Is 1 = 0.9999......

[RossGr]

Originally posted by: zerocool1
not this again

The solution to your problem is simple, Don't open the thread. Such response as yours are counter productive. They simply move the thread to the top keeping it alive. If you truly desired it to die you would let it sink.

 

[MadRat]

I never said it cannot be on the number line, I said it cannot be placed exactly on the line in relationship to definite values. There is no way to compare .999.... to 1 on the line because .999... has no definite form relative to other numbers while 1 does. If someone is going to use measure then definitive segments need to be set. To show position is to set measure.

 

[anxi80]

the self-proclaimed 'thread killer' is here. DIE THREAD DIIIIEEEEEEEEEEEEEEEEEEEEEEEEE!!!!!!!!!!!!

 

[bleeb]

Proof by induction:

Given 0.9 != 1.

then

0.99 != 1.

0.999 != 1.

0.9999 != 1.

0.99999 != 1.

We conclude that

0.9999... != 1.

WHO'S YO DADDY?!

 

[OIKOS]

Originally posted by: bleeb
Proof by induction:

Given 0.9 != 1.

then

0.99 != 1.

0.999 != 1.

0.9999 != 1.

0.99999 != 1.

We conclude that

0.9999... != 1.

WHO'S YO DADDY?!

REPOST!!..... I AM,,....
:|

 

[bleeb]

Originally posted by: OIKOS

Originally posted by: bleeb
Proof by induction:

Given 0.9 != 1.

then

0.99 != 1.

0.999 != 1.

0.9999 != 1.

0.99999 != 1.

We conclude that

0.9999... != 1.

WHO'S YO DADDY?!

REPOST!!..... I AM,,....
:|

jyeah jyeah, but I find the logic in this very simple and understandable by any simpleton, even the morons who think 0.9999... = 1. Bah Humbug!

 

[silverpig]

Originally posted by: MadRat
By the way, if .999... represents a tail of 9's that proceeds to infinity and 1.0 has a tail of 0's that proceed to infinity then its no different that tryng to compare .9 to 1.0, because there is going to be a sliver left. Even when you proceed to infinity the sliver will be there, even if there is not a fixed location. If we take .999... and times it by 10^(infinity) then we end up with an infinity number of nines that extend beyond zero, correct? Basically we shifted the decimal to the right of the nines. Substitute y for (.999... times 10^(infinity)) to simplify the demonstration. If we take 1.0 and times it be 10^(infinity) then we end up with a value that is 1 greater than y, otherwise equal to (.999... times 10^(infinity))+1. Substitute x for (1.0... times 10^(infinity)) to simplify the math and we get a difference of exactly 1. Philosophically it works this way, but whether you math experts want to accept it or not is up to you.

x = 1.0... times 10^(infinity) = (.999... times 10^(infinity))+1
y = (.999... times 10^(infinity))

x - y = 1

10^∞ = ∞
x = 1 * ∞ = .999... * ∞ = ∞ = ∞ + 1 = ∞ + u where u = any number
y = (.999... * &#8734 = ∞
x - y = ∞ - ∞ = ?

infinity - infinity can really equal whatever you want. It's undefined.

 

[Chu]

Originally posted by: bleeb
Proof by induction:

Given 0.9 != 1.

then

0.99 != 1.

0.999 != 1.

0.9999 != 1.

0.99999 != 1.

We conclude that

0.9999... != 1.

WHO'S YO DADDY?!

Not this again. You can't use induction here. The reason is that infinity is a quantum leap past any real number. Induction can proove that it is true for any integer number k, because you can prove it stepwise for 1...k-1. The problem is, you will never get to infinity by simply adding 1 to k a finite amount of times, because of the way infinity is defined.

-Chu

 

[DrPizza]

Originally posted by: bleeb
Proof by induction:

Given 0.9 != 1.

then

0.99 != 1.

0.999 != 1.

0.9999 != 1.

0.99999 != 1.

We conclude that

0.9999... != 1.

WHO'S YO DADDY?!

A proof similar to yours:
1 !=10
1+1 !=10
1+1+1 != 10
1+1+1+1 != 10

you conclude that 1+1+1+1+1+1+1+1+1+1 != 10

Your proof is just as absurd, except you don't recognize it as such.
Proof by induction is a valid method of proving things, and can be used to prove mathematical statements that couldn't be proved by the method of exhaustion (when there are an infinite number of possibilities). However, labeling your proof "proof by induction" does NOT make it a proof by induction, nor does it make it valid.

Here's one that's closer to yours that *maybe* you'll recognize is wrong:
.3 != 1/3
.33 != 1/3
.333 != 1/3
.3333 != 1/3
.33333 != 1/3
.333333 != 1/3
.3333333 != 1/3
.33333333 != 1/3

therefore .333... != 1/3
"proof by not paying close enough attention in math class"

 

[AMCRambler]

this thread = inifinity

 

[crazygal]

Originally posted by: bleeb
Proof by induction:

Given 0.9 != 1.

then

0.99 != 1.

0.999 != 1.

0.9999 != 1.

0.99999 != 1.

We conclude that

0.9999... != 1.

WHO'S YO DADDY?!

If you're serious, I hope you never reproduce

 

[Fudssa]

No it isn't.

 

[Tetsuo]

omg... I can't believe this thread is still going, I think it should go the way of the yagt's...

 

[CyraKrin]

if you look at this from the bank perspective of 1 dollar vs 99 cents:


1 dollar x 1 million = $1,000,000

.99 cents x 1 million = $990,000

Banks dont do repeating decimals either.. i think the max it goes to is 4.. but standard is 2

which would you rather have?

 

[OIKOS]

Originally posted by: CyraKrin
if you look at this from the bank perspective of 1 dollar vs 99 cents:


1 dollar x 1 million = $1,000,000

.99 cents x 1 million = $990,000

Banks dont do repeating decimals either.. i think the max it goes to is 4.. but standard is 2

which would you rather have?

....... CyraKrin..... u made a mistake by posting ur 1st post in this thread....... wat a choice!!
:|

 

[RossGr]

Originally posted by: CyraKrin
if you look at this from the bank perspective of 1 dollar vs 99 cents:


1 dollar x 1 million = $1,000,000

.99 cents x 1 million = $990,000

Banks dont do repeating decimals either.. i think the max it goes to is 4.. but standard is 2

which would you rather have?

This is not a question of bookkeeping or banking. That has nothing to do with the point of the thread. No one that knows that .999... =1 will say that .99 = 1. There is a big difference.

 

[MadRat]

Not really, it is basically the same relative argument.

 

[RossGr]

Originally posted by: MadRat
Not really, it is basically the same relative argument.

To you, who thinks infinity is simply a large finite number, perhaps so.

You might say your limited concept of infinty limits you.

 

[pennylane]

RossGr is right. .999999..... = 1

 

[Looney]

Originally posted by: Tetsuo
omg... I can't believe this thread is still going, I think it should go the way of the yagt's...

Have it's own forum?

 

[Looney]

Originally posted by: DrPizza

Originally posted by: bleeb
Proof by induction:

Given 0.9 != 1.

then

0.99 != 1.

0.999 != 1.

0.9999 != 1.

0.99999 != 1.

We conclude that

0.9999... != 1.

WHO'S YO DADDY?!

A proof similar to yours:
1 !=10
1+1 !=10
1+1+1 != 10
1+1+1+1 != 10

you conclude that 1+1+1+1+1+1+1+1+1+1 != 10

Your proof is just as absurd, except you don't recognize it as such.
Proof by induction is a valid method of proving things, and can be used to prove mathematical statements that couldn't be proved by the method of exhaustion (when there are an infinite number of possibilities). However, labeling your proof "proof by induction" does NOT make it a proof by induction, nor does it make it valid.

Here's one that's closer to yours that *maybe* you'll recognize is wrong:
.3 != 1/3
.33 != 1/3
.333 != 1/3
.3333 != 1/3
.33333 != 1/3
.333333 != 1/3
.3333333 != 1/3
.33333333 != 1/3

therefore .333... != 1/3
"proof by not paying close enough attention in math class"

OWNED!

 

[Hector13]

Originally posted by: MadRat
Not really, it is basically the same relative argument.

yeah.. an incorrect argument.

 

[MadRat]

Originally posted by: RossGr

Originally posted by: MadRat
Not really, it is basically the same relative argument.

To you, who thinks infinity is simply a large finite number, perhaps so.

You might say your limited concept of infinty limits you.

I've already shown you that you can take the argument to any finite place and so even at infinity it is true, too. You reverse polarity of the truth at infinity. I basically deny that the limit of infinity exists, which is the basis in your argument, but if it did then it would not be true anyhow.

 

[bleeb]

Originally posted by: DrPizza

Originally posted by: bleeb
Proof by induction:

Given 0.9 != 1.

then

0.99 != 1.

0.999 != 1.

0.9999 != 1.

0.99999 != 1.

We conclude that

0.9999... != 1.

WHO'S YO DADDY?!

A proof similar to yours:
1 !=10
1+1 !=10
1+1+1 != 10
1+1+1+1 != 10

you conclude that 1+1+1+1+1+1+1+1+1+1 != 10

Your proof is just as absurd, except you don't recognize it as such.
Proof by induction is a valid method of proving things, and can be used to prove mathematical statements that couldn't be proved by the method of exhaustion (when there are an infinite number of possibilities). However, labeling your proof "proof by induction" does NOT make it a proof by induction, nor does it make it valid.

Here's one that's closer to yours that *maybe* you'll recognize is wrong:
.3 != 1/3
.33 != 1/3
.333 != 1/3
.3333 != 1/3
.33333 != 1/3
.333333 != 1/3
.3333333 != 1/3
.33333333 != 1/3

therefore .333... != 1/3
"proof by not paying close enough attention in math class"

Looking at your first argument you see that you are working with numbers which are absolute on the number line. i.e. no decimals after or is exact. If you look at what I'm working with... I begin by working with decimals, and so forth. Your argument isn't quite the same.

Looking at your second argument, you are working with decimals but stop at a certain point. I'm working towards infinity. 0.333333 != 1/3. It still however is an approximation. I do agree that 0.3333.... = 1/3. However, 1 != 0.9999.... .

PWNED!!

 

[bleeb]

Originally posted by: crazygal

Originally posted by: bleeb
Proof by induction:

Given 0.9 != 1.

then

0.99 != 1.

0.999 != 1.

0.9999 != 1.

0.99999 != 1.

We conclude that

0.9999... != 1.

WHO'S YO DADDY?!

If you're serious, I hope you never reproduce

There are many people that shouldn't reproduce. Retarded people and ugly people. I'm neither.