Is 1 = 0.9999......

[ElFenix]

Originally posted by: MadRat
Funny how you say I have no need for mathematics. Quite the contrary. Math is important to my work, thank you very much.

I don't pity you because you believe .999...=1. Actually I pity you because you do not keep to the syntax of what you studied.

you can disprove (or prove) anything by going outside the system, and you keep going outside the system.

 

[RossGr]

Originally posted by: BigNeko
.999...9 does not equal 1.

For
.999...9 + .000...1 = 1
let us agree that you can add as many 9s as you want to .999...9,
and that for every 9 you add, I will add a 0 to .000...1 (somewhere between the decimal and 1)
Then, this equation is ALWAYS true.
Since this equation is ALWAYS true, 1 does not equal .999...9


The shortest distance between two points is a straight line....

except for certain sub-atomic particles

In each of your relationships you are neglecting an infinte tail on 9s.

.999....9... + .000....01 = 1 + .00....0099....


you cannot simply chop off the rest of the 9s they are there.

so you do not have equality by adding 10^-N to .999..... you have 1 + some more

 

[silverpig]

What's bigger?

0.0000...001
0.00...0001
0.000...00....001
0.000...001..0001

 

[JavaMomma]

Come on guys stop picking on CS people

I'm a CS Major and
I believe/know that
0.999... == 1

1.000....
- 0.999....
---------------
0.000...

 

[MadRat]

Originally posted by: josphII

Originally posted by: MadRat

Originally posted by: BigNeko
.999...9 does not equal 1.
For
.999...9 + .000...1 = 1
let us agree that you can add as many 9s as you want to .999...9,
and that for every 9 you add, I will add a 0 to .000...1

Actually, we don't need to depend on a number of 9's or 0's to determine the difference. The whole argument these guys have is that since you cannot halve the difference between 1 and .999... then there is no difference. The problem is that their rules, along with alot of common rules that apply to other numbers, do not work with infinity. We apply stipulations to the use of zero (the "null" value) yet they don't want to use the same intuition to apply relative counter stipulations to infinity.

that is not true at all! the proof begins with the number 0.999... and through the use of algebra and calculus it is shown to equal 1.

BigNeko,

there is a basic flaw with your reasoning that 0.999... + 0.000...1 = 1. the flaw is that 0.000...1 doesnt exist. you can not have an infinite string of 0's, and then have a number after that. infinity, by definition, has no end. so if your infinte string of 0's has no end, then how can a "1" come after it? it cant! this is a common mistake, and one that is easily shown to be flawed.

If your nines can go on forever then our zeroes can go on forever, too. Just because their is an infinite number of zeroes or nines does not mean that the integer at the infinite position must be the same.

 

[MadRat]

Originally posted by: RossGr

Originally posted by: MadRat
That last comment is so funny, Ross. You probably depend on the use of Enlish yet you have terrible writing skills. LOL

I realize that , and have been crippled by it. Yet I struggle along. Difference is I recognize my short commings. You on the other hand are blind to yours.

Before you go about telling others they don't know sh!t perhaps you need some introspective to realize your imperfectly spelled English does not necessarily make you completely illiterate. My focus in mathematics is not on the vocabulary but rather the meaning, just as your focus on communication is not on spelling but rather meaning. The golden rule, bub.

 

[TuxDave]

Originally posted by: MadRat

Originally posted by: josphII

Originally posted by: MadRat

Originally posted by: BigNeko
.999...9 does not equal 1.
For
.999...9 + .000...1 = 1
let us agree that you can add as many 9s as you want to .999...9,
and that for every 9 you add, I will add a 0 to .000...1

Actually, we don't need to depend on a number of 9's or 0's to determine the difference. The whole argument these guys have is that since you cannot halve the difference between 1 and .999... then there is no difference. The problem is that their rules, along with alot of common rules that apply to other numbers, do not work with infinity. We apply stipulations to the use of zero (the "null" value) yet they don't want to use the same intuition to apply relative counter stipulations to infinity.

that is not true at all! the proof begins with the number 0.999... and through the use of algebra and calculus it is shown to equal 1.

BigNeko,

there is a basic flaw with your reasoning that 0.999... + 0.000...1 = 1. the flaw is that 0.000...1 doesnt exist. you can not have an infinite string of 0's, and then have a number after that. infinity, by definition, has no end. so if your infinte string of 0's has no end, then how can a "1" come after it? it cant! this is a common mistake, and one that is easily shown to be flawed.

If your nines can go on forever then our zeroes can go on forever, too. Just because their is an infinite number of zeroes or nines does not mean that the integer at the infinite position must be the same.

There's a difference. For 0.999... we're defining a number where for every 9 we have, there will exist another 9 after that. However for 0.000...001, exactly how do you define it? To get the infinite chain of 0s, we have to say that for every 0, there's another 0 after that. So how do we define the location of the 1? We can't... unless you have a clever way to say it without contradicting the first statement that for every 0, there's another 0 after it.

 

[Sahakiel]

Whew.... after reading through a bunch of posts, I'm getting a lot of material for insults...... if I were still in the second grade, that is...

On the one hand, I understand why 0.9999... + 0.000...0001 = 0.00...000999...
On the other hand, I also understand why electrons passing through the infamous double slit will exhibit a wave-like interference pattern on the other side. Unless, of course, you actually try to watch it happening.

Go figure..

 

[MadRat]

Originally posted by: TuxDave

There's a difference. For 0.999... we're defining a number where for every 9 we have, there will exist another 9 after that. However for 0.000...001, exactly how do you define it? To get the infinite chain of 0s, we have to say that for every 0, there's another 0 after that. So how do we define the location of the 1? We can't... unless you have a clever way to say it without contradicting the first statement that for every 0, there's another 0 after it.

The value at the infinite position, doesn't need follow conventional rules. The idea of infinity is unconventional just as imaginary numbers are unconventional. We are defining a position that by definition has no definitive location. We do know that "infinity" cannot be represented by any integer, which is why we use a symbol. Funny, but this is just as "null" is represented by a zero symbol to signify "nothing"!

How to write it? Many ways. One could use a vinculum over the zero with an "r1" afterwards to represent a "1" integer at the end. One could simply write it like we have been, using .000...1 or .000...r1. All this value represents is the infinitely smallest value that is greater than zero. It doesn't exist in the respect of a normal decimal since none of us have the omnipotence to understand infinity. But just because man cannot fathom infinity doesn't mean we cannot represent its implications.

The mathematical definition of infinity tends to imply "without end" whereas the philosophical definition moreso implies "includes everything". We don't need to get into the idea that philosophically "infinity" does have a limit since we are here to talk math. Anyways, if the .999... number exists closest to "1" than any other number on the number line then a number with the same distance from "0" must also exist, which is a number that the math geniuses here reject even exists.

To say .000...r1 does not exist then implies that .999... also does not exist.

 

[spidey07]

To say .000...r1 does not exist then implies that .999... also does not exist.

If your .000...r1 number exists, please provide some sort of proof that it does. And try to avoid "madrat's number system" please.

 

[TuxDave]

Originally posted by: MadRat

Originally posted by: TuxDave

There's a difference. For 0.999... we're defining a number where for every 9 we have, there will exist another 9 after that. However for 0.000...001, exactly how do you define it? To get the infinite chain of 0s, we have to say that for every 0, there's another 0 after that. So how do we define the location of the 1? We can't... unless you have a clever way to say it without contradicting the first statement that for every 0, there's another 0 after it.

The value at the infinite position, doesn't need follow conventional rules. The idea of infinity is unconventional just as imaginary numbers are unconventional. We are defining a position that by definition has no definitive location. We do know that "infinity" cannot be represented by any integer, which is why we use a symbol. Funny, but this is just as "null" is represented by a zero symbol to signify "nothing"!

How to write it? Many ways. One could use a vinculum over the zero with an "r1" afterwards to represent a "1" integer at the end. One could simply write it like we have been, using .000...1 or .000...r1. All this value represents is the infinitely smallest value that is greater than zero. It doesn't exist in the respect of a normal decimal since none of us have the omnipotence to understand infinity. But just because man cannot fathom infinity doesn't mean we cannot represent its implications.

The mathematical definition of infinity tends to imply "without end" whereas the philosophical definition moreso implies "includes everything". We don't need to get into the idea that philosophically "infinity" does have a limit since we are here to talk math. Anyways, if the .999... number exists closest to "1" than any other number on the number line then a number with the same distance from "0" must also exist, which is a number that the math geniuses here reject even exists.

To say .000...r1 does not exist then implies that .999... also does not exist.

What if I throw in a monkey wrench and concede that 0.000...r1 does exist only if it's equal to zero.

 

[TuxDave]

Originally posted by: Sahakiel
On the one hand, I understand why 0.9999... + 0.000...0001 = 0.00...000999...

Woah... you understand that? Cuz I sure don't.... hehehe

 

[thomsbrain]

i don't have anything to say, but i figured if we're up to 667 posts or whatever this thread is, i better get in on it.

 

[ElFenix]

Originally posted by: MadRat


The value at the infinite position

there is no infinith position. thats why the ellipsis used in the middle of the number means that theres a finite number of decimal places, and that the numbers after that are whats at the end. why is that so hard to comprehend?

 

[RossGr]

If your nines can go on forever then our zeroes can go on forever, too. Just because their is an infinite number of zeroes or nines does not mean that the integer at the infinite position must be the same.

The point is, when we specify an infinte number of nines that is exactly what we mean, that EVERY digit is a nine and continues on that way with no possibility of variation.

When you say an infinite number of zeros then a 1 you have contradicted yourself because there can be no end to infinity.

I have already attempted to convey this information, perhaps if I keep trying certian individuals may start to catch on.

Let D(n) be the digit in the nth position of a Real number. So, for example,

2.000... means that D(n) = 0 for ALL n
This is what is implied when we write 2

.999.... means that D(n) = 9 for All n

I can completly specify the digits of those numbers within the context of real numbers therefore they are valid real numbers.

If I write .00...001 This implies that for some N>0 (pick N to be any integer you choose)

D(n) = 0 for n = 1 to N-1
D(N) = 1
Then
D(n) = 0 for all n > N

So formally for any decimal numbers , x, it is written like this

x = Sum(n=1 to infinity) D(n)* 10^-n

To specifiy x you must specify the D(n), these are the digits of the number.

if x = .000...00100...
Then

x = Sum( n= 1 to N-1)( 0 * 10^-n) + 10^-N + sum (n= N+1 to infinity)( 0 * 10^-n)

All real numbers can be written as this type of sum, if you cannot express the sum then it is not a real number. Therefore it is not a valid real number if you attempt to write

x = Sum(n= 1 to infinity) 0* 10^-n + 10^ - ??? (what do I put in the exponent? we have already committed all integers in the first sum, there are none left for use, we cannot recycle them, a sum to infinity means exactly that, not a sum to some unspecified number. That is how Madrat is treating infinity, he seems to think infinity is the same as unspecified large, this is not the case. A sum to infinity means that ALL integers are used in the sum indexes, none are free to be reused.

 

[Yomicron]

For anyone who is confused by the infinite position of a decimal, there is a wealth of information about it.

 

[RossGr]

Originally posted by: Yomicron
For anyone who is confused by the infinite position of a decimal, there is a wealth of information about it.

LOL, 3 hits is a wealth.~^ , none of which are meaningful! The first links to a parts list, the second is a reference to a switch setting on some piece of equipment, the third is someone elses opinion.

Sorry, there nothing there.

 

[silverpig]

So let me get this straight? You can have an infinite string of nines ENDING with a 4? Riiiiight.


Tell me then madrat, what is 0.000...001 / 2? Isn't that smaller than 0.000...001? Therefore, wouldn't it be closer to 0 than 0.000...001?

Also, please tell me which is the largest of these numbers:

0.000...0001
0.00...00001
0.00...0001
0.000...00..0001
0.000...001...01

 

[OS]

look at what you guys did :Q

 

[AznMaverick]

nm

 

[silverpig]

Why doesn't that link work for me?

 

[silverpig]

Dammit, my DNS is down AGAIN.

 

[silverpig]

Can someone resolve the IP for me? I wanna see

 

[BigNeko]

I think MadRat understands what I am trying to say. Credit to the guy (or gal) back on page one or two who did this.

.9+.1=1
.99+.01=1
.999+.001=1

JosphII said

you can not have an infinite string of 0's, and then have a number after that.

At EXACTLY what point does .000...1 become a less viable number than .999...
If you cannot define that point, then my equation is still correct;

and .999... does not equal 1.


P.S. This is a rather good debate, so let's skip the name calling, even though you did add a nine to your infinite nines, then turn around and say to me that I had one too many zeroes before my one.


















Everything after * j/k

 

[RossGr]

Your nines go on forever. So do my zeroes, with a one at the end.

Make up your mind, the zeros either go on for ever or have an end, can't have both.

The 9's go on forever, no end.