.999~ = 1, Agree?

[ElaineMonster]

I say no.
I also say this belongs in the math forum.

[mrkilla]

no it can't ever reach the #1 thus why its infinite, its trying to be

.9 is not 1; neither is .999, nor .9999999999. In fact if you stop the
expansion of 9s at any finite point, the fraction you have (like .9999
= 9999/10000) is never equal to 1. But each time you add a 9, the
error is less. In fact, with each 9, the error is ten times smaller.

You can show (using calculus or other methods) that with a large
enough number of 9s in the expansion, you can get arbitrarily close to
1

SO in Math it does = 1 , however in reality it doesn't but its accepted that it is. So is the accepted answer its 1 , then yes, is it really no.

[durron597]

There are a billion proofs of this, here's YET another one (op already listed 2)...

Clearly, .99999~ * 2 = 1.9999999~

subtracting .99999~ from both sides gives

.99999~ * 1 = 1
.99999~ = 1

[CallMeIshmael]

[ QUOTE ]
I think the real question is:

What does LMK stand for?

[/ QUOTE ]

I'll be honest, and say the first thing I thought was "Lartin Muther King"


Also.... OMFG THEY ARE EQUAL

[TiK]

[ QUOTE ]
This is silly.

.9999999999 ---> infinity is the number just below 1.

No number or amount is between them, but neither is the same.

[/ QUOTE ]

The OP's proof says otherwise, I think he was asking you to disprove it if you said "No" in the poll.

Personally, 3/3 = 1 = .99999999999...

Even if you do it long division-wise. Try it where instead of just saying 3 goes into 3 once, try it with .9, you'll see it works both ways.

[MrTrik]

I work in a very scientific field. It happens to be aerospace.

Thinkin like this:

Proof 1:
1/3 = .333333...
2/3 = .666666...
1/3 + 2/3 = .999999... = 1.

Will end you doing a Major effing Tom in space (Bowie) rather than landing on the moon.

So, my answer is no.

Sincerely,
Trik.

[graydot]

[ QUOTE ]
no it can't ever reach the #1 thus why its infinite, its trying to be

.9 is not 1; neither is .999, nor .9999999999. In fact if you stop the
expansion of 9s at any finite point, the fraction you have (like .9999
= 9999/10000) is never equal to 1. But each time you add a 9, the
error is less. In fact, with each 9, the error is ten times smaller.

You can show (using calculus or other methods) that with a large
enough number of 9s in the expansion, you can get arbitrarily close to
1

SO in Math it does = 1 , however in reality it doesn't but its accepted that it is. So is the accepted answer its 1 , then yes, is it really no.

[/ QUOTE ]

I dont think you know what infinity is. You cannot just stop a number, it has no end.

[ QUOTE ]

.9 is not 1; neither is .999, nor .9999999999.


[/ QUOTE ]
.9 is not; neither is .999, nor .99999999.
But .9~ is 1.

[mmbt0ne]

[ QUOTE ]
Proof 2:

x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.


[/ QUOTE ]

I'm not positive about this, but wouldn't this be true with infinite decimals?

9.999999999 - .999999999 = 9
9 - .999999999 = 8.0000000001
10x - 2x = 8.000000000001
10x-2x = 8x
8x = 7.9999999992

8x =\= 10x-2x

y=1

10y = 10
10y - 2y = 8
10y - 2y = 8y
8y = 8

8y = 10y - 2y

[CallMeIshmael]

[ QUOTE ]
I'm not positive about this, but wouldn't this be true with infinite decimals?

9.999999999 - .999999999 = 9
9 - .999999999 = 8.0000000001
10x - 2x = 8.000000000001
10x-2x = 8x
8x = 7.9999999992

8x =\= 10x-2x

y=1

10y = 10
10y - 2y = 8
10y - 2y = 8y
8y = 8

8y = 10y - 2y

[/ QUOTE ]

Bold is wrong

[Grivan]

Because of the properties of the real number line there is no number just below one. In the real numbers the notion of "next" does not exist. If I give you any real number and ask you what the next one is there is no valid answer. This is because there is no bijection from the real number line to the natural numbers. Which also means that the real number line is uncountably infinite, but I doubt most people will know what that means.