mathematicians) on this site are false. http://en.wikipedia.org/wiki/0.99

Last edited by -=raska=- (2006-11-29 16:15:51)

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