.999 repeating equals 1

.999 repeating equals 1

prove me wrong faggots
protip: you cant

it equals 9/11

>This number is the same as this different one

observe image

var number = [];
while (i =1; i < 10 ){
var total = .9
number.push(total)
if ( number === 1 ) {
console.log("youre correct");
} else {
dear diary op youre a faggot and wrong
};

Your proof sucks. It implies that after multiplying by 10, you're still left with a remainder of 0.999... with exactly the same amount of infinite repeating digits as before you multiplied by 10 even though you shifted the decimal.

syntax error

multiplications dont leave remainders you mong

9/11 = .818181 repeating

x=0.9999...
10x=9.9999...
10x-x = 9.9999.... - 0.9999....
Fx=9
x=9/F

Lrn2Hex

exactly, syntax error. so !=

Not only did you forget to incriment i, but

>implying a program can account for infinitely repeating digits

The point is if you had 0.999 and multiply it by 10 then you have 9.990 killing off the last spot. Your proof relies on multiplying by 10 which someone can argue will give you 9.999...0. So when you do 10x - x = 9.999...0 - 0.999... then you are left with 8.999...

left off the increment, because it's infinite... derp

there is no last spot, it's (((repeating)))

if you remove one thing from an infinite set it's still infinite

number can't === 1 because it's an array you fucking moron. Also progressing languages can't deal with infinities and they're should be an infinite number of 9s

>while (i = 1; i < 10) {

k

My best proof of this that leaves nothing to chance is using 1/3. If you do the long division yourself, you can easily see that 1 divided by 3 is 0.333...

So you easily know that 1/3 = 0.333... Multiply both sides by 3 and you have 3/3 = 0.999... or 1 = 0.999...

No one argues it's not infinite. The problem is that two things with infinite elements can be proven to not have the same amount of elements. It's why if you have 2 functions with limits that tend to infinity, then doing infinity - infinity is not 0 but is indeterminate form.

So knowing that, someone can argue that the order of magnitude of the infinite set after multiplying by 10 is smaller than the magnitude before multiplying by 10 even though both are infinite.

yeah no

Can't prove it's wrong because its write.

its just another way of writing 1 user. You are right.

>The problem is that two things with infinite elements can be proven to not have the same amount of elements.

correct

>It's why if you have 2 functions with limits that tend to infinity, then doing infinity - infinity is not 0 but is indeterminate form.

doesn't have anything to do with previous sentence

>So knowing that, someone can argue that the order of magnitude of the infinite set after multiplying by 10 is smaller than the magnitude before multiplying by 10 even though both are infinite.

countably infinite set - finite set = countably infinite set

Of course we can't prove you wrong. That's because 0.9 recurring *is* equal to 1 in the reals. Real numbers don't just have a single unique decimal representation.

Just say that it's the sum of a geometric progression. So, for example: 0.99999999... = 0.9 + 0.09 + 0.009 +... to infinity. Then use the result which is a/(1-r) = 0.9/(1-0.1) = 0.9/0.9 = 1

The point is after multiplying 0.999... by 10, then I argue that even though right of the decimal point still has an infinite number of 9's, it no longer has the same amount of 9's as prior to you multiplying by 10 and thus you can't necessarily perform that 9.999... - 0.999... in such a way that it perfectly becomes 9. I argue it will become 8.999... and thus your proof doesn't prove anything. I do know that 1 = 0.999... but I don't think that is a good proof.

>I argue it will become 8.999...

correct, 8.999... equals 9

this

It is "correct" but it obviously proves nothing because that becomes circular logic where you're trying to prove 0.999... equals 1, and yet in order to say 8.999... equals 9 then you have to apply what you're trying to prove in the first place.

except both are true if you use
which isnt circular

1/3=.3333...
2/3=.6666....
3/3=.9999....
3/3=1

OP is correct.

Here is an alternative proof. Say two real numbers a,b are equal if there is no number c between them, that is, there is no c with a

The point I was making was that OP's original image tries to prove it but doesn't do a good job proving it. If you have to apply a different way or proving it then OP's proof isn't a good proof. I prefer the 1/3 = 0.333... then multiplying both sides by 3 proof the best. No one can argue that multiplying an infinite number of 3's would give you anything other than 0.999...

...

>it no longer has the same amount of 9's

This is incorrect. Both have an infinite (but countable) number of 9s. Therefore they both have the same amount of 9s (aleph-null).

i agree here, image isn't a proof, just a demonstration. geometric series argument is tho

if you plot 0.9999... on a graph it falls below the 1 mark

The best proof is the simplest proof. Many of these proofs are relying on applying rules of math that the average high school grad wouldn't know or remember in everyday life. There are better ways or proving which don't rely on you knowing anything other than the basic rules of addition and multiplication without having to have a deep understanding of infinity.

>infinite
>thinking all infinites are equal

All countably infinite sets have the same cardinality m8. Both numbers have the same amount of 9s.

The problem is infinite is not a number and numerical operations get fucked by it.

It's essentially a representation problem.
Draw a circle on your board. You cut it up in 2 halves and you will say the halves are smaller and you can express it in both volume as well as length. Now you can change the representation into points in R2 but here the ammount of points would be infinite and cutting it in half still gives you infinite points. So which one is right? Does the halfed circle contain half of the full one or is it equal like the second example would suggest? Math is fucked as soon as you introduce infinite.

There are also more interesting examples like the sum of N| being -1/12 through likewise fuckery -and this is a value actually used in physics.

1=1

0.9999 = 0.9999

0.9999 =/= 1

.5 + .5 = 1

.5 + .5 =/= 0.9999

All infinities, through positive to negative AND the inbetween infinities always have a set.. that set is "everything", so to take the set and make a new number that isn't within that set is another way of escaping that infinity to the decimals.

In other words, its an axiom of infinity.

...

true
true
false
true
false

Oh I get it. You were just trolling me. Well played, you got me to bite.

I gotta go to bed now though. I have to be up early for my group theory lecture.

You're too dumb to understand the advanced math required to realize this is correct, even if OP's image is a bad example

This works for any base but different numbers, its the whole infinite convergent sums. In base 2, 1 = .1111111111_ #1 10=1.11111111_ #2 10 - 1 = 1.1111_ - .1111111_ (#2 - #1) => 1=1

This
Fuck this thread

its not correct.

the idea of 0.9999999 (repeating) is that there can never be enough 9's to write after that 0.9, it is only endless 9's..... symbolizing the absolute number just before 1, and furthest from 0, but not 1.... because 1 is well defined as being equal to 1.

0 as = to 0.

9 as equal to 9.

0.9 as equal to 0.9.

0.999 as equal to 0.999.

0.999999 repeated as equal to 0.999999 repeated.

1 as equal to 1. Simply the rules and definitions.

0.9999 repeating is not equal to 1. Because 0.9999 repeating is not equal to 1.

> symbolizing the absolute number just before 1
there is no number just before 1

>0.9999 repeating is not equal to 1. Because 0.9999 repeating is not equal to 1.
nice circular logic

Its converges to 1, and thus indistinguishable.

1 = .9999999_ (statement) #1
10 = 9.999999_ (10*#1) #2
10 - 1 = 9.999999_ - .999999_ (#2-#1) -> 9 = 9 -> 1 = 1

Limits prove this as well. L2basiccalculus. Also, any number of digits divided by nine are the decimal of those digits repeating infinitely.

lel top kek m8

Kek take this shit to /sci/ see what they think

The problem is your a fucking retard and syou purposely made one side foul it makes your proff work dont be a fucking mong.

3/3 = 1 fuck the infinate peice of 3.3333... Is suppose to imply that when you take the other 2/3 of it you get 1.

they would agree

learn to type retard

also are you trying to claim that repeating decimals dont exist

wat

no, number places are well defined, there is no converging, there is only a, b, c, d, e, f, g, ad infinitum, none are equal to each other. They increase in equal increments, in the tiniest possible increment if youd like.

the 4th line, where it says 9x = 9

9x = 9 times x.

x = 0.9999

9 times x, or 9 times 0.99999 = 8.991

not 9 as written

it's 0.99.. repeating not 0.99999 terminating

>prove me wrong faggots

Of course we can't...that's the definition of infinitely repeating patterns.

You basically said "1=1. Prove me wrong."

Why are you such a faggot, OP?

>x=0.999....
>(insert penis into a flat surface)
>×=1
>lamao yall try prove my wrong

just imagine having a cube (any material)
now imagine you've removed one atom from the cube

still have a cube? Of course you fucking do you retardos

0.999... = 9/9
9/9 = 1

nice quads