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/sci/ - Science & Math


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11540214 No.11540214 [Reply] [Original]

Consider the sets A and B
A = the partial sums of [math]\sum_{n=1}^{\infty}\frac{9}{10^n}[/math]
B = 1+ the partial sums of [math]\sum_{n=1}^{\infty}\frac{9}{10^n}[/math]
>A:[0.9, 0.99, 0.999, 0.9999, ...]
>B:[1.9, 1.99, 1.999, 1.9999, ...]

Let set C = A union B.
>C:[0.9, 0.99, 0.999, ... , 1.9, 1.99, 1.999, ...]

where is 1 in C?
Can 1 in C be indexed?
Does 1 exist in C?

>> No.11540219

>>11540214
>Does 1 exist in C?
No. That answers the rest.

>> No.11540226

>>11540214
1 is in C but C is uncountable. The union of two countably infinite sets is not necessarily countably infinite.

>> No.11540228

>>11540219
0.999... = 1

>> No.11540229

>>11540226
>The union of two countably infinite sets is not necessarily countably infinite

>> No.11540230

>>11540214
It's not
All distinct so union=union all
If n becomes fractional yes.

>> No.11540231

>>11540214
>A:[0.9, 0.99, 0.999, 0.9999, ...]
>B:[1.9, 1.99, 1.999, 1.9999, ...]
To clarify, are you talking about ordered lists here, or unordered sets? This difference is important, because:

>Let set C = A union B.
This is well-defined only for sets, not lists.
>C:[0.9, 0.99, 0.999, ... , 1.9, 1.99, 1.999, ...]
In particular, this is not a list.

>where is 1 in C?
>Can 1 in C be indexed?
"where" and "indexed" are meaningful only for lists, not sets.

>Does 1 exist in C?
No.

>> No.11540235

>>11540226
C is totally countable.

>The union of two countably infinite sets is not necessarily countably infinite.
Yes it is. You can denumerate them by alternating between the next denumerated item from the two alternating sets.

>> No.11540236

>>11540214
>where is 1 in C?
1 (i.e. 0.999...) is not in C. It is the supremum of A but not in A.

>> No.11540272

>0.999... [math]\neq[/math] 1

>> No.11540286

>>11540214
1 exists in C but cannot be indexed because the set is infinite. as you can see, this is a shitty way to try to answer the does 1 = 0.999... question because you just introduce more quirks related to infinity.

>> No.11540293
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11540293

Consider the partial sums of [eqn]\lim_{n\to\infty} \sum_{1}^{2n} \frac{1}{n} = 2[/eqn]

>> No.11540390

there are a number of unaccounted-for elements in any infinite set, such that [math]\mathbb{N} [/math] is in bijection with [math]\mathbb{R}[/math], utilizing some of the unaccounted elements in N.

There are as many unaccounted elements in N as there are determinable elements, for N contains all of the finite numbers, even the ones nobody has thought of yet, as well as the ones that nobody will ever think of.

>> No.11540802

>>11540214
>the partial sums of
Define partial sum.

>where is 1 in C?
What do you mean by where? Sets do not have an order. Either the definition of partial sum includes 1 in the set C or it doesn't.

>Can 1 in C be indexed?
What do you mean? Do you mean can you form a biection between the members of C and N? Yes.

If C includes 1:

(1,1) (2,2) (3,0.9) (4,1.9) (5,0.99) (6,1.99) etc.

If C does not include 1 then

(1,0.9) (2,1.9) (3,0.99) (4,1.99) etc.

>> No.11540865

>>11540226
>>11540286
Mathematically illiterate retards.
>>11540228
Not relevant.
>>11540214
1 is a limit point of C. 1 is not an element of C.

>> No.11540967

>>11540231
/thread

>> No.11540972

>>11540226
>>11540228
>>11540272
>>11540286
>>11540293
>>11540390
>>11540802
retards

>> No.11541051

>>11540214
>Does 1 exist in C?
Nope

>> No.11541057
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11541057

>>11540802
>Define partial sum.

>> No.11541549

>>11540972
Not an argument.

>>11541057
Depending on your definition of partial sum, it includes the entire sum or it doesn't. Which one is it?

>> No.11541553

>>11540865
>1 is a limit point of C. 1 is not an element of C.
Exactly why 0.9999.... does not equal 1 where lim(0.9999...) does. Try telling the retards here that and they call you schizo.

>> No.11541599

>>11541553
your post implies that you regard 0.999... as a sequence. fair enough, but then for you, the symbol 0.999... is not a real number and the statement

0.999... != 1

is trivially true because the left hand side is a real number and the right hand side is a sequence.

is this what you're trying to say ?

>> No.11541602

>>11541599
swapped RHS and LHS, my bad

>> No.11541681

are you mad that the jannies deleted your poll instantly?

>> No.11541688
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11541688

>>11540214
>>11540219
>>11540226
>>11540228
>>11540229
>>11540230
>>11540231
>>11540235
>>11540236
>>11540272
>>11540286
>>11540293
>>11540390
>>11540802
>>11540865
>>11540967
>>11540972
>>11541051
>>11541057
>>11541549
>>11541553
>>11541599
>>11541602
>>11541681

The inability many people today who view themselves as being "scientific" as well as math fans (left-brain prisoners) have in comprehending that .999 isn't the same as 1, is completely connected to their inability to grasp eternity and infinity. Their minds have basically been programmed to believe eternity and infinity are impossibilites. Over the course of the last 10 or so centuries of indoctrination into an alien world-view, they've "learned" to be unable to comprehend it.

This difficulty they have with eternity/infinity shows up in many different fields, from math to astrophysics.

This mental handicap is inherited directly from the (((Abrahamic))) religions, more specifically Christinsanity for us Westernerns. In it's origin, the inability to understand infinity and eternity is 100% Judaic in thought/philosophy. In contrast, the non-Jewish man; the Pagan man, at least the /European/ Pagan man, never had any problem with infinity and eternity. Christinsanity introduced into the minds of people the idea of life and the world/universe being linear, starting from point A and ending with a point B, whereas in the Native European world-view everything is infinite, a circle.

That's why many people today can't understand that .999 repeating forever will never reach 1 -- they refuse to accept the idea of an infinite/eternal repetition. Saying "it's 1" is their method of escaping from the uncomfortable (and to them insurmountable) challenge which the concept of infinity/eternity is to thier Judaically-induced mental disease.

>> No.11542976

>>11540214
I used to be one of the 0.99999... != 1 people but I've turned the corner, although it may be more about disagreeing about representation than substance.

The idea for the proof is as follows:
0.99999, when continued any amount, let it approach 1.
Now, let the sequence be expanded infinitely. I say that the result is equal to 1. For let it instead be equal to D. Then it were not approaching 1 without limit, but approaching D instead. This is a contradiction.

Sure, you could say that you can't expand something infinitely. But that implies that 0.9999... can't exist as a representation of anything at all(For the "repeating" part of the statement "0.999 repeating" implies an infinity in expansion). By arguing that one or the other representation is right or wrong, you assume that it can represent something in the first place.

>> No.11542982

>>11541688
>Christinsanity introduced into the minds of people the idea of life and the world/universe being linear, starting from point A and ending with a point B
No? In Christianity, your life is eternal. What are you talking about?

>> No.11542990

>>11542976
(Newton in Principia Mathematica uses this proof in a very clear way early on in Book 1. More clearly than I could every phrase it)

>> No.11543012

>>11541553
>he doesn’t know the definition of a decimal expansion
By definition, 0.999... denotes the limit of the sequence 0.9, 0.99, 0.999, 0.9999, ...
This is literally the definition of a decimal expansion.
>>11542655