Is 1 = 0.9999......

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Spencer278

Diamond Member
Oct 11, 2002
3,637
0
0
Originally posted by: MAME
Originally posted by: Spencer278
Originally posted by: MAME
Originally posted by: Hector13
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.

so of .0000...1 exists, you are claiming the number line is quantized and not continous? In that case, what is .00...1 / 2?

Very good Hector. That pretty much proves .000...1 does nto exist


No that just proves that .000...1/2 is equal to .0000...1.



What the hell?

Lets say you have .000...1 of an object (even though the number doesn't exist, but we'll say it does for your sake), lets say it's rice.
So now you take half of the rice away. You claim that when you take half of the rice away, you will still have the same amount?

How can you be serious? That makes no sense at all.


God this thread makes me want to cry


It only makes sense if .0000...1 is equal to 0.
 

Hector13

Golden Member
Apr 4, 2000
1,694
0
0
Originally posted by: Spencer278
Originally posted by: MAME


What the hell?

Lets say you have .000...1 of an object (even though the number doesn't exist, but we'll say it does for your sake), lets say it's rice.
So now you take half of the rice away. You claim that when you take half of the rice away, you will still have the same amount?

How can you be serious? That makes no sense at all.


God this thread makes me want to cry


It only makes sense if .0000...1 is equal to 0.

good, so since 1 - .99999 = .00....1 (or, as you admit, zero), we are back to 1 = .9999r.
 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
Why not just keep it simple and correct. It is impossible to have an infinite string of zeros followed by something. As soon as you place the "something" it must be at a countable location (this by the definition of a decimal expansion), therefore there is not an infinite string of zeros. To have a valid decimal expansion you must have be able to assign an integer to each digit in the representation. When I write .999... I have assigned nines to every integer. The integers represent the power of 10 which multiplies the digit.

When you attempt to create a number with an infinite number of zeros "followed" by something, either you can assign a integer to the the "something" in which case there was not an infinite number of zeros, or you cannot assign an integer to the "something" in which case we are no longer talking about a real number.
 

Spencer278

Diamond Member
Oct 11, 2002
3,637
0
0
Originally posted by: Hector13
Originally posted by: Spencer278
Originally posted by: MAME


What the hell?

Lets say you have .000...1 of an object (even though the number doesn't exist, but we'll say it does for your sake), lets say it's rice.
So now you take half of the rice away. You claim that when you take half of the rice away, you will still have the same amount?

How can you be serious? That makes no sense at all.


God this thread makes me want to cry


It only makes sense if .0000...1 is equal to 0.

good, so since 1 - .99999 = .00....1 (or, as you admit, zero), we are back to 1 = .9999r.


I never questioned the 1 = .99999.... I was just saying that you were wrong about .000...1 not existing.
 

Hector13

Golden Member
Apr 4, 2000
1,694
0
0
Originally posted by: Spencer278
Originally posted by: Hector13
Originally posted by: Spencer278
Originally posted by: MAME


What the hell?

Lets say you have .000...1 of an object (even though the number doesn't exist, but we'll say it does for your sake), lets say it's rice.
So now you take half of the rice away. You claim that when you take half of the rice away, you will still have the same amount?

How can you be serious? That makes no sense at all.


God this thread makes me want to cry


It only makes sense if .0000...1 is equal to 0.

good, so since 1 - .99999 = .00....1 (or, as you admit, zero), we are back to 1 = .9999r.


I never questioned the 1 = .99999.... I was just saying that you were wrong about .000...1 not existing.

I'm not sure I follow. You say .000...1 exists but you think it is the same as zero?
 

Spencer278

Diamond Member
Oct 11, 2002
3,637
0
0
Originally posted by: RossGr
Why not just keep it simple and correct. It is impossible to have an infinite string of zeros followed by something. As soon as you place the "something" it must be at a countable location (this by the definition of a decimal expansion), therefore there is not an infinite string of zeros. To have a valid decimal expansion you must have be able to assign an integer to each digit in the representation. When I write .999... I have assigned nines to every integer. The integers represent the power of 10 which multiplies the digit.

When you attempt to create a number with an infinite number of zeros "followed" by something, either you can assign a integer to the the "something" in which case there was not an infinite number of zeros, or you cannot assign an integer to the "something" in which case we are no longer talking about a real number.


ok we will say .1 but can I claim it is an infitie base thus getting the equalivilnt of 1/infinite or .000...1? Ok if you don't like the .000...1 notation replace it with 1/9999999.... notation.
 

Spencer278

Diamond Member
Oct 11, 2002
3,637
0
0
Originally posted by: Hector13
Originally posted by: Spencer278
Originally posted by: Hector13
Originally posted by: Spencer278
Originally posted by: MAME


What the hell?

Lets say you have .000...1 of an object (even though the number doesn't exist, but we'll say it does for your sake), lets say it's rice.
So now you take half of the rice away. You claim that when you take half of the rice away, you will still have the same amount?

How can you be serious? That makes no sense at all.


God this thread makes me want to cry


It only makes sense if .0000...1 is equal to 0.

good, so since 1 - .99999 = .00....1 (or, as you admit, zero), we are back to 1 = .9999r.


I never questioned the 1 = .99999.... I was just saying that you were wrong about .000...1 not existing.

I'm not sure I follow. You say .000...1 exists but you think it is the same as zero?


you say .99999 exists but you think it is the same thing as one.
 

Hector13

Golden Member
Apr 4, 2000
1,694
0
0
Originally posted by: Spencer278
Originally posted by: Hector13
Originally posted by: Spencer278
Originally posted by: Hector13
Originally posted by: Spencer278
Originally posted by: MAME


What the hell?

Lets say you have .000...1 of an object (even though the number doesn't exist, but we'll say it does for your sake), lets say it's rice.
So now you take half of the rice away. You claim that when you take half of the rice away, you will still have the same amount?

How can you be serious? That makes no sense at all.


God this thread makes me want to cry


It only makes sense if .0000...1 is equal to 0.

good, so since 1 - .99999 = .00....1 (or, as you admit, zero), we are back to 1 = .9999r.


I never questioned the 1 = .99999.... I was just saying that you were wrong about .000...1 not existing.

I'm not sure I follow. You say .000...1 exists but you think it is the same as zero?


you say .99999 exists but you think it is the same thing as one.


.999r exists, and yes, I do say it is the same as one.

I am just trying to confirm that you are saying that .00...1 exists and that it is equal to zero. At the same time you claim that 2 x .00...1 = .00..1. This "number" seems to defy most of our basic mathematical operations.
 

Spencer278

Diamond Member
Oct 11, 2002
3,637
0
0
Originally posted by: Hector13
Originally posted by: Spencer278
Originally posted by: Hector13
Originally posted by: Spencer278
Originally posted by: Hector13
Originally posted by: Spencer278
Originally posted by: MAME


What the hell?

Lets say you have .000...1 of an object (even though the number doesn't exist, but we'll say it does for your sake), lets say it's rice.
So now you take half of the rice away. You claim that when you take half of the rice away, you will still have the same amount?

How can you be serious? That makes no sense at all.


God this thread makes me want to cry


It only makes sense if .0000...1 is equal to 0.

good, so since 1 - .99999 = .00....1 (or, as you admit, zero), we are back to 1 = .9999r.


I never questioned the 1 = .99999.... I was just saying that you were wrong about .000...1 not existing.

I'm not sure I follow. You say .000...1 exists but you think it is the same as zero?


you say .99999 exists but you think it is the same thing as one.


.999r exists, and yes, I do say it is the same as one.

I am just trying to confirm that you are saying that .00...1 exists and that it is equal to zero. At the same time you claim that 2 x .00...1 = .00..1. This "number" seems to defy most of our basic mathematical operations.


0 * 0 = 0 what is wrong with that?
 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
It is all about what you mean by equality. In the Real Number system it is not quite as easy as it is with integers. I think that is where a lot of the troubles come from with this thread. People not familiar with the basic properties of the Real Number system cannot separate themselves from the integers.

In order to discuss equality on the Real Number line we must think only of the position a number occupies, not how it looks. The basic concept is, that if there is a "distance" between 2 points on the Real Line they are not equal, if there is no "distance" then they are equal. It does not matter how you represent a point on the line, it is all about where the point is. As was mentioned above we can represent points on the number line in an infinite number of different ways, but all representations of the the same point are equal.

I have shown in the link in my sig that .999... and 1 both exist in the intersection of an infinite number of nested sets. This is one way to DEFINE a single point on the number line. Since both 1 and .999... exist in this intersection, which can contain only one point they have to be the same point.

 

MAME

Banned
Sep 19, 2003
9,281
1
0
Spencer, there is no such thing as .000...1
You CANNOT have an infinite repeating number followed by a 1. You have no comprehension of what infinity means.
 

Spencer278

Diamond Member
Oct 11, 2002
3,637
0
0
Originally posted by: MAME
Spencer, there is no such thing as .000...1
You CANNOT have an infinite repeating number followed by a 1. You have no comprehension of what infinity means.


Hey you started using the .0000...1 notation before I did. All I said was that the proof to show .000...1 didn't exist was flawed.

if you really dislike .000...1 notation just use
1 + sum( n = 0 to infinity )( -1.0 * 10^(-n) + 1.0 * 10^-(n+1) )

That will be equal to .0000...1 right?
 

RossGr

Diamond Member
Jan 11, 2000
3,383
1
0
1 + sum( n = 0 to infinity )( -1.0 * 10^(-n) + 1.0 * 10^-(n+1) )= .000...1

LOL! NOT!

Refresh youself in basic exponetial arithematic!

10^-(n+1) - 10^(-n) = 10^-n ((10^-1) -1 ) = -.9*10^-n
 

Spencer278

Diamond Member
Oct 11, 2002
3,637
0
0
Originally posted by: RossGr
1 + sum( n = 0 to infinity )( -1.0 * 10^(-n) + 1.0 * 10^-(n+1) )= .000...1

LOL! NOT!

Refresh youself in basic exponetial arithematic!

10^-(n+1) - 10^(-n) = 10^-n ((10^-1) -1 ) = -.9*10^-n

for sum up to 0
1 + -1.0^(-0) + 1.0*10^-(0+1) = .1
upto 1
.1 + -1.0^(-1) + 1.0*10^-(1+1) = .01
upto 2
.01 + -1.0^(-2) + 1.0*10^-(2+1) = .001
I will let you finish the sum.

If you don't like the sum you could think of .000...1 as
1/1000...
 

MAME

Banned
Sep 19, 2003
9,281
1
0
Originally posted by: cmaMath
Name,

Yes, 0.999... does equal 1.

Yay! I have faith in our schools.

Spencer, I never used .000...1
I only mentioned it in responce to the people trying to use it.

Why not ask someone with a PhD in math if 1 = .999.... and/or if .000...1 even exists.

Then you can come back and tell us how wrong you are
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Originally posted by: Spencer278
Originally posted by: RossGr
1 + sum( n = 0 to infinity )( -1.0 * 10^(-n) + 1.0 * 10^-(n+1) )= .000...1

LOL! NOT!

Refresh youself in basic exponetial arithematic!

10^-(n+1) - 10^(-n) = 10^-n ((10^-1) -1 ) = -.9*10^-n

for sum up to 0
1 + -1.0^(-0) + 1.0*10^-(0+1) = .1
upto 1
.1 + -1.0^(-1) + 1.0*10^-(1+1) = .01
upto 2
.01 + -1.0^(-2) + 1.0*10^-(2+1) = .001
I will let you finish the sum.

If you don't like the sum you could think of .000...1 as
1/1000...

You have a progression whose limiting value is zero.

.9 + .09 + .009 + ... doesn't actually EQUAL anything, however the LIMIT of this series is 1. Your summation doesn't equal anything as it's a progression. It's limiting value is 0.
 

Spencer278

Diamond Member
Oct 11, 2002
3,637
0
0
Originally posted by: silverpig
Originally posted by: Spencer278
Originally posted by: RossGr
1 + sum( n = 0 to infinity )( -1.0 * 10^(-n) + 1.0 * 10^-(n+1) )= .000...1

LOL! NOT!

Refresh youself in basic exponetial arithematic!

10^-(n+1) - 10^(-n) = 10^-n ((10^-1) -1 ) = -.9*10^-n

for sum up to 0
1 + -1.0^(-0) + 1.0*10^-(0+1) = .1
upto 1
.1 + -1.0^(-1) + 1.0*10^-(1+1) = .01
upto 2
.01 + -1.0^(-2) + 1.0*10^-(2+1) = .001
I will let you finish the sum.

If you don't like the sum you could think of .000...1 as
1/1000...

You have a progression whose limiting value is zero.

.9 + .09 + .009 + ... doesn't actually EQUAL anything, however the LIMIT of this series is 1. Your summation doesn't equal anything as it's a progression. It's limiting value is 0.


So you agree that .000...1 exsits and is equal to zero?
 
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