If 0.999...=1 then does 1.888...=2?

DEATH eVADER · Jun 16, 2006

bbson_john said:
ahhh. We mean 0.999... with infinite 9.

Same thing, it would just take infinite time to change those infinite row of .9s into a 1

Solaris · Jun 16, 2006

Oh stabby, f**k off.
I was just reading about how 0.999...=1 yesterday so I was just wondering what 0.999x2 =. Sorry if I'm not a master of maths, jesus. I like maths it's fun, but cummon, no need to be a prick simply becuase I didn't understand the concept.

Could we delay the lock until ikerous gets here please, I just want to understand it a little more.

Mr Stabby · Jun 16, 2006

calm down, it's only the internet

Dan · Jun 16, 2006

Qonfused · Jun 16, 2006

*Looks at thread name...*

Solaris · Jun 16, 2006

Oh sh*t lol.
I'm so obviously wrong.....

Please forgive my stupidity D: D: D:

;(

jondy · Jun 16, 2006

We forgive, but we never forget. Or forgive.

Raeven0 · Jun 16, 2006

DEATH eVADER said:
Generally speaking, in maths 0.999 is not equal to one.

0.999 is nine hundred ninety-nine thousandths. It is obviously less than one, because

0.999 + 0.001 = 1

The magical number to which we refer is called point nine recurring.

POINT NINE RECURRING IS EXACTLY THE SAME NUMBER AS 1.
THERE IS NO DIFFERENCE BETWEEN THEM.

http://www.qntm.org/pointnine
http://www.qntm.org/pointnine
http://www.qntm.org/pointnine
http://www.qntm.org/pointnine
http://www.qntm.org/pointnine
http://www.qntm.org/pointnine
http://www.qntm.org/pointnine
http://www.qntm.org/pointnine
http://www.qntm.org/pointnine
http://www.qntm.org/pointnine

TheSomeone · Jun 16, 2006

Dan said:
The problem with what you're doing is that 9.999... - 0.999... doesn't equal 9. It equals 8.999... Because when you multiply by 10 you are moving the decimal over. You now have infinity minus one numbers following the decimal place

If 9.999... - .999... = 8.999..., haven't you just proved that .999... = 1?

Yes you have, yes you have.

JellyWorld · Jun 17, 2006

oh god not another one

http://en.wikipedia.org/wiki/Proof_that_0.999..._equals_1

Read especially the order proof.

bbson john · Jun 17, 2006

Come on, Let this stupid thread diezzzz.

The Monkey · Jun 17, 2006

Also, if one number is greater than another, there will always be a number between them. Can you find a number between 0.999... and 1?

Steven · Jun 17, 2006

Everyone knows that 1.999999999999999999999 = 2, there just trying to make you angry.

The Brick · Jun 17, 2006

2.777... = 3 ?

3 = white shoe, so how can white shoe be 3 and 2.777... ? Your theory is flawed.

bbson john · Jun 17, 2006

The Brick said:
2.777... = 3 ?

3 = white shoe, so how can white shoe be 3 and 2.777... ? Your theory is flawed.

No! 3.666666= 4? 1.999=2 is wrong :|

:frog:

Steven · Jun 17, 2006

1 = 2

Mr Stabby · Jun 17, 2006

0.99999... x 3 = 2.999999...9997 which = 3
0.99999... x 4 = 3.999999...9996 which = 4

JellyWorld · Jun 17, 2006

Mr Stabby said:
0.99999... x 3 = 2.999999...9997 which = 3
0.99999... x 4 = 3.999999...9996 which = 4

0.999999999999... x 4 is not 3.9999999996

You can't have a final 6, because there's no final digit, it's an infinite recurring decimal.

0.999999 * 4 = 3.999999 (or 4, whichever way you like it).

Mr Stabby · Jun 17, 2006

there is a final digit, it is just never reached

JellyWorld · Jun 17, 2006

Mr Stabby said:
there is a final digit, it is just never reached

NO

There is no final digit.

There is no final element in an infinite sequence. You have to let go of your intuitive notions, there is no such thing as a digit that is never "reached".

Mr Stabby · Jun 17, 2006

I disagree

15357 · Jun 17, 2006

2 + 2 = 5

INGSOC.

War is Peace.

DreadLord · Jun 17, 2006

Mr Stabby said:
I disagree

You don't have that right in Mathetmatics.

Kyorisu · Jun 17, 2006

It's a flaw in our number system get over it. This is what happens when mankind developes a system to understand things? Unlike say physics which doesn't purley rely on stupidity.

Ikerous · Jun 17, 2006

JellyWorld said:
0.999999999999... x 4 is not 3.9999999996

You can't have a final 6, because there's no final digit, it's an infinite recurring decimal.

0.999999 * 4 = 3.999999 (or 4, whichever way you like it).

Actually, 0.99999... x 3 = 2.999999...9997 is a perfectly valid statement

Although 0.999... x 3 = 2.999...n where n is any integer (In the above case 9997)

JellyWorld · Jun 17, 2006

Kyo said:
It's a flaw in our number system get over it. This is what happens when mankind developes a system to understand things? Unlike say physics which doesn't purley rely on stupidity.

It's not a flaw in any way.

JellyWorld · Jun 17, 2006

Ikerous said:
Actually, 0.99999... x 3 = 2.999999...9997 is a perfectly valid statement

Although 0.999... x 3 = 2.999...n where n is any integer (In the above case 9997)

'tis not.

Kyorisu · Jun 17, 2006

Positive infinity, negative infinity and zero for great justice!

Ikerous · Jun 17, 2006

JellyWorld said:
'tis not.

Do i really have to write up a proof? :|

JellyWorld · Jun 17, 2006

There's a proof in my analysis book (that it's not possible), but i'm lazy to type it out.

Ikerous · Jun 17, 2006

A proof for what? That .000...n != 0?

JellyWorld · Jun 17, 2006

A proof that 0.000000000000....n can't even exist... Actually it's not really a proof, it's an immediate consequence of the Archimedean property of the real numbers.

Also, 0.000000000...n cannot even be defined, since by definition a decimal number of the form

0.x1x2x3x4x5x6....

is equal to the limit of the sequence

{0.x1, 0.x1x2, 0.x1x2x3,0.x1x2x3x4,...}

(damn i hate not having LaTeX).

Kyorisu · Jun 17, 2006

Alternatively you could give the great number debate up. I myself find this hilarious but still.

Ikerous · Jun 17, 2006

JellyWorld said:
A proof that 0.000000000000....n can't even exist... Actually it's not really a proof, it's an immediate consequence of the Archimedean property of the real numbers.

Also, 0.000000000...n cannot even be defined, since by definition a decimal number of the form

0.x1x2x3x4x5x6....

is equal to the limit of the sequence

{0.x1, 0.x1x2, 0.x1x2x3,0.x1x2x3x4,...}

(damn i hate not having LaTeX).

But couldnt you just change the limit series proof of .999 = 1 to

Code:

                 n=N
          = lim (Σ  9·10-n) + c·10^-(N+1)
            N→∞ n=1

          = lim 1 - 10-N + c·10^-(N+1)
            N→∞

          = 1

In the first equation with N=1 and c=7 you get .97, with N=2 you get .997, with N=3 you get .9997. So as N goes to infinity, couldnt it be represented by '.999...7'?

I mean, it wouldnt be the best way to represent the series, but it seems to make sense.

Raeven0 · Jun 17, 2006

Kyo said:
Unlike say physics which doesn't purley rely on stupidity.

Quote of the year.

Yuri · Jun 17, 2006

Raeven0 said:
Quote of the year.

Due to the irony of the statement?

TheSomeone · Jun 17, 2006

Erestheux said:
Due to the irony of the statement?

I hope. That was the dumbest shit I've heard in this thread.

DreadLord · Jun 17, 2006

Erestheux said:
Due to the irony of the statement?

I would hope so.

"This forum requires that you wait 30 seconds between posts. Please try again in 1 seconds." Hahaha.

Raeven0 · Jun 17, 2006

Erestheux said:
Due to the irony of the statement?

Quite.

Yuri · Jun 17, 2006

Just making sure

If 0.999...=1 then does 1.888...=2?

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