Is 1 = 0.9999......

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silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: Kyteland
Originally posted by: thraashman

Actually you did violate something. 9.9999..... - 0.99999...... is technically undefined. As you cannot do set mathematical actions with infinites. Infinity minus infinity is not zero, it's still infinity. So you cannot perform the mathematical action of 9.9999...... - 0.9999.......

So 0.99999...... != 1, though they are logically equivalent, they are not equal.


And please let this thread DIE!!!!!!!!!!

I will now prove that you can do set mathematical actions with infinites using proof by contradiction:

Assume that you cannot do set mathematical actions with infinites
0.666..... is an infinite number (2/3)
0.166..... is an infinite number (1/6)

0.666... - 0.166... = .5

Since subtraction is a set mathematical action and I performed it successfully with infinites the hypothesis has been contradicted proving that you can do set mathematical actions with infinites.



And if you want the thread to die why don't you try NOT POSTING IN IT!!!!!!!!!!one

Furthermore, 1 = 1.0000000... and 2 = 2.00000.... and 3 = 3.00000.....

1 + 2 = 1.0000... + 2.0000.... = 3.0000... = 3

All numbers are really infinite strings, we just don't bother writing the zeros that don't affect the value.
 

TuxDave

Lifer
Oct 8, 2002
10,571
3
71
Originally posted by: OIKOS
is 1.0000000000000000000000000000000000000000000000000000000001 = 1?

No.... I win, do I get a cookie?

My question:

Is 0.00000.........0001 a number that exists?
 

dighn

Lifer
Aug 12, 2001
22,820
4
81
Originally posted by: OIKOS
is 1.0000000000000000000000000000000000000000000000000000000001 = 1?

of course not.

same reason that 0.9999999999999999999999999999999999999999999999999 is not 1, but 0.999... is
 

naddicott

Senior member
Jul 3, 2002
793
0
76
OK, OK, ATOT majority rule isn't good enough for you?

Anyone who is playing teacher's pet with a Math prof want to have them discuss this excerpt?

Cliffnotes:
"1 = 0.99.... + *" (by definition)
In that "non-standard" system of reals (not that "non-standard" is necessarily wrong), 1 > 0.99...
* - is nicknamed the "dark number"
X/0 is still undefined, but X/* is defined. - The author claims this enables the solution of unsolvable problems in string theory, and other advanced fields.

Your choice, cling to classical axioms because you feel insecure without them, or challenge the fundamentals of the theory of real numbers and get a number system that is more useful for modern science. I prefer the useful option. (One more option is just to say "huh?" and move on).

Anyone with access to a University library and time to waste should read some works by dutch mathematician Luitzen Brouwer (1881-1966) and come back and add to the discussion.
[edit: specifically, check out his paper titled, "Does every Real number have a decimal expansion?" - Brouwer answers "No"]
 

abracadabra1

Diamond Member
Nov 18, 1999
3,879
1
0
Originally posted by: Kyteland
We're having a debate at work. Is 1=0.99999..... repeating. I say that this holds but one of my coworkers claims that multiplication breaks down for an infinitely repeating number.

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

What do you think?

The bolded part is the flaw. You can' t subtract the two numbers because the stream of 9s is continuous (the same reason we can't subtract infinity from infinity). It's really just a flaw of algebra that produces a flaw in logic.
 

RIGorous1

Platinum Member
Oct 26, 2002
2,053
0
71
Originally posted by: abracadabra1
Originally posted by: Kyteland
We're having a debate at work. Is 1=0.99999..... repeating. I say that this holds but one of my coworkers claims that multiplication breaks down for an infinitely repeating number.

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

What do you think?

The bolded part is the flaw. You can' t subtract the two numbers because the stream of 9s is continuous (the same reason we can't subtract infinity from infinity). It's really just a flaw of algebra that produces a flaw in logic.

oh god why did you revive this thread! ...
:disgust:
 

TechnoKid

Diamond Member
Feb 12, 2001
5,575
0
0
This thread is really l o n g..........................................................................................................................................................................
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: abracadabra1
Originally posted by: Kyteland
We're having a debate at work. Is 1=0.99999..... repeating. I say that this holds but one of my coworkers claims that multiplication breaks down for an infinitely repeating number.

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

What do you think?

The bolded part is the flaw. You can' t subtract the two numbers because the stream of 9s is continuous (the same reason we can't subtract infinity from infinity). It's really just a flaw of algebra that produces a flaw in logic.

So you can't do 2 -1 because you can't account for all the zeros?
 

TuxDave

Lifer
Oct 8, 2002
10,571
3
71
Originally posted by: naddicott
OK, OK, ATOT majority rule isn't good enough for you?

Anyone who is playing teacher's pet with a Math prof want to have them discuss this excerpt?

Cliffnotes:
"1 = 0.99.... + *" (by definition)
In that "non-standard" system of reals (not that "non-standard" is necessarily wrong), 1 > 0.99...
* - is nicknamed the "dark number"
X/0 is still undefined, but X/* is defined. - The author claims this enables the solution of unsolvable problems in string theory, and other advanced fields.

Your choice, cling to classical axioms because you feel insecure without them, or challenge the fundamentals of the theory of real numbers and get a number system that is more useful for modern science. I prefer the useful option. (One more option is just to say "huh?" and move on).

Anyone with access to a University library and time to waste should read some works by dutch mathematician Luitzen Brouwer (1881-1966) and come back and add to the discussion.
[edit: specifically, check out his paper titled, "Does every Real number have a decimal expansion?" - Brouwer answers "No"]

I remember a proof that it is theoretically impossible to make a list containing all the real numbers in decimal form. I dunno if he used that to prove that there exists a real number that does not have a decimal expansion.

 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
Originally posted by: abracadabra1
Originally posted by: Kyteland
We're having a debate at work. Is 1=0.99999..... repeating. I say that this holds but one of my coworkers claims that multiplication breaks down for an infinitely repeating number.

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

What do you think?

The bolded part is the flaw. You can' t subtract the two numbers because the stream of 9s is continuous (the same reason we can't subtract infinity from infinity). It's really just a flaw of algebra that produces a flaw in logic.


What you have pointed out is the reason this DEMONSTRATION does not constitute a mathematical proof. BUT the questionable operation does NOT negate the result. It can be proven without resort to such manipulations.


Here is such a proof,
 

OIKOS

Banned
Mar 29, 2000
1,669
0
0

Originally posted by: dighn
Originally posted by: OIKOS
is 1.0000000000000000000000000000000000000000000000000000000001 = 1?

of course not.

same reason that 0.9999999999999999999999999999999999999999999999999 is not 1, but 0.999... is





#%@$!&%$$@.......:light:
 

TuxDave

Lifer
Oct 8, 2002
10,571
3
71
I think it's pretty much pointless to argue this anymore. To standard mathematics as we know it, I'm sure the rules will point to 0.999... = 1. However, as someone quoted above, I can pretend to be smart and start devising a new number system where anything is possible.

I can say 0.000....0001 is a number despite the fact that there are an infinite number of zeros.

I can claim that there are two end points (or one) to an infinitely large line.

So either way, there's no way to convince those who are truly stubborn.
 

crazygal

Senior member
Feb 26, 2002
469
0
0
Originally posted by: abracadabra1
The bolded part is the flaw. You can' t subtract the two numbers because the stream of 9s is continuous (the same reason we can't subtract infinity from infinity). It's really just a flaw of algebra that produces a flaw in logic.

You have got to be kidding me. You have no problem subtracting if it looks like this:
3/3 - 1/3 = 2/3

but you say that you can't do it like this?
.999... - .333... = .666....

Wow. Just....wow...
 

amnesiac

Lifer
Oct 13, 1999
15,781
1
71
Originally posted by: crazygal
Originally posted by: abracadabra1
The bolded part is the flaw. You can' t subtract the two numbers because the stream of 9s is continuous (the same reason we can't subtract infinity from infinity). It's really just a flaw of algebra that produces a flaw in logic.

You have got to be kidding me. You have no problem subtracting if it looks like this:
3/3 - 1/3 = 2/3

but you say that you can't do it like this?
.999... - .333... = .666....

Wow. Just....wow...

Did the scores of people insisting on letting this thread die somehow escape your glance? The post had been forgotten for a whole 4 days and now you have to dredge it back up again.
 

bleeb

Lifer
Feb 3, 2000
10,868
0
0
Originally posted by: amnesiac
Originally posted by: crazygal
Originally posted by: abracadabra1
The bolded part is the flaw. You can' t subtract the two numbers because the stream of 9s is continuous (the same reason we can't subtract infinity from infinity). It's really just a flaw of algebra that produces a flaw in logic.

You have got to be kidding me. You have no problem subtracting if it looks like this:
3/3 - 1/3 = 2/3

but you say that you can't do it like this?
.999... - .333... = .666....

Wow. Just....wow...

Did the scores of people insisting on letting this thread die somehow escape your glance? The post had been forgotten for a whole 4 days and now you have to dredge it back up again.


Somehow this thread lives on and on.
 

Kyteland

Diamond Member
Dec 30, 2002
5,747
1
81
Originally posted by: bleeb
Originally posted by: amnesiac
Originally posted by: crazygal
Originally posted by: abracadabra1
The bolded part is the flaw. You can' t subtract the two numbers because the stream of 9s is continuous (the same reason we can't subtract infinity from infinity). It's really just a flaw of algebra that produces a flaw in logic.

You have got to be kidding me. You have no problem subtracting if it looks like this:
3/3 - 1/3 = 2/3

but you say that you can't do it like this?
.999... - .333... = .666....

Wow. Just....wow...

Did the scores of people insisting on letting this thread die somehow escape your glance? The post had been forgotten for a whole 4 days and now you have to dredge it back up again.


Somehow this thread lives on and on.

You know, you never did add this to your list of things to do before you died.
40. Prove 1 != 0.9999......
Sorry, I couldn't resist!
 

bleeb

Lifer
Feb 3, 2000
10,868
0
0
Originally posted by: Kyteland
Originally posted by: MegaloManiaK
Lock it on up

Please list one good reason why this thread should be locked.

"Because I want to" isn't a good reason.

This is one of those threads that people agree or disagree on and have fun on. It has created a fondness in the hearts of many and is something that will hopefully go on, and on, and on... beyond the 2000th post. Its kind of like the enduring American spirit, post 9-11.
 
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