0.999... = 1

[byates5637]

Come at me *******s

[jono28613]

you almost made it to the store = you made it to the store.


strong logic OP.

******* as usual.

[BeastlyChange]

Okay

[dobsqc]

Op boyfriend almost touched his balls with his balls = Op boyfriend made balls contact

OP is a *******

[DragonRyan]

0.999... = 1

seems legit

[babrah]

Come at me *******s

= =/= ≈

.999...≈1

[Mosho]

you almost made it to the store = you made it to the store.


strong logic OP.

******* as usual.

OP is correct. 0.999... with infinite 9's is 1, this is as true as 1+1 = 2

strong comparison to making it to a store though, retard

= =/= ≈

.999...≈1

sorry, this is false. thanks for trying to sound smart though.



ITT calculus not even once

[DavidJr74]

you almost made it to the store = you made it to the store.


strong logic OP.

******* as usual.

brb doesn't know sh*t about decimal expansion. Let me give you a example buddy:

1/3 = .333 forever
2/3 = .666 forever

1/3 + 2 / 3 = 1

Therefore 1=0.999 repeating forever.

[byates5637]

nah, equals bro, equals. As in the same exact fuking thing.

[DasWunderBrah]

limits not even once

[Archie824]

brb doesn't know sh*t about decimal expansion. Let me give you a example buddy:

1/3 = .333 forever
2/3 = .666 forever

1/3 + 2 / 3 = 1

Therefore 1=0.999 repeating forever.

this

[Jecoma]

So you're saying lim x=>1 (x) is 1?
yeah.

[shawty321]

lmao i watched a 30 min video on this a couple years ago when i was high as fuuuuuu

mind was blown

[BbaLLbrAH]

Proofs:

1/9 = 0.111111111...
9*1/9 = 9 * 0.111111111...
1 = 0.999999999...

Or
x = 0.999999999...
10x = 9.999999999...
10x - x = 9.999999999... - 0.999999999...
9x = 9
x = 1

I should change my name to MatHbRah.

[Nathan17]

seems legit

made you green feelsgoodman

[jb4476]

I can't believe people actually argue about this. You stupid ****s actually know math right?

The expression "0.999..." is shorthand for "the limit of the infinite sequence {0.9, 0.99, 0.999. 0.9999,...}". The nth term in the sequence is \sum_{i=1}^n(9*(10^(-i))). The distance between 1 and the nth term in the sequence is just 10^(-n). So for any epsilon>0, you can find an N such that for all n>N, the nth term of the sequence is within epsilon of 1. Therefore, 0.999... is 1. Because that is the definition of the limit of a sequence. This is math. You don't get to argue about trivial ****. 0.999... is equal to 1 because 0.999... is defined to be the limit of a sequence and it is easy to show that the limit of that sequence (by the definition of a limit) is 1.

[SigmundFreud]

.999.... is NOT equal to 1. it is .0...001 with an infinite amount of 0's with a 1 on the end.

So it is equal to 1. But it is a hipster 1, you nomsayin?

[Lucretius]

Try to find a number between 0.999... and 1. You can't. So they're the same number.

[kdelamont]

OP, pls go.

[xtraviolent]

you almost made it to the store = you made it to the store.


strong logic OP.

******* as usual.

It isn't "almost", he made to the store. As written, "..." denotes that he arrived already. So it's actually:

he made it to the store = he made it to the store

Which is true.

[M1LD_S]

.999.... is NOT equal to 1. it is .0...001 with an infinite amount of 0's with a 1 on the end.

So it is equal to 1. But it is a hipster 1, you nomsayin?

Oh lawd stick to your phallus studies

[thqmas]

1:59 isnt 2:00 same as 0.999 isnt 1

[Rayzor84]

Come at me *******s

According to calc 2 ur right brah...

[xtraviolent]

.999.... is NOT equal to 1. it is .0...001 with an infinite amount of 0's with a 1 on the end.

So it is equal to 1. But it is a hipster 1, you nomsayin?

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

You were probably not being serious but this should /thread it.

[awasay]

old news, anyone who disagrees, do you even math?

[iiradman]

I can't believe people actually argue about this. You stupid ****s actually know math right?

The expression "0.999..." is shorthand for "the limit of the infinite sequence {0.9, 0.99, 0.999. 0.9999,...}". The nth term in the sequence is \sum_{i=1}^n(9*(10^(-i))). The distance between 1 and the nth term in the sequence is just 10^(-n). So for any epsilon>0, you can find an N such that for all n>N, the nth term of the sequence is within epsilon of 1. Therefore, 0.999... is 1. Because that is the definition of the limit of a sequence. This is math. You don't get to argue about trivial ****. 0.999... is equal to 1 because 0.999... is defined to be the limit of a sequence and it is easy to show that the limit of that sequence (by the definition of a limit) is 1.

You're just using the workings of a method as proof or evidence that 0.999... = 1.

I always think back to asymptotes. There will be lines in a graph which will NEVER hit a certain number, based on the equation. It will always be 1.9999 or 5.9999 but NEVER 2 or 6.
The fact that these infinite sequences will never develop into whole unit numbers dictates their ultimate fate...of staying a decimal.

Fuk yea. Na but i actually don't know. Both arguments make sense?





EDIT: It's all good and dandy until the universe goes full potato and throws an '8' in there or something. Then what??

THEN WHAT!!??

[poundfive2]

.9999.... is indeed = to 1.

i believe you learn this in the first 2 weeks of taking calc 1 when the teacher goes over basic limits.

[Efre]

but

limit =/= actual value

you can try approaching 1 all you want, .9999 will never be actually 1

[Nkukk]

.999.... is NOT equal to 1. it is .0...001 with an infinite amount of 0's with a 1 on the end.

So it is equal to 1. But it is a hipster 1, you nomsayin?

how the flying **** can you have an inf amount of a number

then have another number at the end?

[poundfive2]

but

limit =/= actual value

you can try approaching 1 all you want, .9999 will never be actually 1

does .6666666.... = 2/3?