Is 1 = 0.9999......

[jinduy]

once and for all...

1 != .9999999999999999999999999......9

>

 

[flexy]

Originally posted by: RossGr
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?

common sense... ?

And no, 0.99999.. does not equal 1

Btw . i never doubted the 'existence' of eg. 0.999999... in a strict mathematical context.
So is 0.333...+0.333..+0.333..=0.999... also legitimate as a mathematical formula.

But you WILL (and this thread of ~1900 posts is proof for it) run into problems when you mix and/or compare real numbers and numbers which are only an approximate.

0.999... is the best possible approximate to 1, but it IS not 1...otherwise you must be kidding me

That's pre-school knowledge....

 

[flexy]

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.

Can you write this 'fixed' number down, here, in this thread with all its digits ? (And not the ABSTRACTION of it !) I guess not

 

[opticalmace]

What a great thread.

 

[flexy]

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

what's 0.333... then ? or any other periodic number for that matter ? 0.333... can obviously NOT be 0.333[inf]4...so why should 0.999... be 1 ?

 

[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?

 

[DrPizza]

Originally posted by: flexy

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.

Can you write this 'fixed' number down, here, in this thread with all its digits ? (And not the ABSTRACTION of it !) I guess not

Actually, I can. Here it is:
(drum roll)

1

Ta dahhhhhh!

Or, did I misinterpret you. You want us to write down .999999999... with 9's going on forever and forever? And, since we can't, do you believe that you've provided proof that .9999... doesn't equal 1? Well, in that case, I've got a square that's exactly 3 inches on each side. And, since the diagonal isn't changing its length, it's length must be a fixed number. Could you write down all the digits of the fixed number for me? I guess not. Oh wait, that must mean that the diagonal can't exist for a square?!

Actually, the diagonal would be 3 times the square root of 2. The square root of two is a non-terminating decimal. It goes on forever and forever and forever. It would be ridiculous for me to think that I could stick a 1 or a 2 or a 3 at the end of it, after it repeats forever, wouldn't it? Hopefully that points out how retarded it is to think that I could stick a 1 at the end of a string of zeros (0.000000000.......) that repeat forever. And, as a bunch of people keep incorrectly insisting, the difference between .999999... and 1 is a bunch of zeros with a 1 at the end, hopefully you might now realize they're wrong.

 

[cpals]

Dang this is a long thread....

*exits quietly.*

 

[Einz]

Ok people, I don't know if this counts or if you guys care or not, but I asked a few math profs here at MIT and they have all told me that .9 repeating is equal to 1. Now if you guys are going against MIT profs.. then yeah...

 

[TGregg]

Originally posted by: ElFenix
gah! people who know nothing of number theory trying to bend their minds around it is horrible! this is as bad as the "you're on a game show and you've shown 3 doors and behind 2 are nothing and behind the other is a million bucks" question

I figured out a good way to convince 50/50 folks of the right answer on that one.

Me: "OK, let's play. I'm the contestent and you are the host, and I'm gonna get em all right."
50/50: "OK"
Me: "Except, we'll play with a billion doors. First, you pick the door (a number between one and a billion). Then, I'll guess a number between 1 and a billion and you open all the doors except mine and one other (and you cannot open the door you picked). Then I'll switch to the unopened door."
50/50: "OK. I picked my door."
Me: "I pick door number 435,988,242."
50/50: "So, door number 283 and . . . uh . . . the one you picked are still closed."
Me: "I switch."
50/50: "You got it right."
Repeat a few more times.

Of course, the only way I can lose by switching is if I somehow guess the same number as the 50/50 guy picks, a literally one in a billion chance.

EDIT: And for the record, 1 does indeed equal 0.999...

 

[WhiteKnight]

Originally posted by: Shanti
Everyone who's taken calculus should know that it approaches one but isn't equal to one.

And everyone that's gotten through calculus and then taken analysis should know that it does equal one.

Edit: From The Way of Analysis by Robert S. Strichartz.

On in finite decimal expansion: "Because of the familiarity of infinite decimals, this proposal is quite appealing. However, it has two technical drawback. The first is that the decimal expansion is not unique: .999... and 1.000... are the same number. This is usually met by the ad hoc requirement that the decimal cannot end in an infinite string of zeros. The second drawback is that it is somewhat awkward to define addition and multiplication, because long carries could change earlier digits."

 

[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

 

[MAME]

Originally posted by: Joker81

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

Joker, with a comment like that it is OBVIOUS you are not qualified to argue about math. There is no such thing as infinity - 1. There is no such thing as 1.000...1.

 

[MAME]

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 1.000...1 does not exist

 

[cmaMath13]

Being a mathematics teacher, I think it's awesome that this many people are discussing the value of 0.999...

 

[MAME]

Originally posted by: cmaMath
Being a mathematics teacher, I think it's awesome that this many people are discussing the value of 0.999...

And, according to you, are they equal?

 

[xEDIT409]

5=7

cuz see

is 5+2=7
and 7-2=5

therfore

5=7

hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah hahaahah

 

[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.

 

[Hector13]

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.

Are you saying the two numbers are equal?

just to clarify the notation, say that .000..1 = x (if it were to exist). Are you saying x/2 = x? Or are you somehow saying that x/2 < x (which constradicts the fact that x exists to begin with).

 

[Spencer278]

Originally posted by: Hector13

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.

Are you saying the two numbers are equal?

just to clarify the notation, say that .000..1 = x (if it were to exist). Are you saying x/2 = x? Or are you somehow saying that x/2 < x (which constradicts the fact that x exists to begin with).

I'm saying that the two numbers below are equal
.0000...1
.0000....1/2
in fact the two numbers below are equal for any x and y.
.0000....x
.0000....y

 

[RossGr]

I'm saying that the two numbers below are equal
.0000...1
.0000....1/2
in fact the two numbers below are equal for any x and y.
.0000....x
.0000....y

So according to your logic .001 = .0005

Please enlighten us.

 

[Hector13]

Originally posted by: Spencer278

Originally posted by: Hector13

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.

Are you saying the two numbers are equal?

just to clarify the notation, say that .000..1 = x (if it were to exist). Are you saying x/2 = x? Or are you somehow saying that x/2 < x (which constradicts the fact that x exists to begin with).

I'm saying that the two numbers below are equal
.0000...1
.0000....1/2
in fact the two numbers below are equal for any x and y.
.0000....x
.0000....y

so what happens if x (or why) in your example is 0? Are you saying then that .00...1 = 0?

 

[conjur]

Jeezus....let this thread archive already!!!

 

[Spencer278]

Originally posted by: RossGr

I'm saying that the two numbers below are equal
.0000...1
.0000....1/2
in fact the two numbers below are equal for any x and y.
.0000....x
.0000....y

So according to your logic .001 = .0005

Please enlighten us.

I didn't come up with the retard idea of writing numbers as . then repeating zeros terminated with a number but if that even made senses they would be equal if the numbers where looked at in the base of the highest number as .00000...1 and .0000...1 .
like if we have .0000....1 and .000...5
in base 2 that would be
.0000...1
.0000...01
Because there is an infinit number of zeros infitite + 1 is still equal to infinite.

 

[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