Does .999... equal 1?

does it?

pwned

UR NOT A STUDENT OF MATHEMATIC(SINGULAR) YOU ARE A STUDENT OF MATHEMATICS(PLURAL)

MATHS MOTEHRFUCKAR

>>82 IS BRITFAG

THIS THREAD IS FUCKING DQN.

idont get it, is 0.999... equal to 1 or not?

The longest threads are ones in which you don't have to think to post.

YES .999... = 1. END OF STORY

THIS THREAD HAS PEACEFULLY COME TO AN END. THANK YOU.

>>86
you seem vexed, everything ok?

>>85 is troll

No, it's not

0.999~ not = 1.

Why, You ask? Take calculus.

It's called limits, as 0.999~ approaches infinite digits, we can declare it as 0.999 infinite, being that there are infinite nines.

Now, if 0.999 really equaled 1, then, and number subtracted from itelf is 0.

1 - 1 = 0

Anyone care to disagree?

But...

1 - 0.999~ = 0.0..01 as it approaches infinite.

You cannot properly multiply radicals 0.33~, 0.66~, 0.99~ by integers without using radians and the sort.

Also...

If done properly, 0.333~ * 3 is actually 1 because there are an infinite amount of three's there.

>>90
0.000...0001 = lim(x->inf, 1/x) = 0

>>90

0.999... = 1.

Why, you ask? Take analysis.

Real numbers are equivalence classes of cauchy sequences of the topological completion of rational numbers.

the rational sequences 0.9,0.99,0.999,... and 1.0,1.00,1.000 converge on the same number. therefore, they are the same.

qed

Anyone care to disagree?

Who gives a shit? I hate numbers.

>>90

You really shouldn't talk about infinity when you don't understand its most fundamental concept.  There is no "last" 9 in an infinite string of digits which will result in the magical transfinite ...01. 

1 - 0.999~ = 0.000~ = 0

Also...

Infinite strings of digits, such as 0.333~, are not approximations of the rational notation, such as 1/3.  They are exactly equivalent and describe exactly the same number.  0.333~ * 3 does equal 1, and does equal 0.999~, because 1 = 0.999~.

so does it?

>>93
and they hate you too

it does

This really should be common knowledge.

For the definitive answer, .999~ is equivalent to 1.

Look, does 1/3 = 0.333... or not?

If YES, then it only stands to reason that:

1/3 = 0.333...
2/3 = 0.666...
1/3 + 2/3 = 0.333... + 0.666... = 0.999... = 1

It was always true that 0.999... = 1.  Deal with it, bitches!

test
[tex]\prod_{n}^{1}\cdot cx^{n-d}[/tex]
\prod_{n}^{1}\cdot cx^{n-d}

go away.

I think you can all agree that:
0.999... = 0.9 + 0.09 + 0.009 + ... = the sum from i=1 to n of (9*10^(-i))

Let me propose that: sum from i=1 to n of (9*10^(-i)) = 1 - 10^(-n) for all natural n

Show for n=1:
sum from i=1 to 1 of (9*10^(-i)) = 9*10^(-1) = 9/10 =
1 - 10^(-1) = 1 - 1/10 = 9/10

Assume true for n=k, show for k+1 etc:
sum from i=1 to (k+1) of (9*10^(-i))
= ( sum from i=1 to k of (9*10^(-i)) ) + 9*10^(-(k+1))
= 1-10^(-k) + 9*10^(-(k+1))
= 1-10^(-k) + (10-1)*10^(-(k+1))
= 1-10^(-k) + 10^(-k) - 10^(-(k+1))
= 1-10^(-(k+1))

Proved that sum from i=1 to n of (9*10^(-i)) = 1-10^(-n)

Now consider this when n -> infinity

lim (1-10^(-n)) as n -> infinity
= lim (1-1/10^n) as n -> infinity
= 1 - lim(1/10^n) as n -> infinity
= 1 - 0
= fucking 1

You can split a circle into 3 perfectly even parts but you can't divide the number 1, 10, or 100 by 3. Interesting.

>>103
you can avoid the infinity shit if you express it as the ratio 1/3 or if you use a different base, one that is divisible by 3, such as 9.

1/3 = 1.0/3 = 0.3 using base 9.

YA

dependes on the way you build your g-adic algorithm... could be 1.

>>103

You can divide whatever the fuck you want by 3, you complete fucking moron.

By uniqueness theorem for decimal representations they cannot be equal.

No theorem required: here is counter-proof of 0.999...!=1
http://pastebin.com/0x35eiWn