Discuss

Discuss.

While it does not equal one exactly, there is a point when any rational thinker says "fuck it, close enough" and rounds it to 1.

prove it

when you round it, it is 1, but technically it is not 1

Well since you show the end of the sequence as 9999 with no ... after it, it's actually a terminating sequence and so, doesn't equal 1. I think.

indeed

1/3*3 = 1
But, 1/3 = 0.3333....
and 0.33333....*3=0.999999....
If 1/3*3 = 1, then 0.99999... = 1

.999 isn't 1
3/3 is 1

0,99...9=x
10x-x=9x
9,99...9-0,99...9=9
9=9x
X=1
Fuck off uneducated cuck

Zeno's paradox by any other name...

op is correct


you guys are idiots

his guy is correct

using maths is being idiot?
sure thing buddy

decimals do not properly represent numbers

Wrong.

1 is not 2. and these are dubs.

No, but 0.999... is

It's all good using maths unless your maths doesn't apply because OP hasn't posted a recurring decimal, just a very long one which eventually terminates.
p.s. I'm not person you replied to

In what way?

So how many .9999 would it take to be = 1?

fuck your snowflake mentality.

Fuck off trump you have a country to be ruining

0.999... = x
99.999... = 100x
99 = 99x
1 = x

Only correct answer here
Can't assume 1/3=0.3333... to prove the same idea with 0.9999...

Infinitely many

This is how the Jews turned the death of a few into the Holocaust

/thread

So what is the difference from OP's number, there are also infinitely many.... and you said that OP was right..... like, wtf ??

infinite.
probably op joke is that his 0.999....999 have a finite number of 9, but math-nerds are proud that they know something and want to explain it to feel smarter. so proud that they didn't noticed the joke.

recurring decimals don't work like that, mate.

0.999...999
and
0.999...
are not the same thing. The first sequence terminates at some point, the second does not.

>Can't assume 1/3=0.3333... to prove the same idea with 0.9999...
It's not an assumption, though. 1/3=3/10+1/30=33/100+1/300=333/1000+1/3000 and so on.

(2/9)= .222
(3/9)= .333
...
...
(7/9) = .777
(8/8) = .888
(9/9) = 1 = .9999

OP's number ends with a 9, it doesn't have a '...' sequence after the last 9. This means there is a finite number of 9's

It actually isn't 1. Human math is incorrect in assuming it is 1.

It is definitely not 1 at the quantum level.

You all are fucking stupid. it is 1. let me prove it, faggots.

.99bar = .33bar x 3

.33bar = 1/3

1/3 x 3 = 1

Therefore, .99bar = 1

The fuck are you even talking about

RED ALERT, IT'S HAPPENING. 20+ DEAD IN BOMBING AT SUPREME COURT:

>his 0.999....999

Not a joke, not at all. The last three 9's are the very last existing 9's in the whole universe, you can't get ant infiniter....... and if you can, those last 9's are the last of what you found............... So still no fucking difference.

1/3 is not 0.333333
its some number that can't be written in decimal form that is between 0.333333 and 0.333334

1 would be 1.

And 0.99999999999 etc would just be very close to 1.

Case closed.

Your species will learn eventually

0.99999.....9999 is NOT = 1
0.99999... is NOT = 1 either.

Discussing this is just being creative with maths, but doesn't make it less wrong. Infinite numbers don't share the same properties as finite ones. Everything else is bullshit.


/thread

There's nothing to discuss you fucking retard. If you had 0.period9 it'd be equal to one. This is not, because 0.99999...9999 + 0.00000...00001 equals one.

Learn2trole and lurk moar newfag.

Holy shit this thread. When did this board get loaded with so many fucking brain dead faggots? Fuck it, I'm out

But is there a number that is even more than 9?

What about pi and 1/3, you mindless otter

...

My dick is 1 inch long and yours dick is 0.999...9 inch long
I've longer dick.

You're assuming that the number is finite (0.999....999), which means that 0.999...999 does NOT equal 1 since you can always find another real number between 0.999...999 and 1 ,e.g. (0.999....999+1)/2

0.999.... with an infinte number of nines is debatable. You cannot fit a single real number, not even an infenitesimal between 0.999... and 1 which implies that they are equal. Technically there is no difference between the two numbers even though it appears that they are.

"Proof":

0.999...= 9*(1/10)+9*(1/10)^2+9*(1/10)^3+...=
using the properties of the geometric series:
= 9*(1/10)*1/(1-(1/10))=
=9*1/(10-10/10)=
=9* 1/9=
=1

what about them, retard?

They do mate

>Not a joke
>The last three 9's are the very last existing 9's in the whole universe, you can't get ant infiniter
this is not math, sry.
(you should really act like you were joking to not look like a retard, but do what you want idc)

you are correct. there is a way to prove 0.9999.... is in fact equal to 1 but not 0.99999....9999 because the sequence is finite.

You're right.
But 0.99999(9) is equal to 1

all of these 0.3333 * 3 = 0.9999 are all bs

this is just a side effect of our number system

if we use a number system that is in base 12 something like this:
0 1 2 3 4 5 6 7 8 9 a b
where 10 equals 'twelve' and doesn't equal 'ten'
then 1/3 = 0.4 2/3 = 0.8 3/3 = 1

1/3 doesn't equal 0.3333 its some number that is between 0.3333 and 0.3334 that can't be represented in base 10

Fuck, why is it so hard for you humans to understand the concept on infinity? People build theories and a bunch of crap study this matter, as it is something else than a simple idea taken out of proportion.

There is no practical use for uncountable infinity. Reduced infinity, however, as a concept is used a lot, but you won't see someone doing arithmetical operations to it. The limits of the concept is what is studied.

0.999... =/= 1
It's just 0.999...
is that so hard to understand?

You can prove 1=2 if you want to use syntax wrong.

>1/3 doesn't equal 0.3333 its some number that is between 0.3333 and 0.3334 that can't be represented in base 10
It can, that's literally what the ellipsis at the end is for.

Glad you figured that one out OP

>is that so hard to understand?
No, it's merely wrong. 1-0.999...=0, therefore 1=0.999...

Case closed. Now fuck off faggots.

yes they do.
faggot.

wtf, how can 1-0.999... be 0?

The only correct thing is that infinity can't be used.
It's just like dividing by 0

>wtf, how can 1-0.999... be 0?
If not zero, then what IS the result?

>The only correct thing is that infinity can't be used.
No, that's not correct at all. We DO have methods for dealing with numbers such as these, it's only idiots who get confused.

This

1/3 = 0.333333333333333333333
3 * 1/3 = 1 = 3 * 0.33333333.. = 0.9999999...
so
1 = 0.99999999999999999................

no you stupid cunt everything what repeats like 0.543543(543) is 543/999

We have methods to apply limits to the concept of infinity, and the operations are based around those limits. Not just plain recurring numbers like in this context.

>If not zero, then what IS the result?
Why should there BE a result? The operation is just incorrect on it's own.

If 1/0 is not 0, then what IS the result?

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

and you fucking autistic idiot virgins still dare to lie about having 200iq

>We have methods to apply limits to the concept of infinity, and the operations are based around those limits. Not just plain recurring numbers like in this context.
Recurring numbers are just artefacts of the number system. There isn't anything special about them.

>Why should there BE a result? The operation is just incorrect on it's own.
No it isn't. There isn't anything incorrect about subtracting a real number from another real number. Shit, even if it WAS infinite, subtracting an infinite number from a real number is still a valid operation. 1-infinity=-infinity.

Close enough.

1/3 = .33333
2/3 = .66666
3/3 = .99999

you cant put anything after infinity

>an infinite number of 9 has and end

It depends. If you're working with infinite decimal numbers, it's not. However if you'd work with integer numbers, 0.999~ would round down. Meaning that it would be 0.

However in regular math it could (if it needs to) be rounded up since usually the rounding of numbers goes up if it's .5 or higher in decimal value.

x=.9999...;

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

also, consider this;

x=.9999....
1-x=.0000....and what?is there a theoretical 1 at the end of all the 0s? there can't be, the 0s are unending. in other words, their cardinality = א null. you'd need to represent the position of the final 1 as an ordinal larger than that, Ω. but you can't add these types of numbers -- numbers belonging to א and numbers belonging to Ω. they exist outside of each other. they are different sets.

thus 1-x=.0000000000000... forever, and if the difference between two somethings is indistinguishable from 0, then those somethings must be the same value.

Mathematician here.
0.9(p) = 1

There is nothing to discuss in this retarded board. Just look up any analysis book you want.

The fact that there is a number infinitely repeating is what makes it different, because inifity is not rational.

1/3 = .33333
2/3 = .66666
3/3 = .99999 or 1

this is actual mathematical fact, you literally can not deny it

>The operation is just incorrect on it's own.
good lord when you have no idea about math and how decimal fractions are actually defined, just shut the fuck up

try to put another number between .99999.... and 1 i dare you

3/3 is not .99999

0.99999.....99999 != 1

limit n -> infinity Sum n=1-> infinity 9/10^n =1 != 0.9999....99999

but you've conceded that your number has finitely many 9s. .999...999=/=.999...

>The fact that there is a number infinitely repeating is what makes it different, because inifity is not rational.
Nope. Being recursive means that it is a rational number. Not infinity, despite being infinitely long in some notation. 1/3 is a perfectly normal rational number, and yet in decimal system it is ALSO an infinitely long, recursive number. The two are not exclusive in any way or form.

follow the pastern.

if 1/3 is .33333 then 2/3 is .33333+.33333 which equals 6. so following that

3/3+ .33333 + .33333 + .33333 = .99999 + 1

prove me wrong

>pastern
>.33333+.33333 which equals 6
>3/3+ .33333 + .33333 + .33333 = .99999 + 1

literally what

infinity is an idea, not a number. .999... has infinitely many zeros; its cardinality belongs to the set of real numbers, but the cardinality of the number of zeros does not. same with 1/3 and 2/3, actually. you can add numbers of the same cardinality, so something like 1+1/3 works out just fine. however, you can't make meaningful statements about the length of the trail of zeros using numbers with cardinality less than א null.

.3333 + .3333 is not 6 it's .6666
autism

You are fucking retarded if you don't see the mistake in your "educated" calculation.

Irrelevant.

Well, 1+2+3+4+5+... to infinity is -1/12, so anything is possible

it's completely relevant. you've made the argument that .999... is somehow equivalent to an infinity, but it's not. what's infinite about it is the number of trailing 0s, which hold no bearing on the number's value

.99999.... + 0.00...9

here you go

Wat? That's the OPPOSITE of the argument I've made, idiot.

I accidentally printed just 6 there which is an honest mistake, not autism. but because you just ignored everything else let me lay this out for you.

1*3=3
2*3=6
3*3=9

If you cant agree on that then you are just denying basic principals of math. lets continue

1*33=33
2*33=66
3*33=99

1*0.3=0.3
2*0.3=0.6
3*0.3=0.9

1*0.33=0.33
2*0.33=0.66
3*0.33=0.99

if you add an infinite amount of digits you will fiind that 3*0.33333 does equal .99999. even though we know that 0.33333=1/3 and 3/3 equals 1

you can put that nine at the end of infinity because that does not exsist

0.99999.....9999 is not 1, but 0.999... is 1.

0.99999.....9999 has an end, 0.999... does not.

True, only because OP is a giant faggot and is not using correct notation. 0.99...9 just means the nines continue for an unspecified amount, then end with a 9. If it was 0.999..., then it would be exactly equal to 1:

0.999... is the result of mathematical operations (e.g. resolving fractions, or division, since they are the same thing). There are fractions that do not have decimal equivalents, such as 1/3, which are represented as "0.333..." when displayed in a decimal form.

Everyone knows and accepts that 1/3 * 3 = 1. But for some reason, people don't make the connection that 0.333... * 3 = 1, and NOT 0.999....:
>1/3 = 0.333..., then 0.333... * 3 = 1.

Try another:
>1/9 = 0.111..., 0.111... * 9 = 1.

The ellipsis is simply a mathematical tool for showing a decimal that does not end, which is ALWAYS a decimal equivalent to a much simpler fraction.

0.333... ALWAYS means EXACTLY 1/3. Therefore 0.999... ALWAYS means EXACTLY 1.


Alternatively:
>0.999999...
>equivalent to: 9/10 + 9/100 + 9/1000 + 9/10000 + ...
>equivalent to: 9/10 + (9/10)(1/10) + (9/10)(1/10)^2 + ... + (9/10)(1/10)^n as the integer n goes forever
>can be represented by the series summation( (9/10)(1/10)^n) with n going from 0 to infinity, meaning: (9/10)(1/10)^0 + (9/10)(1/10)^1 + ...
>the sum of this series is the first value in the series divided by 1 minus the multiplier ratio.
>(9/10)/(1-1/10) = (9/10)/(9/10) = 1

Someone should thread this, and we can be on our way.

>1/3=3/10+1/30=33/100+1/300=333/1000+1/3000 and so on
did you know you're a retard?