Is 1 = 0.9999......

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MadRat

Lifer
Oct 14, 1999
11,965
279
126
Originally posted by: RossGr
Originally posted by: MadRat
I want to see .999... represented on a number line and then the number line broken down to intervals where we can compare .999... to any other fraction, like .5 or even .9 for that matter. No matter how many slices are taken out between 1 and whichever fraction is chosen, we will never arrive at .999... along said interval to be able to make the direct comparison of values between .999... and 1.

Consider it done.

Take a piece of paper, draw a number line showing the interval between 0 and 1, break it into subintervals corresponding to .1 .2 etc Then number each of the boxes 0 -9

Should look like this

_____________________________________________
...|.0.|.1.|.2.|.3.|.4.|.5.|.6.|.7.|.8.|.9.|

...0...1...2....3...4...5...6....7...8...9...1

The bottom row off number represent the most significant digit .1 .2 etc the number in the box represents the next digit of all number in that box. For example in the first number box the digit 0 appears all numbers in that box start 0.0 that would be all numbers x such that 0 < x < .1.

Now, go to the 9th box, repeat this procedure. when that is complete go to the 9th box repeat the procedure. Repeeat....

to say that .999.... does not exist is to say that this process has some forced ending point. Getting tired of writing does not count.

Unfortunately your procedure never arrives at an interval upon which .999... can be place. That would make a direct comparison therefore impossible.

 

Brutuskend

Lifer
Apr 2, 2001
26,558
4
0
Ok this thread has been around for some time and I never posted.

I also didn't read all the replys so if this has been said before, eat me!

Is really dark grey black or is it really dark grey?
 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
Originally posted by: MadRat
Originally posted by: RossGr
Originally posted by: MadRat
I want to see .999... represented on a number line and then the number line broken down to intervals where we can compare .999... to any other fraction, like .5 or even .9 for that matter. No matter how many slices are taken out between 1 and whichever fraction is chosen, we will never arrive at .999... along said interval to be able to make the direct comparison of values between .999... and 1.

Consider it done.

Take a piece of paper, draw a number line showing the interval between 0 and 1, break it into subintervals corresponding to .1 .2 etc Then number each of the boxes 0 -9

Should look like this

_____________________________________________
...|.0.|.1.|.2.|.3.|.4.|.5.|.6.|.7.|.8.|.9.|

...0...1...2....3...4...5...6....7...8...9...1

The bottom row off number represent the most significant digit .1 .2 etc the number in the box represents the next digit of all number in that box. For example in the first number box the digit 0 appears all numbers in that box start 0.0 that would be all numbers x such that 0 < x < .1.

Now, go to the 9th box, repeat this procedure. when that is complete go to the 9th box repeat the procedure. Repeeat....

to say that .999.... does not exist is to say that this process has some forced ending point. Getting tired of writing does not count.

Unfortunately your procedure never arrives at an interval upon which .999... can be place. That would make a direct comparison therefore impossible.


Sure it does. .999... is in every interval in the boxes contianing a 9 in the first, second etc postions. It is an ifinite number of intervals.
 

DT4K

Diamond Member
Jan 21, 2002
6,944
3
81
Everyone who's taken calculus should know that it approaches one but isn't equal to one.
 

DT4K

Diamond Member
Jan 21, 2002
6,944
3
81
Originally posted by: Brutuskend
I also didn't read all the replys so if this has been said before, eat me!
:Q
What?
How dare you post in a thread without reading the ~ 1800 previous posts.

 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
Originally posted by: Shanti
Everyone who's taken calculus should know that it approaches one but isn't equal to one.

Everyone who has gotten beyond Calculus and taken a Real Analysis Course know that it is equal to 1. Once again, .999... is a fixed number, it does not APPROACH anything, it is a fixed number, just like 1 does not approach anything.
 

MadRat

Lifer
Oct 14, 1999
11,965
279
126
Originally posted by: RossGr
Originally posted by: Shanti
Everyone who's taken calculus should know that it approaches one but isn't equal to one.

Everyone who has gotten beyond Calculus and taken a Real Analysis Course know that it is equal to 1. Once again, .999... is a fixed number, it does not APPROACH anything, it is a fixed number, just like 1 does not approach anything.

I took it he was talking about your slicing up the number line in intervals.
 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
Originally posted by: MadRat
Originally posted by: RossGr
Originally posted by: Shanti
Everyone who's taken calculus should know that it approaches one but isn't equal to one.

Everyone who has gotten beyond Calculus and taken a Real Analysis Course know that it is equal to 1. Once again, .999... is a fixed number, it does not APPROACH anything, it is a fixed number, just like 1 does not approach anything.

I took it he was talking about your slicing up the number line in intervals.

and your point is?
 

spidey07

No Lifer
Aug 4, 2000
65,469
5
76
Originally posted by: MadRat
Originally posted by: RossGr
Originally posted by: Shanti
Everyone who's taken calculus should know that it approaches one but isn't equal to one.

Everyone who has gotten beyond Calculus and taken a Real Analysis Course know that it is equal to 1. Once again, .999... is a fixed number, it does not APPROACH anything, it is a fixed number, just like 1 does not approach anything.

I took it he was talking about your slicing up the number line in intervals.

number line isn't calculus. A number approaches something???
 

DT4K

Diamond Member
Jan 21, 2002
6,944
3
81
Originally posted by: spidey07
Originally posted by: MadRat
Originally posted by: RossGr
Originally posted by: Shanti
Everyone who's taken calculus should know that it approaches one but isn't equal to one.

Everyone who has gotten beyond Calculus and taken a Real Analysis Course know that it is equal to 1. Once again, .999... is a fixed number, it does not APPROACH anything, it is a fixed number, just like 1 does not approach anything.

I took it he was talking about your slicing up the number line in intervals.

number line isn't calculus. A number approaches something???

If .999... exists, it is not equal to one. If it was, it wouldn't be .999...., it would be 1.
When I said approaches, I was thinking in terms of continously adding 9's to the end.

The real question is what is .999.....
If it is a fixed number, it is certainly not the same as one.
just as 1/3 does not equal .32
 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
The real question is what is .999.....

It is indeed a fixed number and it is indeed 1.

you are correct .32 is not equal to 1/3 nor is .99 is not equal to 1 but there is a big difference between .32 and .333... just as there is a big difference between .99 and .999...

Shanti, have you taken the time to read and consider the 2 proofs I link to in my sig?
 

Joker81

Golden Member
Aug 9, 2000
1,281
0
0
they prove you that .9999 = 1 but infact it isn't because the proofs are based on mathmatical functions. Which infact they are proving that these functions fail at certain levels in math. So in theory I shouldn't have to take my math classes because the mathematical functions are obviously not correct.


if .9999999 repeat exists doesn't .00000repeat(1) also exist.
 

bleeb

Lifer
Feb 3, 2000
10,868
0
0
Originally posted by: Joker81
they prove you that .9999 = 1 but infact it isn't because the proofs are based on mathmatical functions. Which infact they are proving that these functions fail at certain levels in math. So in theory I shouldn't have to take my math classes because the mathematical functions are obviously not correct.


if .9999999 repeat exists doesn't .00000repeat(1) also exist.

With that said, if 0.000000....1 exists, then this means there is a number that is the difference of 1 - 0.9999...., which then therefore proves that 0.9999.... != 1. PROVING ALL ALONG WHAT I SAID that 0.9999... != 1.
 

flexy

Diamond Member
Sep 28, 2001
8,464
155
106
(0.9999999... eventually converges to 1)

When ? Date/Time/Location ?

I say: it never converges !!! (Or it does, but so far away in the future that it's irrelevant...)

 

Joker81

Golden Member
Aug 9, 2000
1,281
0
0
I bet you in cave man times it started out that .9=1 but with time it has increase to .999999(infinite) = 1
 

MAME

Banned
Sep 19, 2003
9,281
1
0
Originally posted by: Joker81
they prove you that .9999 = 1 but infact it isn't because the proofs are based on mathmatical functions. Which infact they are proving that these functions fail at certain levels in math. So in theory I shouldn't have to take my math classes because the mathematical functions are obviously not correct.


if .9999999 repeat exists doesn't .00000repeat(1) also exist.

1.000...1 does not exist. How can you have INFINITE 0's and then add a 1 to the end? This is a major pitfall.

.999... is EACTLY equal to 1
 

MAME

Banned
Sep 19, 2003
9,281
1
0
Originally posted by: Joker81
I bet you in cave man times it started out that .9=1 but with time it has increase to .999999(infinite) = 1

ok, now you're being stupid
 

MAME

Banned
Sep 19, 2003
9,281
1
0
Originally posted by: bleeb
Originally posted by: Joker81
they prove you that .9999 = 1 but infact it isn't because the proofs are based on mathmatical functions. Which infact they are proving that these functions fail at certain levels in math. So in theory I shouldn't have to take my math classes because the mathematical functions are obviously not correct.


if .9999999 repeat exists doesn't .00000repeat(1) also exist.

With that said, if 0.000000....1 exists, then this means there is a number that is the difference of 1 - 0.9999...., which then therefore proves that 0.9999.... != 1. PROVING ALL ALONG WHAT I SAID that 0.9999... != 1.

find me a math professor on this planet that agrees with "1.000...1" existing
 

flexy

Diamond Member
Sep 28, 2001
8,464
155
106
i think a number like 0.999999999999... contradicts math itself.....and you're right..the mistake is that these constants are rather 'assumptions' or, better, 'the nearest best we have'...ergo no real number....only some abstract 'something' which somehow has n place in a mathematical formula which cotains 'real' numbers......so also the comparison of a 'real' number and an abstract 'not able to define' number does not make sense.... (eg. 1 <> 0.999..)...or only makes sense if you bring in some philosophy..

IMHO...

boring thread..i hate math....i always did and still do even as a programmer

 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
flexy,
Since when is admitting you don't know anything about a subject qualify you to make calls on either the existence or equality of .999... and 1.

I guess you are guessing, right?
 

Joker81

Golden Member
Aug 9, 2000
1,281
0
0
Originally posted by: MAME
Originally posted by: Joker81
they prove you that .9999 = 1 but infact it isn't because the proofs are based on mathmatical functions. Which infact they are proving that these functions fail at certain levels in math. So in theory I shouldn't have to take my math classes because the mathematical functions are obviously not correct.


if .9999999 repeat exists doesn't .00000repeat(1) also exist.

1.000...1 does not exist. How can you have INFINITE 0's and then add a 1 to the end? This is a major pitfall.

.999... is EACTLY equal to 1

fine then infinity-1 zeros. then a 1
 
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