Is 1 = 0.9999......

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bleeb

Lifer
Feb 3, 2000
10,868
0
0
Originally posted by: xirtam
The real question is, if we can show that there's no such thing as .9999999... due to finite states of reality, does that mean that there is no "1"?

that presumes that 0.9999... = 1. but we all know that 0.9999... != 1. Therefore showing that 0.9999... doesn't exist doesn't mean that 1 doesn't exist.
 

Kyteland

Diamond Member
Dec 30, 2002
5,747
1
81
Originally posted by: bleeb
nice dp Kyteland.. you prick!!

I used to think "How the hell can people double post? They must be stupid or something!" Now I've gone and done it 3 times.

Edit: Oh, and postcount++

Perhaps if I doubleposted more I could catch the top posters by the end of the month.

Or not.

Edit 2: Holy sh*t! To do that I would have to post once every 30 seconds until December was over. :Q:Q:Q:Q

No sleep fo me.
 

xirtam

Diamond Member
Aug 25, 2001
4,693
0
0
Originally posted by: bleeb
Originally posted by: xirtam
The real question is, if we can show that there's no such thing as .9999999... due to finite states of reality, does that mean that there is no "1"?

that presumes that 0.9999... = 1. but we all know that 0.9999... != 1. Therefore showing that 0.9999... doesn't exist doesn't mean that 1 doesn't exist.

We don't all know that .9999... != 1. I believe that it does for the inferrential reason that there is no difference between the two mathematical entities. I'm persuaded by that argument. It's simple. The only logical refutation I can see is that there's no such thing as .9999..., but that there is such an entity as 1.
 

MadRat

Lifer
Oct 14, 1999
11,965
278
126
Exactly. To accept .999... as real then one must accept the presence of a .000...1, thereby making 1 <> .999... and 1-.999...=.000...1. To deny .999... as real then one must accept that either .999...<>1 or that 1 does not exist.

Personally I see no conflict between the ideas of infinity and null/zero (as in "none"), and one (as in "all") and 'irrational one' (the "anti one" so to speak) as positional values being that these terms all compliment each other to form rudimentary ideals in mathematics that do not necessarily correllate with the physical world as its presently known. Limiting one's logic to the physical world would have never allowed Archimedes to postulate the basic tenants of calculus.
 

MAME

Banned
Sep 19, 2003
9,281
1
0
Originally posted by: MadRat
Funny how we're to believe that .999... = 1 when no other infinite pattern of digits in a decimal would. We are to believe the digit nine is a magic number in this way.

I for one still believes an infinitely small remainder is the difference between 1 and .999... making 1 <> .999... as true. There has not been one solid proof to show otherwise.

you failed math didn't you?
 

MisterPants

Senior member
Apr 28, 2001
335
0
0
Originally posted by: spidey07
, then does .9999.....98 = .99999
You can not arbitratrilly manipulate real numbers like you just stated. There is not such real number as .9999...98 in the real number system. If so please express it as a fraction.

I'm not arguing with your 1st point, I couldn't think of a valid way to define .999999999999...8 but Transcendentals are still real.
 

MAME

Banned
Sep 19, 2003
9,281
1
0
.999...9998 is not a real number, you can't have an infinite amount of 9's and then at an 8 to it. It's pointless arguing about this with people who refuse to look at the proof already provided by people who are much smarter than us
 

MovingTarget

Diamond Member
Jun 22, 2003
9,002
115
106
Originally posted by: MAME
.999...9998 is not a real number, you can't have an infinite amount of 9's and then at an 8 to it. It's pointless arguing about this with people who refuse to look at the proof already provided by people who are much smarter than us

Hmmm...can you provide a link? Just curious as to what you have seen on the subject....thanks.
 

MadRat

Lifer
Oct 14, 1999
11,965
278
126
Originally posted by: MAME
.999...9998 is not a real number, you can't have an infinite amount of 9's and then at an 8 to it. It's pointless arguing about this with people who refuse to look at the proof already provided by people who are much smarter than us

I find people that claim to be smarter than everyone else or depend on people telling them that they are smarter than everyone else to be retarded. So far there has been no proof to show that .999... exists. If .999... exists then it is plausible that .999...8 exists, so don't be so sure it a pointless argument.
 

MadRat

Lifer
Oct 14, 1999
11,965
278
126
Eventually it will die. Fusetalk 5.x threads are limited to something like 63999 posts.
 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
Originally posted by: MadRat
Funny how we're to believe that .999... = 1 when no other infinite pattern of digits in a decimal would. We are to believe the digit nine is a magic number in this way.

I for one still believes an infinitely small remainder is the difference between 1 and .999... making 1 <> .999... as true. There has not been one solid proof to show otherwise.

9 is the magic digit in decimal
1 is the magic digit in binary (.111... = 1 (Binary)
2 is the magic digit in ternary (.222...=1 (Ternary)
3 is the magic digit in base 4 (.333...=1 Base 4)

need I continue?

The proof does not exist for you simply cuz you are incapable of comprehending it. Your preconceptions limit you.
 

Imaginer

Diamond Member
Oct 15, 1999
8,076
1
0
Well in my point of view, such simplifications will greatly reduce some headaches thus eliminating messy numbers.

So yes 1 is similar to .99999999... and i will make that swap. Hell, what about the .9999988 some?
 

MadRat

Lifer
Oct 14, 1999
11,965
278
126
Originally posted by: RossGr
9 is the magic digit in decimal
1 is the magic digit in binary (.111... = 1 (Binary)
2 is the magic digit in ternary (.222...=1 (Ternary)
3 is the magic digit in base 4 (.333...=1 Base 4)

need I continue?

You never began anything meaningful here except to persist in your argument that .999...=1 in other base numbering systems. Your point?

Originally posted by: RossGr
The proof does not exist for you simply cuz you are incapable of comprehending it. Your preconceptions limit you.

Ad hominum was dejected early on in this argument.

You've yet to define the simple fraction to represent .999... other than to say 1/1. So in effect the only preconceived limit here is the one which you've set for your own argument.
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: MadRat
Exactly. To accept .999... as real then one must accept the presence of a .000...1, thereby making 1 <> .999... and 1-.999...=.000...1. To deny .999... as real then one must accept that either .999...<>1 or that 1 does not exist.

Personally I see no conflict between the ideas of infinity and null/zero (as in "none"), and one (as in "all") and 'irrational one' (the "anti one" so to speak) as positional values being that these terms all compliment each other to form rudimentary ideals in mathematics that do not necessarily correllate with the physical world as its presently known. Limiting one's logic to the physical world would have never allowed Archimedes to postulate the basic tenants of calculus.

No, you absolutely do not have to accept 0.000...1. It's completely stupid.
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: MovingTarget
Originally posted by: MAME
.999...9998 is not a real number, you can't have an infinite amount of 9's and then at an 8 to it. It's pointless arguing about this with people who refuse to look at the proof already provided by people who are much smarter than us

Hmmm...can you provide a link? Just curious as to what you have seen on the subject....thanks.

Yeah, this thread.
 
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