Let's say I have 0.9 repeating, eg 0.99999999999.....so on.
0.9999999.... times 10 equals 9.9999999.... right? Therefore, wouldn't 0.9999999.....times 9 equal 9? But 1 x 9 = 9. Therefore 0.999.... = 1.
Let's say I have 0.9 repeating, eg 0.99999999999.....so on. 0.9999999.... times 10 equals 9.9999999.... right?
what is about decimal representation that rapes people's brains so much
but am i right though my good anon
yes 0.999... is 1, they're decimal notation for the same number (assuming you define an infinite decimal expansion as the limit of a series)
infinitely long decimal expansions involve limits even though it doesn't say limit on the page.
I prefer the nested interval definition of infinite decimals.
1.414... is the unique real number in [math][1, 2] cap [1.4, 1.5] cap [1.41, 1.42] cap [1.414, 1.415] cap ldots[/math]
yeah, turns out its pretty silly to write then entire limit of a series out every single time when anybody who passed calculus (a tremendously large number of people) knows what is meant by the symbol
Only midwits are bothered by stuff like
. It's a waste of time and intelligence to contemplate.
This might be true for any integer, but multiplying by 10 just moves the decimal to the right. You can imagine infinite zeroes behind the decimal of an integer.
1/9= .1111111111...11112
8/9
=0.888888888888888...889
so with 9/9, at some point the 9 decimal becomes a 10 an so the 9 decimal before that becomes a 10, until the last decimal becomes a 10 and so the number becomes 1.0
therefor .999....9 =/=1 but 9/9=1
For 1/9 = 0.111.. the last number would not be 2 but 1/9th of a number
2/9 = 0.222.. 2/9
8/9 = 0.888... 8/9
9/9 = 1
The last number would be like a vertical bar that fills lol (thrembo????)
>wouldn't 0.9999999.....times 9 equal 9?
0.9x9=8.1
8.1 + 0.81 + 0.081 + ...
so 8.999...
But anon
0.999.... x 10 = 9.9999...
0.999.... x 9 = (9.9999....) - (0.999....)
therefore 0.999... x 9 = 9
But anothnon
0.999... x 10 = 9.999... ..90
but that's wrong
you gonna multiply 9 x 10 and you are happily ignoring the other part of the answer which is 0 but happily use the other part that is 9
brilliant
no, that's silly and makes no sense anon, if you have ten instances of 0.999.... that is 9.999....
proof?
>no, that's silly and makes no sense anon
Said anon and equated an infinite steps into finite
9.999... - 0.999... = 8.999...1
wrong tho
9.999... - 0.999 = 9
9.999... - 0.999... = 9
Thats true, however
0.999... x 10 = 9.999... ..90
Which is not the same as 9.999...
>0.999... x 10 = 9.999... ..90
it's unironically actually 10
Anything multiplied by 10 ends in 0. This is basic, anon
1/3 * 10 = 3.333333333...
>it’s true because my teacher said so!
So 1/3 does not end in 0?
>10π
lol, frick you
π = 1
(in base π)
no it isn't you fricking moron
Yes it is you gay homo.
it's 10 you shitstain smoothbrain
Yes if you pretend the base π happens inside base 10
holy shit you're stupid
what a waste of cum
Maybe you are confused so let me put it clearly
π (1)
ππ (2)
πππ (3)
>clearly
lol
your brain is mush, give up
Math is hard
Medsnow.
medium tier bait, not bad
1/9 = 0.111...
2/9 = 0.222...
3/9 = 0.333...
4/9 = 0.444...
5/9 = 0.555...
6/9 = 0.666...
7/9 = 0.777...
8/9 =0.888...
9/9 = 0.999...
9/9 = 1
0.999... = 1
>assuming what you’re proving
sheep
No. You have 1/9 = 0.111... thanks to the division algorithm. Then, you can do
9 * (1/9) = 9 * 0.111... = 0.999...
The only thing he is assuming is certain continuity properties of the operations involved, but that's acceptable.
i would say 1/9 = 0.111... is not entirely correct
you cannot do fractions into decimals totally without some tricks like 0.111... ..1/9
>i would say 1/9 = 0.111... is not entirely correct
then I'd call you a fricking moron with no proper argument
>you cannot do fractions into decimals totally without some tricks
long division is not a "trick" LOL
yeah but you never 'finish' the division. there's always a little bit more work to do until you're done. As an algorithm it approximates continuously better what the actual answer would be if it could be represented in decimal, but it never gets to the right answer.
>let's pretend finite is infinite
you're just pretending that infinite is finite.
>i can't read
shocker
what is nine divided by nine?
.99999999 ≈ 1
What?
Obviously, no.
First, you can't do math with infinite numbers like that.
To understand, let's imagine it's finite.
A = 0.9999...9999
10 * A = 10 * 0.9999...9999
10 * A = 9.9999...9990
(10*A) - A = 9.9999...9990 - 0.9999...9999
(10*A) - A = 8.9999...9991
So, no, 0.999... is not equal to 1.
>First, you can't do math with infinite numbers like that.
You can. /thread
Hum...
Then, if A = 0.999... and B=1
What's (A+B)/2 ?
It's both A and B?
yes
I understand the way of thinking, but I still dislike way it's used...
Maybe I will see the light sometime...
Strange how a finite number can be expressed as infinite too.
No, I am just trying to understand.
I was always told that this reasoning was false because it mixes finite operators and infinite objects.
So I try to understand, that's all.
Great, so A<C<B.
But, as A is infinite, it's infinitely close to B.
So, like a limit, if you find a gap, you didn't go far enough.
Lim of A toward infinite is 1, and so is limit of C.
Ho well, it's possible, both just get closer and closer together I believe...
>I was always told that this reasoning was false because it mixes finite operators and infinite objects.
0.999... is literally equal to 1 so any operation that can be performed on integers can also be performed on 0.999...
Well these are two different numbers, so being able to say they are equal... Well it took some time to me to understand it.
I see no benefits of learning dumbly if I don't understand what's the inner working behind it.
But very well then, I will try to wrap my mind around the fact that integer operators can be used on infinite numbers...
Thank you very much for your compassion Anon.
I work on myself to better understand what I learnt, and I am destroying what i think I new to learn new things.
No need to be edgy about it.
Yeah, that's what I have a hard time understanding.
You put it nicely.
See?
This confusion is what I mean by "you can't use integer operators on infinite numbers".
0.999... is infinite. You will never ever have anything else than a 9 after the 9.
Theorically, multiplying 0.999... by 10^infinite will give you infinite, because anything multiplied by infinite is infinite.
But by using integer's operators, 0.999... * 10^infinite gives ...999.999... which is not the same at all.
Hence my confusion.
>But very well then, I will try to wrap my mind around the fact that integer operators can be used on infinite numbers...
The "+" in 0.999... + 1 isn't an operator on integers, it's an operator on real numbers.
There are many equivalent ways of defining what real numbers are and what the operations on real numbers are. One of my favorites is the nested interval model. I can make some posts explaining how it works.
You know how you can't find a rational number that squares to 2, but if you assume a number exists that squares to 2, you can figure out that it has to be between 1 and 2? You can go further and figure out that it has to be between 1.4 and 1.5, and then that it has to be between 1.41 and 1.42, and so on. You can carry this process out forever.
Whenever we have a infinite sequence of statements like
[math]1 le x le 2[/math]
[math]1.4 le x le 1.5[/math]
[math]1.41 le x le 1.42[/math]
and so on forever,
then each of the statements is a "closed interval" telling us where to find the number,
and we say that this series of statements defines a real number provided that
(1) the endpoints of the intervals are rational numbers
(2) the intervals are nested, meaning the endpoints of any interval are included in all the intervals that come before it
(3) the intervals converge to zero width, meaning that for any fraction 1/n, n being a positive integer, there is an interval in our sequence whose width (second endpoint minus first endpoint) is smaller than 1/n.
If we have two infinite sequences of nested intervals like this, and they overlap at each stage, we say the two infinite sequences define the same real number.
When we write an infinite decimal like
[math]pi = 3.14...[/math]
what it means is that
[math]3 le pi le 4[/math]
[math]3.1 le pi le 3.2[/math]
[math]3.14 le pi le 3.15[/math]
and so on forever.
We can add [math]sqrt{2}[/math] and [math]pi[/math] to get another real number.
[math]1 le sqrt{2} le 2[/math] and [math]3 le pi le 4[/math], therefore [math]4 le sqrt{2} + pi le 6[/math].
[math]1.4 le sqrt{2} le 1.5[/math] and [math]3.1 le pi le 3.2[/math], therefore [math]4.5 le sqrt{2} + pi le 4.7[/math].
[math]1.41 le sqrt{2} le 1.42[/math] and [math]3.14 le pi le 3.15[/math], therefore [math]4.55 le sqrt{2} + pi le 4.57[/math].
Other operations can be defined similarly.
It's a great way to approximate results.
so, A = 0.999...
0 ≤ A ≤ 1
0.9 ≤ A ≤ 1.0
0.99 ≤ A ≤ 1.00
So:
9.99 ≤ 10*A ≤ 10.00
(9.99 - 0.99) ≤ (10*A) - A ≤ (10.00 - 1.00)
9.00 ≤ (10*A) - A ≤ 9.00
Kek, funny.
...
Shit I have to re-learn everything now..
(x^2)^0.5 does not equal (x^0.5)^2
(x^n)^(1/n) is only negated with odd numbers of n.
>higher dimensional 'space time shit' requires even dimensions to explain curvature
kek math
0.99 ≤ A ≤ 1.00
and
9.99 ≤ 10*A ≤ 10.00
would actually give us
(9.99 - 1.00) ≤ (10*A) - A ≤ (10.00 - 0.99)
8.99 ≤ (10*A) - A ≤ 9.01
But we still have 9*A = 9 since
8 ≤ 9*A ≤ 10
8.9 ≤ 9*A ≤ 9.1
8.99 ≤ 9*A ≤ 9.01
...
overlaps with
9 ≤ 9 ≤ 9
9 ≤ 9 ≤ 9
9 ≤ 9 ≤ 9
...
at every stage in the sequence, making the numbers equal.
The version that uses addition instead of subtraction (pic related) works out a bit easier.
Of course we can also simply note that
0 ≤ A ≤ 1
0.9 ≤ A ≤ 1.0
0.99 ≤ A ≤ 1.00
...
overlaps with
1 ≤ 1 ≤ 1
1 ≤ 1 ≤ 1
1 ≤ 1 ≤ 1
...
at every stage, making A = 1.
And both overlap with the sequence of intervals defining B = 1.000...
1 ≤ B ≤ 2
1.0 ≤ B ≤ 1.1
1.00 ≤ B ≤ 1.01
...
at every stage.
So we have 0.999..., 1, and 1.000... all equal to each other.
What is an equation that might come up in the wild wherein .999.... Is the result, and if .999... = 1, why did it show up in the .999... Form and not as 1?
What kind of equation would result in: .999999999999999999(many many 9's)99999999999999...8
And the next number after that is simply 1?
If you take an apple and cut it in 3 seperate parts;
.333...
.333...
.333...
If you cut the same apple in 4 equal parts; .25, .25, .25, .25
So this is a situation with odd numbers;
That is due to? Something about the number system, of decimals?
1/3, 1/3, 1/3 = 3/3 = 1
Apparently there 3 parts can perfectly equal 1.
But due to the decimal system it doesn't work so smoothly?
Because no 3 equal numbers equals 1.
The closest you can get to getting 3 equal numbers to equal 1, is throwing a bunch of .3s on the end of .3
Theres no way to put some series of numbers after .333333
That when added to 2 others, gives you 1. All the candidates or 3 numbers to give you 1, are .333
It would just need a single .1 to push it over the edge to 1, but we run into the same problem we are trying to solve;. How to evenly build that 1? It can be done.
If we sum 0.333... + 0.333... + 0.333... carefully, we also get the result of "0.999... or 1.000...".
Let x = 0.333..., meaning
[math]0 le x le 1[/math]
[math]0.3 le x le 0.4[/math]
[math]0.33 le x le 0.34[/math]
[math]0.333 le x le 0.334[/math]
and so on.
Then multiplying by 3, we obtain
[math]0 le 3x le 3[/math]
[math]0.9 le 3x le 1.2[/math]
[math]0.99 le 3x le 1.02[/math]
[math]0.999 le 3x le 1.002[/math]
and so on.
At the second step, we have enough information to conclude that
[math]0.9 le x le 1.0[/math] or [math]1 le x le 2[/math]
meaning our result in decimal form starts with either 0.9 or 1.
At the third step, we know that
[math]0.99 le x le 1.00[/math] or [math]1.0 le x le 1.1[/math]
meaning our result in decimal form starts with either 0.99 or 1.0.
At the fourth step, we know that
[math]0.999 le x le 1.000[/math] or [math]1.00 le x le 1.01[/math]
meaning our result in decimal form starts with either 0.999 or 1.00.
I just realized that 10 is a placeholder for 0, as 11 starts as the 1 again. So the idea of 0 is kept up and up.
I stead of
5
6
7
8
9
11
12
13
14
15
16
17
18
19
21
This is kinda what is being implied;
That 9 goes to 1.
Or 10. Which is ""0"".
It's 10 spots from 0 to 10.
9 spots 1 to 10.
Well it's literally "0 units" and 1 group of what didn't fit before. Let's call it tens.
1 ten, and 0 units.
What if you got 9 tens and 9 units, but need to add 1 unit again?
well, easy, you make units go back to 0 as usual, and...
Ho...
We need a new name to do the same thing with tens.
Let's call it hundreds.
1 hundred, 0 tens, 0 units.
And so on until you invent scientific notation...
>this is a situation with odd numbers
it's a situation with numbers that don't like base 10(That is, they never divide evenly with 2^x*5^x). 5 is odd and 1/5 is 0.2, simpler than 1/4.
Oh, is it about 3 equally odd numbers cannot get you an even?
1 is an odd.
3 equally odd numbers can get you an odd number.
3, 3, 3 is 9.
.333, .333, .333 is .999
Can 3 equal odd numbers equal 10? Or 100?
No it's the same problem;
3 + 3 + 3 = close to 10 as can get.
33 + 33 + 33 = close to 100 as can get.
.333 .333 .333 close to 1 as can get;
With 3 equal odd numbers.
?
>3 equally odd numbers can get you an odd number.
>3, 3, 3 is 9
>333 .333 .333 close to 1 as can get;
1 is an odd number.
An even number of odds or an odd number of evens can make it up.
.5 .5
.2 .2 .2 .2 .2
Yaaaasssssss?
>.333 .333 .333 close to 1 as can get;
Now it's interesting that;
9 goes to 10.
While .999 goes to 1.
As 10 is an even number.
And 1 is an odd number.
That's not that intersting, consider .999 goes into all odd and even numbers; as in;
6.999 goes into 7
And 7.999 goes into 8.
For the difference between 7 and 8 is 1.
As 7 is 7; 1's and 8 is 8; 1's
And the whole, 6.1... 6.2... 6.3... 6.4....6.8... 6.9... 7
Is just counting from 0 to 10; creating the 1 that is the difference between 6 and 7
Every number is made of 1('s) and every 1 is made of 10.
it's not from oddness, it's from it not dividing a power of 2 or 5. All even things do, but some odd things(Anything divisible by 5) also do.
if you look at how long division works with 1 as the numerator, then it's pretty clear. Either the division is completed, which means that some power of 10 is divisible by the number(And so, some power of 10 has all the factors of the number) or it isn't completed, and so no matter how high the power of 10 is, it won't share factors with that number.
10 is 2*5, so any number that consists of factors 2^x*5^y will eventually divide 10, and any that have prime-numbered factors other than 2 or 5 won't.
>will eventually divide 10,
will eventually divide (a power of) 10,
>even things do, but some odd things(Anything divisible by 5) also do.
Because 5 may as well equal 2, or 1/2
If 10 is 4
5 is 2
This whole situation may seem to be due to the 1d ideal point like nature of Numbers.
And then trying to finitely demarcate mesurement distances to them, and then also trying to squeeze infinitesimals at every point between any number.
>Because 5 may as well equal 2, or 1/2
>
>If 10 is 4
>5 is 2
I'm not sure I follow. Is this about ratios?
It seems to me that 'if 10 is 4, 5 is 2' is true under a certain ratio (4/5ths), but I'm not sure what that means to you. If you're speaking in the context of that, you introduce 2 as 0.8, and when you scale everything down you should find that '0.8 or 2' would have the same effect, although it complicates things for a couple reasons.
The reason it's anything divisible by 5 is because 5 is one of the factors of 10, the other being 2, and the completion of division is based on the numbers having common factors, when raised to some power of 10.
>This whole situation may seem to be due to the 1d ideal point like nature of Numbers.
>And then trying to finitely demarcate mesurement distances to them, and then also trying to squeeze infinitesimals at every point between any number.
This is an interesting idea. Do you have demonstrations of this in mind?
>So what does this say about the .999... quandry
>.333... + .333... + .333... = 1
it depends on what one means by the '...'.
It's worth asking someone involved in this conflict whether they consider '.333...' in base 10 to be the same as 0.1 in base 3. If so, then 0.333... = 1/3, so 3*.333... = 3/3 = 1, but if they don't(Because it is a decimal division that never ends, and therefore can't be equal to an endpoint that cannot exist), then 3*.333... should be understood to be 'infinitesimally short' of 1.
>(4/5ths)
Sorry, I meant 2/5ths.
I think I was pondering, in base 10, 10 is the limit of a row.
5 is an odd number; but evenly in the center making 2 equal halves up out of 10.
Imagine 10 equal bricks in a line on the ground. There is a line split down the middle with 5 on one side and 5 on the other.
If you start from 1 on the left and count right, the 5th brick will be a certain brick.
From 10th brick counting to the left, a different brick would be the 5th brick.
But when counting from 0:
0..1..2..3..4..5..6..7..8..9..10
5 is perfectly in the middle.
I think there is something to this that may be relevant to the .999... Quandry, where numbers don't line up evenly
>>This whole situation may seem to be due to the 1d ideal point like nature of Numbers.
>>And then trying to finitely demarcate mesurement distances to them, and then also trying to squeeze infinitesimals at every point between any number.
>This is an interesting idea. Do you have demonstrations of this in mind?
Well first off, it starts with considering that 1 is a perfect idea of a single whole, the idea of 1 is the foundation of it all, a singular perfectly equal object; by which if you imagined another exactly exactly like it, then you would have 2 of the exact exact same objects.
583 is then just 583 of those objects.
Then to organize the number system and I realized this nessecity due to convienience;. Imagine 500 Different words to label the numbers between 1 and 500; ebauh and kfjd and keirj and ueud and 300 numbers later, the number eoifjd and eidurjd and eirje and ofjej
No. We have twenty, thirty, forty
Twen...two..... Thir....three, third.....for....four
200 and 30
700 and forth four
So base ten is a labeling system convienience. To reuse the same terms, to make a pattern to not have every single new addition of +1 +1 +1 +1 be a brand new word term.
So all quantity is rooted in 1. As opposed or in relation to 0, the absense of quantity.
(Now I dont know how related to our experience in reality this came about by, we have 3 apples, 1 and 1 and 1, but nature is not perfectly ideal so the apples are not equal so these 1's are not equal; but anyway, you can cut an apple in half, so the 1 was actually made of more than 1 the whole time; the ideal version of 1 would be a 1d point; but then that would imply indivisibility)
Imagine a string of mountain peaks like ^^^^^^^^^^^^^^^
It's like each tippy 1d pin prick point is a number, and in-between the peaks are the equal distance between the numbers;
Cntd pondering
Cntd
I mean yeah I don't know, .999.. is like getting near the tippy tip peak, and then getting a ladder, and keeping climbing straight up to the moon, or perpendicular to the line of peaks, once you get to .9.
As it's easy enough imagining 1 2 3 4 5 6 7 8 9
It's easy to imagine
.1 .2 .3 .4 .5 .6 .7 .8 .9
But in the journey from peak to peak, when you arrive at .9 which is like to 1 what .1 is to 0;
To consider .99 or .999 or .999999999999999999
There comes a point where it doesn't make sense that you could still possibly be moving closer to the peak at all.
This is the mixing of ideal and physical reality, ideally anything that can be imagined is valid. In ideals sense is not required, nothing is held or holding back, whatever is said goes.
I can say .999..., But I don't know the meaning of this in reality.
1 is an ideal concept and a real concept, 1 pebble and 1 pebble are not equal, but more similar than 1 pomegranate, and 1 pomegranate contains 100 seeds and a billion atoms, in fundamental physics it's trying to get to the parts that are indivisible; hopefully a quark and electron cannot further be divided in parts; of the pomegranate these would be more akin .00000001s than 1s, that make the pomegranate 1.
See the number 1 can be viewed as a single pomegranate, and the number 1 can be viewed as being full of infinite atoms.
So in this sense, maybe it goes back to, 0 and 1;
0 is the idea of there existing absolutely No reality;
The ultimate idea of 1 as opposed to 0 is the Whole that is the Uni(one)Verse
This 1 ultimate fact of collection total existence; is composed of near infinite parts.
So the 1 is at once looked at as the most possibly finite. And the most possibly infinite.
1 is composed of infinite decimal parts.
Each of the infinite decimal parts is precisely 1 occurance of that part.
The symbol and idea of 1, is the symbol and idea of singularity, singular, a reference to refer to uniqueness
So what does this say about the .999... quandry
.333... + .333... + .333... = 1
>What is an equation that might come up in the wild wherein .999.... Is the result, and if .999... = 1, why did it show up in the .999... Form and not as 1?
It's common for a result to be "either 0.999... or 1.000...".
Here's an example with "either 1.999... or 2.000..."
First, take the square root of two by successive decimal approximations,
[math]1 le sqrt{2} le 2[/math]
[math]1.4 le sqrt{2} le 1.5[/math]
[math]1.41 le sqrt{2} le 1.41[/math]
[math]1.414 le sqrt{2} le 1.415[/math]
and so on.
Then multiply the result by itself, obtaining
[math]1 le sqrt{2}^2 le 4[/math]
[math]1.96 le sqrt{2}^2 le 2.25[/math]
[math]1.9881 le sqrt{2}^2 le 2.0164[/math]
[math]1.999396 le sqrt{2}^2 le 2.002225[/math]
and so on.
At the second step we can say
[math]1 le sqrt{2}^2 le 2[/math] or [math]2 le sqrt{2}^2 le 3[/math]
meaning that the decimal representation of our result starts with either 1 or 2.
At the third step we can say
[math]1.9 le sqrt{2}^2 le 2.0[/math] or [math]2.0 le sqrt{2}^2 le 2.1[/math]
meaning that the decimal representation of our result starts with either 1.9 or 2.0.
At the fourth step we can say
[math]1.99 le sqrt{2}^2 le 2.00[/math] or [math]2.00 le sqrt{2}^2 le 2.01[/math]
meaning that the decimal representation of our result starts with either 1.99 or 2.00.
Yes, that's what I understand.
But to my knowledge, the '+' on infinite number didn't have the same rules as the '+' on integers of real numbers.
That's what I have to wrap my mind around.
>nested interval model
Like in this post
?
If so, I understand and it can be elegant.
But I am not used to work with these.
Of the integers? 2
Of the rationals? I don't know
Of the reals? I don't know
Of the infinite? Great questions. Probably 2-0.999...
>Strange how a
It's almost as if 0.9...=1
I think the fundamental thing to understand is that real numbers are defined in such a way as to ignore infinitesimal differences between quantities. "Infinitesimal" meaning having an absolute value smaller than 1/n for any positive integer n.
I know I am 10 days late, but what gets me is that the same people that would gladly accept there is some infinitesimal difference, so that 0.99... =/=, are the same ones that would refuse to believe in different sized infinities, but you need both for surreal numbers.
Its mostly just the same thing like with black holes like who the f defines them to be HOLES and not BALLS..
Same thing with different "sizes" of infinities
Infinite is infinite (never ending) and thats it
Well, the average of 1 and 1 is 1 so
Not sure what kind of gotcha moment you were expecting there.
That would be B-A = C
C+A=your answer
take your pills asap
>A = 0.9999...9999
a=1
>10 * A = 10 * 0.9999...9999
10a=10
>10 * A = 9.9999...9990
10a=10
>(10*A) - A = 9.9999...9990 - 0.9999...9999
10a-a=10-1
>(10*A) - A = 8.9999...9991
10a-a=9
>to prove my point about infinite numbers, let's imagine it's finite lmao
Ah, yes, the average 4chinz mathematician
>First, you can't do math with infinite numbers like that.
>To understand, let's imagine it's finite
The definition of infinite is: that which does not and cannot exist; that which is continually changing/growing informationally.
>definition of infinite
An unbounded quantity that is greater than every real number.
But this is also infinite .111...
Come onz your camp knows there is more than one meaning of infinite.
But all meanings of infinite include the characteristic: incomplete, unfinished, unfinishable, unstoppable
TIL that infinity is a bit less than 0.12
ty anon, your Fields medal is in the mail
There are different infinities silly. But thanks for the medal anyway.
Finite means descrete, of a specific extent, complete.
Nonfinite means, not that.
I will forward your opinion to the highest authorities in the field.
Thank you for your contribution.
Math has nothing to do with authority, therefore:
>highest authorities in the field.
Not a thing
Froghomosexual I will disembowel you
>0.999... decimal
>1 integer
morons.
What about 1.000...?
That would be
>1.000...
then wouldn't it?
What number is 1.000... equal to?
What does
>.000...
represent?
Hopefully 0. Though I wonder what the [math]0.999...ne 1[/math] would say.
>what is the smallest number greater than 1
2
Unless you want a real number. Then there's no such number.
so then 2 = 1.99.. = 199999999.../ 9999999.... = inf/inf ?
1 = 1/1 = 1000/1000 = 10... / 10... = inf/inf
Pretty silly. Your point being?
>there is a fundamental flaw in math operations
>a hammer is very bad at driving screws
>the obvious conclusion is that the hammer is flawed
>the irony of implying math is the hammer without realizing it
ngmi
Not math. The division operator. Imagine using the division operator outside of its domain and being smug that it doesn't work.
Even if you extend the reals and allow for operations on infinities, infinity over infinity is still undefined.
>0.9999999.... times 10 equals 9.9999999.... right?
Wrong its
>0.999....0
You don't just "create" a new number out of nothing; after the infinite set of 9s there will be a 0 at the end.
>after an infinite set of 9s
Say that again but slowly
muh meaningless formal manipulations
but worse since neither of you understand you can't write down an infinite series of 9s in the first place
You can't but that doesn't stop you from knowing that an infinite series does not have a last element so the last element of the supposed series of 9s cannot be 0 or whatever.
There actually will be last element
0.999...
since its repeating 9's there cannot be 0,1,2,3,4,5,6,7 or 8 and as those options are completely used the last element will be 9 but the size would still be infinite and you dont know the position of that last element
>wrong
i dont care
There is no last number. It's not at an unknown position, it's at no position.
Assume there is a last number and its position is nth digit after the comma. The number has an infinite number of decimal digits so there must be more digits than any number n. Therefore it can't be in any position.
If there is no last number it cannot equal to 1 also
If there were last number it would be 9
Anyway im not here for any arguments, im just discussing
it's cope to justify representing 1/3 as 0.333...
0.9 repeated = 1
0.9 repeated * 9 = 1 * 9 = 9
0.9 repeated * 10 = 9.9 repeated = 10
Not sure what the issue here is
0.99999 can't be 1 because it takes more than a number to write it down.
1 = 1
0.9999.... = 0.9999....
A continuous infinite decimal multiplied by an integer would give you another continuous infinite decimal, generally except if you have placed some specific rules down, then you'd follow those.
kinda funny that infinite repetition and infinitely complex both uses "..."
3.141...
0.999...
totally different meanings
which is why math books prefer
[math]
0.overline{9}
[/math]
I was wondering if there was simpler way like calling:
3.14... (2 numbers, 3 dots)
and
0.999.. (3 numbers, 2 dots)
but thats just stupid in technical terms, both would need their own symbolics imho, like 0.999''' and 3.141...
That upper dash works but idk how someone without a math backround would view it without any context
The upper dash notation only shows infinite repetition. You can't write [math]pi[/math] in terms of a repeating decimal expansion because that would make it a rational number. You just write [math]pi[/math]
If 1/3 = o.33333.. If 1/3 exists and we are in a simulation do you think it would have a rounding error or a buffer overflow? Otherwise 0.3 repeating doesnt exist because it would be rounded up by the simulation.
>1/3 = o.33333..
1 / 3= .3... + .0...1 / 3
(.3... + .0...1 / 3) * 3 =.9... + .0...1 = 1
.3... * 3 = .9...
.9... < 1
just add 0.0....1 because it makes my point make sense what are you a physicist? Just keep adding stuff until it works.
1/3 = 3/10 + 1/30
= 0.3 + 1/30
= 0.33 + 1/300
= 0.333 + 1/3000
:
= 0.3... + 1/inf
= 0.3... + 0
= 0.3...
I geuss fractions brush away the issue, fractions are an ideal idea,
3 Cavemen could have come across a snake to eat and said, 'we each get one even piece', 1/3 1/3 and 1/3.
But with no tape measure, and even if they had one we would see this pesky difficulty, they would only aproximate to the best of their ability. Or find a single stick that fit across the snake body 3 even times, or a blade of grass, to use to proportion equally.
Using a unit of 1 we hope everything can be evenly measured and proportioned by it. But because our macro scale, unable to see the minut distances of the atoms of the snake, it's hard to demarcated perfectly evenly.
So why is splitting 1 into 3 equal pieces less perfect than splitting 1 into 4 equal pieces?
Because the proportional balence of the base 10 number system, has the relation between 0 and 10,
It is like laying bricks; easier to evenly stack 4 bricks 2 on top of 2, making sure their edges are level; than 3:
2 on the bottom, and 1 in top perfectly absolutely equally in-between the 2
>wall-of-text schizo
errytime
Quote the parts you didn't get or don't agree with
wall
The .999.. situation may be due to:
If you have one (1) rectangle cake, and you cut it into 3 even pieces (this is to your physics reality question) 1/3, 1/3, and 1/3
.3, .3, .3
.333..., .333.., .333...
The missing amount is removed from the equation! By the need to remove volume from the whole complete 1, by the act of cutting with the knife, crumbling a blades width of material.
For real though, what changes between the perfect working representation of 1/3 1/3 1/3 .. 3/3
And the decimal? Is it something with 0 and base 10 that make this .999.. awkwardness?
I geuss after all this thread I would agree that .999.... Is not a possible number in any sense of the concept, and that it is just the signifying of a 1.
Decimals aren't real, it's realtard schizophrenia
Only fractions exist
>Decimals aren't real
Proof it
You don't have to proof anything. 0.999... equals 1 by definition. They're just different notations of the same number.
1 = 0.999... = 9/9 = e^2πi = etc
Playing algebraic games doesn't prove anything because it's all tautological.
I swear you can recognize 80% of the recurring threads just by lurking for a single month
>0.999
>Schizo ramblings
>3 coinflips
Just off the top of my head
0.111... = 9
Checkmate, math chuds.
0.999... = 0.999...
Why can't we just leave it like this?
Why do we need to do additional math frickery just to create a proof that triggers a few mathtards every now and then?
having 0.333... = 1/3 is pretty important dude
What really is the difference between .9 and .999999999999999?
If you showed that difference distancely, spatially, on a ruler, how large of a difference would that be?
And then how much more distance would .999999999999999999999999999999999999999 be?
The more you add the less difference in distance it becomes right?
So at some point the difference in distance practically stops
.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
I don't know, just never allow the term .999... To be written, only allow 1 to exist in its place.
1/3 + 1/3 + 1/3 does not equal .999... It equals 1
[math]
1 = dfrac{3}{3} = 3 cdot dfrac{1}{3} = 3 cdot 0.bar{3} = 0.bar{9}
[/math]
1/3 ≥ 0.333...
Nice try tho, almost fooled me
Isn't the whole point of decimals showing how the system of decimals is not equally equal to the system of fractions?
How do you prove that
3 x 1/3 = 3 x 0.3....?
How is the decimal system proven and grounded? I know us anons say a lot about this topic, but what did the best mathematicians of history say about this .999... Funny business?
1/3 = 0.3...
so does this mean that
1/2 = 0.4999... ?
or do you not realize how fricking stupid that notation is
that is exactly the point
0.a(9) = a.(9) / 10 = (a + 1) / 10
if a =4, then we have 0.4(9)=(4 + 1) /10 = 5/10=1/2=0.5
See https://en.wikipedia.org/wiki/Repeating_decimal#Converting_repeating_decimals_to_fractions
If you want to define 0.999... as undefined, you can. But I would say
is a simpler and more useful way to define what an infinite decimal means, and it says that 0.999... = 1.
A proper computation of 3 * 0.333... gives you 0.999... or 1.000... as in
>A proper computation of 3 * 0.333... gives you 0.999... or 1.000... as in
Let x = 0.333..., meaning
[math]0 le x le 1[/math]
[math]0.3 le x le 0.4[/math]
[math]0.33 le x le 0.34[/math]
[math]0.333 le x le 0.334[/math]
and so on.
Then multiplying by 3, we obtain
[math]0 le 3x le 3[/math]
[math]0.9 le 3x le 1.2[/math]
[math]0.99 le 3x le 1.02[/math]
[math]0.999 le 3x le 1.002[/math]
and so on.
At the second step, we have enough information to conclude that
[math]0.9 le x le 1.0[/math] or [math]1 le x le 2[/math]
meaning our result in decimal form starts with either 0.9 or 1.
At the third step, we know that
[math]0.99 le x le 1.00[/math] or [math]1.0 le x le 1.1[/math]
meaning our result in decimal form starts with either 0.99 or 1.0.
At the fourth step, we know that
[math]0.999 le x le 1.000[/math] or [math]1.00 le x le 1.01[/math]
meaning our result in decimal form starts with either 0.999 or 1.00.
or more properly: 0.999... or 1.00...
When we say 1/3 = 0.333... we do not mean some gay metaphysical shit about infinity, we mean that you can approximate 1/3 to whatever precision you need with 0.3 followed by a sufficient number of 3's. The notation 0.333... represents the unique number which can be approximated this way. Likewise, when we say 0.999... we mean the number that can be approximated to whatever precision you need with 0.9 followed by sufficiently many 9's. This number is 1.
>approximate
your dad is approximate
0.3... is exactly 1/3
Yes. 0.333..., meaning the number that you can approximate to any precision desired by 0.3 followed by a sufficient number of 3's, is exactly 1/3.
Oh I think I get it.
Base 10 is really base 9.
The 10 is the begining of the next series.
So 1/3 + 1/3 + 1/3 = 3/3
1/3 = .3
.3 + .3 + .3 = .9
.9 = 1
0 1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18 19
20 21 22 23 24 25 26 27 28 29
30
>0 1 2 3 4 5 6 7 8 9
>10 11 12 13 14 15 16 17 18 19
>20 21 22 23 24 25 26 27 28 29
I think there is something to this; 9 is the end of the road. So reaching 9 is a full set of the first row of base 10 numbers.
10 is the "0" place holder of the next row.
When 9 is reached; 1 count of base has been occured and reached.
I called it out a bit ago, this is some: the 15th century is the 1400s funny business
Is this the solution to the quandry?
I can't tell what
and
are trying to express.
.9 = 1
Because in base 10:
0 1 2 3 4 5 6 7 8 9 = 1 row of the base.
10 11 12 13 14 15 16 17 18 19 = 1 row ontop of the first base
1 is not the first number;
Because one itself requires the distance from 0 to reach 1, so we so don't start counting At 1, we start counting at 0 to 1, the distance space inbetween 0 to 1 is the first number. And once you begin to get on the other side of 9, you enter 10 space.
Because the 10th number is made of the stuff between 9 and 10
Okay this is better than
but you should work on using words to mean the same things other people understand them as and using different words if you need to express a different concept. For example what you are calling the "10th number" you ought to call an interval rather than a number.
To write down the correct explanation I think you are getting at:
Infinite decimals starting with 0.9 represent numbers in the space starting at 0.9 and ending at 1.0. We call a space like this an "interval."
Infinite decimals starting with 1.0 represent numbers in the interval starting at 1.0 and ending at 1.1.
We can denote the interval starting at 0.9 and ending at 1.0 as [0.9, 1.0]. The square brackets indicate the starting and ending numbers are included. If we instead wrote (0.9, 1.0), that would indicate that 0.9 and 1.0 were not included. We can also write things like [0.9, 1.0), which indicates that 0.9 is included but 1.0 is not included.
In this notation, an infinite decimal starting with 0.9 represents a number in the interval [0.9, 1.0].
An infinite decimal starting with 1.0 represents a number in the interval [1.0, 1.1].
The number 1, being at the boundary point, is in both intervals. It is at the end of [0.9, 1.0] and at the beginning of [1.0, 1.1]. It is the last number which can be represented by an infinite decimal starting with 0.9 and the first number that can be represented by an infinite decimal starting with 1.0. So it has two infinite decimal names, 0.999... and 1.000... .
In general, every number that can be represented exactly as a terminating decimal has two infinite decimal names. For example, 21.624 has the names 21.623999... and 21.624000... .
A picture to show how infinite decimals work.
1=1/1
1=(3*1)/(3*1)
1=3/3
1=(1/3)+(1/3)+(1/3)
1=0.33...+0.33...+0.33...
1=0.99...
What was the process by which the system of decimals was proven to have it's relation to fractions?
1/3 = .3
Says who, how? Anything ca be written and by convention check out, what are the axioms that prove this?
1/3 [math]neq[/math] .3
1/3 [math]approx[/math] .3
specifically .3 < 1/3 < .4 making |1/3 - .3| < 1/10
more specifically
1/3 - .3 = 1/3 - 3/10 = 10/30 - 9/30 = 1/30
1/3 - .33 = 1/3 - 33/100 = 100/300 - 99/300 = 1/300
1/3 - .333 = 1/3 - 333/1000 = 1000/3000 - 999/3000 = 1/3000
What's the number between 0.9 repeating and 1? There isn't any. So they're the same
>source: real analysis
missing 0.00000000...1
0.00000000...1 meaning what?
meaning 0.9999... does not equal to 1
>missing 0.9999... does not equal to 1
I missed it because it's not true
off by 0.00000...1
What is 0.00000...1?
better question is why would you use decimals?
use fractions
[math]frac{1}{3} neq 0.3333... \
frac{1}{3} approx 0.3333...[/math]
0.3333... means the number which 0.3, 0.33, 0.333, 0.3333 ... approximate to any desired precision, which is exactly 1/3.
Decimals are used because they make arithmetic calculations easier.
1/3 = 0.333...
2/3 = 0.666...
3/3 = 1
There is no such thing as 0.999...
It doesn't exist
https://en.wikipedia.org/wiki/0.999...#Algebraic_arguments
Hey, there it is
So there's a theorem that says if there are two numbers with no other possible numbers between them then they are the same number. So with that, .9999...=1 and 9x1=9
So all the numbers are the same number?
huh?
>if there are two numbers with no other possible numbers between them then they are the same number.
Literally every number would then fit this description
..8 being both ..7 and ..9, and so forth
huh?
For that proof to make sense you must first show that every real number has a decimal representation and no two different numbers have the same decimal representation.
also that comparing two infinite decimals by the usual method gives the same results as comparing the numbers that have those decimal representations
How can it equal one? If 0.999... equals one that means that 0.000...1 is of no value because one minus 0.000...1 has no effect. 0.999... is less than one because 0.9 is less than one, so what's 0.999...? Should be less than one, but we're told it's one, and that cannot be.
>How can it equal one?
because 1/inf=0
Once you accept that, the rest is just bookkeeping
Fricking newbies. Go back.
found the idiot
0.999... is a pointless notation because any equation that tries to output that will just equal 1
Easiest example is 3/3
It doesn't equal 0.999..., it equals 1
Same with 1/3 + 1/3 + 1/3, it just equals fricking 1
"0.000...1" doesn't mean anything
0.1=10^-1
0.01=10^-2
0.001=10^-3
:
0.0...1=10^-inf=0
so 0.000...1 = 0
By the well-ordering theorem there must be a smallest (last) digit in .999…..
Let's say a cake was 10 inches long, and you wanted to cut it into 3 even pieces;
You take out your tape measure; what are the 2 numbers you would make your cuts on?
Im geussing 3.3 inches, 6.6 inches
The tape measure perfectly on the left side is on 0. Perfectly on the second side 10 inches.
The intervals are:
0...........3.3
3.3........6.6
6.6.......10.0
Is any of this situation due to;
Starting at 0
3.3 used by 2 seperate intervals
6.6 used by 2 seperate intervals
Oh and here might be another relavant idea;
Imagine a number is like a footstool, you walk to that footstool and to stand on it is occupy that numbers value.
But, to walk toward it, and not stand on it, but just touch your foot against it's side is also to touch that numbers value?
In the first cut at 3.3 inches;
We approach from the left side in the first interval, we cut right in top of 3.3, making an interval that ends at 3.3 and one that begins at 3.3
These intervals are not standing on top in the center of 3.3, but touching the 3.3 footstool from either sides?
>0...........3.3
>3.3........6.6
>6.6.......10.0
>Is any of this situation due to;
>Starting at 0
>3.3 used by 2 seperate intervals
>6.6 used by 2 seperate intervals
This seems interesting.
Because in the 1/3 1/3 1/3
It implies you need to use the same accounted for number more than once...?
0___ 1___ 2 __3.3
3.3 __4___ 5 __6.6
6.6___ 7___8 ____9.9
Or does it imply?
0.01___ 1___ 2 __3.3
3.34 __4___ 5 __6.6
6.67___ 7___8 ____9.9("/10")
please translate into non-autistic
One cut should be between 3.3 and 3.4 (closer to 3.3). The other cut should be between 6.6 and 6.7 (closer to 6.7).
>One cut should be between 3.3 and 3.4 (closer to 3.3). The other cut should be between 6.6 and 6.7 (closer to 6.7).
So the first segment is:
0-----------3.3
Does the second segment start at 3.3 or 3.34?
>One cut should be between 3.3 and 3.4 (closer to 3.3).
Why in-between 3.3 (albeit close) and 3.4 and not directly directly on top of it?
If you cut it at 3.3 and 6.6, the first two slices are 3.3 inches long, and the last is 3.4 inches long. That said, if you do this IRL you should be fine unless your guests are the kind of people to raise a fuss about 1/30 of an inch of cake.
>That said, if you do this IRL y
Nah just tryna conceptualize the quandry of this thread and the nature of this proportion
So is the first segment;
0-------3.3
Second segment
3.34------------6.6
Third
6.67------------10
?
It's just like,if you cut exactly on the 3.333
Is it valid for the first segment and the second segment to both possess the same number?
To cut a cake into three equal pieces you would cut it like
0-------------3 1/3
3 1/3-------------6 2/3
6 2/3-------------10
>--3 1/3
>3 1/3-------------6 2/3
>6 2/3--
How can the same singular number be used twice on 2 seperate things?
Cakes aren't made of points; they're made of molecules, and the molecules lie between the specified boundary points.
If you wanted to cut up a set of numbers instead of a cake, then you would have to choose which set each of the boundary points went to. For example, you could partition [0, 10] into [0, 3 1/3], (3 1/3, 6 2/3], and (6 2/3, 10].
>For example, you could partition [0, 10] into [0, 3 1/3], (3 1/3, 6 2/3], and (6 2/3, 10].
Yes and the quandry of this thread is tryih to understand the meaning of doing this with decimals and the infinities thereof and the apparent inequality and equality of 2 seperate but equal numbers
Hmm maybe ?
1/3 = 0.333...
2/3 = 0.666...
3/3 = 0.999...
1 = 0.999...
QED
nobody ever says that 3/3 = 0.999... you fricking moron
might as well say 1/1 = 0.999...
it's called a pattern you ridiculous ass
1/3 = 0.333...
wrong
prove it
0.333... x 3 = 0.999...
0.999... ≤ 1
0.999... / 3 = 0.333...
1/3 ≥ 0.333...
>0.999... ≤ 1
also 0.999... = 1
So do you see it now
≤ is ( <, = )
.333... will end in 3 which means it's not a perfect 1/3
>.333... will end in 3
The meaning of the '...' is that it will not end.
WTF I left this place months ago, you guys are still talking about it?
We need to ban this discussion topic. The white race is dying and we can’t be spending our time rehashing this over and over.
Howbout browneyed race and hairy butthole race
>none ever talks about me
frick off leave
your proof sucks, but you are right
Let x = 0.999...
Then 10x = 9.999...
So 10x - x = 9.999... - 0.999...
Then 9x = 9
x = 1
Therefore 0.999... = 1
>10x = 9.999...
????????????????????????????
I don’t get this problem
>Hey guys lets say i have A and A approaches B, does that mean that A=B?
Are you moronic?
what are "A" and "B" supposed to be in this case?
Is something which approaches something else; that something?
Usually no. What's your point? How is that question relevant to the thread?
"0.999..." means the number being approached.
The decimal system was a mistake.
All numbers have 3 aspects.
The tippy top
The left side
The right side
1 is the tippy top of 1
.999... Is the left side of 1
1....0001 is the right side of 1
the number itself
the left halo
the right halo
https://en.wikipedia.org/wiki/Monad_(nonstandard_analysis)
you can also call it a monad if you're gay for Leibniz
>1....0001
nah, just use 1.000...
>>1....0001
>nah, just use 1.000...
That would equal 1 would it not?
Were not gonna start the whole trouble again by saying a .000...1 is magically gonna pop up at the end of a 0.
That's going to the right.
We need to start from the right, and travel infinitesimally to the left, to reach the minimum right side or the 1.
_1_
Meaning
1.000...
Would equal 1.00
1
The tippy top of 1.
.999... Is not the tippy top of 1. 1 is a mountain, .999... Is the base of the mountain.. or maybe a step away from the tippy top peak, a step into the peak is 1, a step beyond the peak is....
1.000...1?
Applying
[math][1, 2] cap [1, 1.1] cap [1, 1.01] cap cdots[/math]
to the hyperreals or any other system that extends the reals with infinitesimals would give you the number itself plus the right halo.
About this halo relation to, I just picked a few small numbers but like:
3, 4, 5
5 + 3 = 8
4 is half
11, 12, 13
11+ 13 = 24
12 is half
27, 28, 29
27+29= 56
28 is half
But anyway;
So what is the absolutely first number after 1? If .999... Is the absolutely first number before 1, to the degree that it """practically is 1"""
1.000...1?
1.000... is just 1.0000000000000...
Is just 1.0
Is just 1.
Unless you are suggesting the 0.000... turns into 0.00000000....00001 eventually
In which case I geuss that's equally hard to understand how a gureenteed repeated 9 turns into (1)0 as gurenteed repeating 0 turns into 1.
>So what is the absolutely first number after 1?
obviously none exists
>1.0000000000000... Is just 1.0
not under
definition if we allow infinitesimals
if you prefer another definition then state it
>So what is the absolutely first number after 1?
>obviously none exists
How can no number exist after 1?
Ok now I'm seeing something again I think. Forget everything after 1.... Because it's nothing essentially novel..
It's just the same idea of 0 to 1, repeating.
So the first number after 1, would be the same idea essentially, as the first number after 0.
Once you get to 1 it's game over, that's all you really need. And! Maybe! The "numbers" in-between 0 and 1 are not "numbers" but only ratios.
Ways of dividing/splitting 1.
Which.... Maybe...I geuss... Can be done infinitly?
>How can no number exist after 1?
Infinitely many numbers exists after 1. But there is no first such number. If there was a first such number x, (x+1)/2 would be a number after 1 but before x, contradicting the assumption x was the first such number.
Then there is no first number before 1.
Which makes sense because .999...9 is meaningless to right,
As much as, .999...8 is,
Or maybe since I don't know I will ask, what is the number before .999...?
Maybe we solve this but saying, decimals are not numbers! They are ratios.
They are not existing quanta, they are the endless fractal comparison of 2 or more quanta
>ratios
Well that is what real numbers originally were, long before they were expressed as decimals. Read Book 5 of the Elements.
>Magnitudes are said to be in the same ratio, the first to the second, and the third to the fourth, when equal multiples of the first and the third either both exceed, are both equal to, or are both less than, equal multiples of the second and the fourth, respectively, being taken in corresponding order, according to any kind of multiplication whatever.
Well it's obvious, because this thing starts by:
1/3
So to say 1/3 = .333...
Is to say, .333... Is not a number, it is not a quantity, it is a ratio, namely the ratio, 1/3
All the confusion is from thinking, 1 and 2 and 5 and 8 are numbers, and .333... Is a number, but it's not a number, it's the ideal endless computation of a ratio.
But now, .384 and .373733 are numbers, even they also are likely ratios, but it's the infinitude ... that suggests the non numberness,
The only thing then is that infinity is numberness, because all though the end can't be counted to, we know it at least contains 1 and 2 and 3 and 4 some 5 and 8448838484884 etc. Which are numbers.
But then too, the same thing, .9 is a number, is a quality is it?
If I want to cut something 10 inches long in 10 even pieces one of those pieces will be labeled the .9th piece.
So too, .999.... Is an endless computation, a ratio, but we know it does contain numberness
>So to say 6/3 = 2
>Is to say, 2 Is not a number, it is not a quantity, it is a ratio, namely the ratio, 6/3
this is how stupid you sound
Does the number 2 = 6/3 ?
I have a pair of shoes, 1 shoe, 2 shoe, I have 2 shoes. Do I have 6/3 shoes?
This is silly.
>>1.0000000000000... Is just 1.0
>not under
[math][1, 2] cap [1, 1.1] cap [1, 1.01] cap cdots[/math]
to the hyperreals or any other system that extends the reals with infinitesimals would give you the number itself plus the right halo. # definition if
It's one thing to be beaten over the head into accepting maybe in some twisted way of looking at things, .999.... Could eventually turn into 1.
But much harder to imagine ONLY 0's, turning into 1.
So it's not that, 1.000..... Auto turns into.
1.000....1
In this case it seems it must be specified, 1.000...1
But also the very idea is absurd, because the 1 at the end would never be reached.
So to piggy back on this post
Trying to totally adjust my perspective.
Not to view numbers and the number line as nessecerily needing to logically go continumnously, from number to number,
But Just axiomatically saying.
0 is the idea of no quantity.
1 is the idea of quantity.
All possible ratios are contained in the distance between them.
It is not a physical scaffolding that can be constructed from 0 to sturdily hoist up 1.
It is just that 1 can be infinitly divided linfinitly ways. None of those divisions have to be practically existing or possible. It is just possible to state a code function, that references the idea of the impossible, and this reference refers to a general but specific location in the sequence of all those possible/impossible theoretical ideas.
Ultimately the numerical rules may not refer to any ultimate true physical law.
Physics and physical laws are not at any given time infinite.
So the pure ruler (math, quantity, numerics, comparisons) can proportion and ratio itself to any physical data, and run it's functions until the nature of the physics is computed.
>1.000...1
>But also the very idea is absurd, because the 1 at the end would never be reached.
Pardon me for joining the talk
The logical conclusion from 0.999... is that there would be only 9's following one after another.
That means that if you move the infinite size of those 9's to the center of the caculation you would end up with 0.999... ..9 and nothing really changed
Then someone joins and says 2 - 0.999... ..9 and you end up with 1.000... ..1
So surely it wont be reached but when the size is infinite and you do the calculation you are gonna end up with that answer
Its only a different perspective and not absurd at all (when talking about infinites)
>Therefore, wouldn't 0.9999999.....times 9 equal 9?
Utterly risible
...and this thread comfirms, IQfy is full of complete and utter morons with way too much time on their hands.
> wouldn't 0.9999999.....times 9 equal 9?
...no
1/9 = 0.111...
+
8/9 = 0.888...
=
9/9 = 0.999...
I like that proof, it's simple, I like simple things.
the trick here is where’s the difference. watch how no one can tell you the difference
The difference between .999... And 1 is:
1 tells you 1 is 1, it is a finite object with exact extent and boundry.
.999... Tells you, whatever you do, never stop adding another 9 on the end.
These activities appear to be different, thus they are.
If by numbers you mean something that behaves like integers (example: for every integer there is a first integer after it), then real numbers are certainly not numbers.
My bad, integer is what I meant.
0 and 1 are the only integers that exist. (5 is just five 1's, 564 is just five hundred sixty four 1's); fractions and decimals are ratios between integers.
How many 1's go into how many 1's.
Four 1's go into eight 1's 2 times
One 1 goes into ten 1's ten times,
How many times does three 1's go into ten 1's evenly?
You need to take an extra 1, besides the three, and break it into many pieces, and give a little to each group of three 1's, till the three groups of three 1's approached, ten 1's total.
>You need to take an extra 1, besides the three, and break it into many pieces, and give a little to each group of three 1's, till the three groups of three 1's approached, ten 1's total.
And you cannot do this because....
Think in terms of measuring a recipe;
The recipe calls to make 3 pieces of food of perfectly equal grams, with a 10 gram mixture.
A mixing bowl has the mixture of perfectly exactly 10 grams.
You have 3 bowls;
Bowl A gets 3 grams
Bowl B gets 3 grams
Bowl C gets 3 grams
The mixing bowl has 1 gram left.
You split that gram in 3 and put it in the 3 bowls.
The mixing bowl has .1 gram left.
You split that .1 gram into 3 equal parts, and spread them among the 3 bowls.
There is .03 grams left in the mixing bowl.
You split that into 3 equal groups to spread among the 3 bowls.
You are left with .01 grams in the mixing bowl.
You seperate that into 3 equal pieces, to spread among the 3 bowls.
You are left with .003333333 grams in the mixing bowl.
You divide it by 3 to spread among the 3 bowls.........
Respond to this. Think it puts things in good perspective
>There is .03 grams left in the mixing bowl.
>You split that into 3 equal groups to spread among the 3 bowls.
>You are left with .01 grams in the mixing bowl.
Hmmm really?
If you have .03 grams in a mixing bowl, and evenly distribute it to 3 bowls, you will have .01 left in the mixing bowl? That doesn't seem right.
If you have .03 grams in a mixing bowl, and evenly distribute it between 3 bowls, will you not give .01 to each bowl, and be left with 0 grams in the mixing bowl?
Answer I'm confused
>You have 3 bowls;
>Bowl A gets 3 grams
>Bowl B gets 3 grams
>Bowl C gets 3 grams
>The mixing bowl has 1 gram left.
>You split that gram in 3 and put it in the 3 bowls.
>The mixing bowl has .1 gram left.
>You split that .1 gram into 3 equal parts, and spread them among the 3 bowls.
>There is .03 grams left in the mixing bowl.
>You split that into 3 equal groups to spread among the 3 bowls.
>You are left with .01 grams in the mixing bowl.
>You seperate that into 3 equal pieces, to spread among the 3 bowls.
>You are left with .003333333 grams in the mixing bowl.
>You divide it by 3 to spread among the 3 bowls.........
>If you have .03 grams in a mixing bowl, and evenly distribute it between 3 bowls, will you not give .01 to each bowl, and be left with 0 grams in the mixing bowl?
Guys explain this: how come it seems you can perfectly evenly split 10 grams between 3 bowls?
>>You split that gram in 3 and put it in the 3 bowls.
>>The mixing bowl has .1 gram left.
>3 - 3 = 0.1
I realized later that when I did division; it wasn't resulting me with the number left over, but the number of times the divided can occur..
So yeah I messed up.... soooooooo
Yep, never mind, .1 divided by 3 is .0333....
Which yes proves my statement wrong, but proves the thread wrong, because 1 can never be perfectly divided into 3 equal parts.
But WAIT ACTUALLY I don't get why you wouldnt keep doing this dividing, until you were at the atom scale and left with how many atoms? Could you be left with 3 atoms?
How many atoms are in 1 gram of _______?
Could it be after dividing into 3 and dividing into 3 and dividing into 3, eventually there are only 3 atoms left?
This would only suggest that unlike the infinitude of abstract symbolic numbers, reality does always have ground quanta limits.
But does it even make sense, or would it just imply the substance could not have accurately been referred to as a unit of 10, if after being many times divided by 3, it resulted in 3 atoms remaining?
Stop shilling, does not matter if we use base9 in math, then the problem would be dividing with 2
You just change base and the problem goes to other divisions
Base 1 you could do easily everything and always get a result
1/111 = 0.01
1/11 = 0.1
1/1 =1
Its only a base thingy
>1/111 simply means 1/3 in base 1
Logic is basically base1 math, and thats why computers works so efficiently.. There is no useless crap there, but they translate everything to our crappy needs lightning fast
Ok so in base 1.
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
Yeah I don't get it. How do you make 3 even piles outt of 10.
1
1
1
1
1
1
1
1
1
You have 1 left over. Divide it in 3
.3
.3
.3
.1 left over
Divide it in 3.
.03
.03
.03
.01 left over
Divide it by 3
.003
.003
.003
.001 left over.
How does base 1 solve this?
Oh you mean that, well i guess you can also make 3 even piles out of 5, 7, 8, 10, 11, 13, 14, 16 etc.
>does not see the problem
It's about 3 equal piles. Of 10.
That's why the 1/3 = .333...
Enters the picture.
3 of them are .999....
But.
How can 3 equal piles ACTUALLY TRUTHFULLY be made of 10.
They cannot, thats why 0.999... is not 1
1/3 > 0.333...
2/3 > 0.666...
3/3 = 1 ( or > 0.999... )
>You split that gram in 3 and put
That, gram. 1 gram. Split in 3.
.3 grams in one bowl
.3 grams in other bowl
.3 grams in another bowl
.1 left over.
That's where you get the rest from to tack on each bow.
.03
.03
.03
But we see how without grounding in physics reality, this abstract process repeats forever with no signs of equalling 0.
There for no signs of in totally having 3 equal parts that equal 1
>>You have 3 bowls;
>>Bowl A gets 3 grams
>>Bowl B gets 3 grams
>>Bowl C gets 3 grams
>>The mixing bowl has 1 gram left.
>you have 4 bowls
>>The mixing bowl has .1 gram left.
>>You split that .1 gram into 3 equal parts, and spread them among the 3 bowls.
is .03 grams left in the mixing bowl.
There's no hope for IQfy.
I can see all perspectives, I'm just trying to understand
Decimals are gay and moronic
Why do they start teaching decimals and calculators to kid at such an early age?
You shouldn’t be allowed to even think about decimals until university level mathematics, only fractions in middle and high school
Quick rundown
"0.999..." is a finite number.
The measure of redacting continuity is for the purpose of practical applications of assuming prerequisit accuracy.
For example, PI is continuous, however it isn't valuable to use a trillion digits of PI in order to use PI to get an appreciable result.
It is as useful to use "0.999..." whether there are 11948 nines or unlimited.
Jokes aside, it depends.
In the normal number system, 0.999...= 1,
in the Hyperreal number system 0.999... = 1- 0.000...1
I mean normally if something starts with 0 it usually is less than something starting with 1 no ?
We are nearing the end of the thread.
What did you learn, what has been concluded?
0.9... = 1
because 1/inf=0
>because 1/inf=0
1 divided infinitly times is the infinite real decimals between 0 and 1
0.999999.... / inf = .0000000...1
>.0000000...1
[math] displaystyle
0. bar{0}1
= lim_{n to infty} 0. underbrace{0 dots 0}_{n ~ text{times}}1
= lim_{n to infty}
left [
left (
sum_{k=1}^n dfrac{0}{10^k}
right )
+ dfrac{1}{10^{n+1}}
right ]
=0
[/math]
x = 0.99
10x = 9.99
10x-x = 9.99 - 0.99
9x= 9
x= 1
don't you learn this stuff in middle school?
>x = 0.99
>10x = 9.99
10x is 9.9
Anything multiplied by 10 has 0 at the end of it
1/3 x 10 = 3.3
10/30 if you are intellectual like me
There are fundamental smallest possible particles.
So in reality you can nearly count up to infinity, but there is no infinitesimal.
If you keep dividing eventually you reach a fundamental smallest unit.
If you have 100 quarks and you want to make 3 even piles, it is impossible