math q


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Post Friday, 26th August 2016, 17:28

Re: math q

twelwe wrote:It's like Blink, but you end up drowning.

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Post Saturday, 27th August 2016, 04:54

Re: math q

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Post Saturday, 27th August 2016, 04:56

Re: math q



finitists are also lunatics, though.
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Post Saturday, 27th August 2016, 05:07

Re: math q

Arrhythmia wrote:

finitists are also lunatics, though.


They certainly like making things harder for themselves. I think the extent to which you can do mathematics with finitist methods is interesting though.
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Post Saturday, 27th August 2016, 05:19

Re: math q

Zeilberger is very good and cool.
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Post Wednesday, 31st August 2016, 20:58

Re: math q

So, lets say I have negative 9 m&ms, and I offer you the square root of the amount of M&Ms I have. How do I give you 3i M&Ms?

It doesn't make sense, obviously, therefore you cannot take the square root of a negative number. Sorry mathematicians, back to the drawing board, if I can't do it in M&M's it's wrong.

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Post Wednesday, 31st August 2016, 21:32

Re: math q

dowan wrote:So, lets say I have negative 9 m&ms, and I offer you the square root of the amount of M&Ms I have. How do I give you 3i M&Ms?

give 3 skittles

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Post Thursday, 1st September 2016, 07:16

Re: math q

Even if complex numbers don't model any physical structure well, which parenthetically isn't true, there are things they do model well, it's still not a good reason to not study them. I would sincerely be baffled if large cardinals modeled anything whatsoever, but they give some truly fascinating results about logic and ZFC.
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Post Thursday, 1st September 2016, 08:08

Re: math q

Suppose I have a capacitor with an impedance of 2/(i \omega) ohms,
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Post Thursday, 1st September 2016, 16:04

Re: math q

CanOfWorms wrote:
dowan wrote:So, lets say I have negative 9 m&ms, and I offer you the square root of the amount of M&Ms I have. How do I give you 3i M&Ms?

give 3 skittles

TIL: skittles are imaginary m&ms, taste the rainbow, baby!
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Post Thursday, 1st September 2016, 23:55

Re: math q

4Hooves2Appendages wrote:Of course being flat doesn't mean it's infinite. It just means that as it expands it doesn't in a weird way fold back onto itself. Given that the universe is expanding, and that many objects are moving away from us at relatively faster than the speed of light, it is actually not possible for us to reach the (current) edge, based on what we know now. We'd never catch up.

There really aren't any fixed distances as soon as we start thinking about stuff outside of our own solar system. Everything is moving, and most of it away from us (or we from it depending on your perspective).
objects (like galaxies) are not moving, they are in same position where they were like one billion years ago, they are moving away because space is expanding
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Post Friday, 2nd September 2016, 08:26

Re: math q

What you don't know I'm going to tell you right now is the object and the space is the same thing, yo, I'm the spaceman.
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Post Friday, 2nd September 2016, 20:55

Re: math q

Here's an example of infinity having to act like a number to yield correct results:

Lets say I have a line segment of length 1. There are infinite points on this line segment, so the odds of any given point being picked are 1 out of infinity.
Now lets say I have an overlapping parallel line segment of length 2 that completely overlaps the first line segment. What are the odds that the point picked on line 1 is also on line 2?

If you think about it, it's obviously 50%. Line segment 1 has infinite points. line segment 2 has infinite points, but it also has double the points of line 1 (this has to be true, if they're both 1 out of infinity then there's 100% chance of getting a point on line segment 2, which obviously isn't true).

So to make sense of this, we have to find the ratio between the number of possible points on line 1 (infinity) and the number on line 2 (infinity, but also double of line 1). The correct equation to give the known correct result of 50% then is:

1/infinity / 1/(2*infinity)
Then we end up with infinity over 2 * infinity, which allows us to cancel out the infinities, giving the correct answer of 1/2
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Post Saturday, 3rd September 2016, 00:55

Re: math q

dowan wrote:Here's an example of infinity having to act like a number to yield correct results:

Lets say I have a line segment of length 1. There are infinite points on this line segment, so the odds of any given point being picked are 1 out of infinity.
Now lets say I have an overlapping parallel line segment of length 2 that completely overlaps the first line segment. What are the odds that the point picked on line 1 is also on line 2?

If you think about it, it's obviously 50%. Line segment 1 has infinite points. line segment 2 has infinite points, but it also has double the points of line 1 (this has to be true, if they're both 1 out of infinity then there's 100% chance of getting a point on line segment 2, which obviously isn't true).

So to make sense of this, we have to find the ratio between the number of possible points on line 1 (infinity) and the number on line 2 (infinity, but also double of line 1). The correct equation to give the known correct result of 50% then is:

1/infinity / 1/(2*infinity)
Then we end up with infinity over 2 * infinity, which allows us to cancel out the infinities, giving the correct answer of 1/2


This idea immediately will fall apart when you consider more complex sets though, like the Cantor set.
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Post Saturday, 3rd September 2016, 03:12

Re: math q

dowan wrote:Here's an example of infinity having to act like a number to yield correct results:

Lets say I have a line segment of length 1. There are infinite points on this line segment, so the odds of any given point being picked are 1 out of infinity.
Now lets say I have an overlapping parallel line segment of length 2 that completely overlaps the first line segment. What are the odds that the point picked on line 1 is also on line 2?

If you think about it, it's obviously 50%. Line segment 1 has infinite points. line segment 2 has infinite points, but it also has double the points of line 1 (this has to be true, if they're both 1 out of infinity then there's 100% chance of getting a point on line segment 2, which obviously isn't true).

So to make sense of this, we have to find the ratio between the number of possible points on line 1 (infinity) and the number on line 2 (infinity, but also double of line 1). The correct equation to give the known correct result of 50% then is:

1/infinity / 1/(2*infinity)
Then we end up with infinity over 2 * infinity, which allows us to cancel out the infinities, giving the correct answer of 1/2

this is what we call a broken clock

let's apply your reasoning in a different way

what are the odds that a randomly chosen number between 0 and 2 is smaller than 1? well, the odds of picking a random number between 0 and 1 is 1/infinity

but if you have a way of randomly picking a number between 0 and 1, well you also have a way of picking a random number between 0 and 2 (just double the number you picked.) so the odds of picking any specific number between 0 and 2 is also 1/infinity

so the odds that a random number between 0 and 2 is smaller than 1 is (1/infinity)/(1/infinity) = 100%

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Post Saturday, 3rd September 2016, 03:26

Re: math q

dowan wrote:Here's an example of infinity having to act like a number to yield correct results:

Lets say I have a line segment of length 1. There are infinite points on this line segment, so the odds of any given point being picked are 1 out of infinity.
Now lets say I have an overlapping parallel line segment of length 2 that completely overlaps the first line segment. What are the odds that the point picked on line 1 is also on line 2?

If you think about it, it's obviously 50%. Line segment 1 has infinite points. line segment 2 has infinite points, but it also has double the points of line 1 (this has to be true, if they're both 1 out of infinity then there's 100% chance of getting a point on line segment 2, which obviously isn't true).

So to make sense of this, we have to find the ratio between the number of possible points on line 1 (infinity) and the number on line 2 (infinity, but also double of line 1). The correct equation to give the known correct result of 50% then is:

1/infinity / 1/(2*infinity)
Then we end up with infinity over 2 * infinity, which allows us to cancel out the infinities, giving the correct answer of 1/2
This is circular reasoning. You're assuming that infinity acts like a number ("odds of any given point being picked are 1 out of infinity") in order to demonstrate that infinity acts like a number.
The probability of a given point being picked isn't 1 out of infinity. It's 0.

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Post Saturday, 3rd September 2016, 19:34

Re: math q

dowan wrote:An infinitesimally small number would be the difference between 0.333333333333 repeating * 3 vs 1/3*3.
In other words, 1 - 0.9999999 repeating.
It's equivalent to 0, clearly, but equally clearly it's not quite the same. The difference between those two numbers is 0.0000 repeating, followed by 1. That number is "one infinitith" or the result of 1 divided by infinity.

Let's say you have a line extending to infinity in both directions. Clearly there are infinite points on this line (2 * infinite points, if we consider + and - directions). Lets say you pick a point on this line at random. What are your odds of any given point being chosen? It's 1 out of infinity. Yet, it's not quite 0, because a point will be picked, and any given point has a chance of being picked.

No, you can't write this number, you can't represent it just like you can't write the number that i represents, nor can you write the number pi represents for that matter. The closest you can come to writing it is:
  Code:
  _
0.01


[b]But that' not considered a valid number, real or unreal, rational or irrational. Yet it still seems to exist.[\b]



Actually, it's a hyperreal number: https://ncatlab.org/nlab/show/nonstandard+analysis. The details of nonstandard analysis are unfortunately rather technical, but the basic idea is as follows: we replace numbers by sequences of numbers. For instance, the number 1 is replaced by an infinite sequence of 1's (1,1,1,1,...), 0.9999 repeating is replaced by the sequence (0.9,0.99,0.999,...) and their difference is (1-0.9,1-0.99,1-0.999,...) = (0.1,0.01,0.001,...), which informally is "0.0000 repeating, followed by 1." Now there are problems with this scheme - for instance, which number is bigger, (1,0,1,0,1,0,...) or (0,1,0,1,0,1)? - but with the help of something called an ultrafilter you can answer all of these questions in a logically consistent way. Note however that (1) you still can't divide by 0 (although (0.1,0.01,0.001,...) /= 0 and so 1/(0.1,0.01,0.001,...) = (10,100,1000,...) is a sensible, infinite number) and (2) that there are not one but many "infinities" (for instance (10,100,1000,...) \= (1,2,3,4,...) yet both are infinite).

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Post Saturday, 3rd September 2016, 21:04

Re: math q

ManMan wrote:
dowan wrote:An infinitesimally small number would be the difference between 0.333333333333 repeating * 3 vs 1/3*3.
In other words, 1 - 0.9999999 repeating.
It's equivalent to 0, clearly, but equally clearly it's not quite the same. The difference between those two numbers is 0.0000 repeating, followed by 1. That number is "one infinitith" or the result of 1 divided by infinity.

Let's say you have a line extending to infinity in both directions. Clearly there are infinite points on this line (2 * infinite points, if we consider + and - directions). Lets say you pick a point on this line at random. What are your odds of any given point being chosen? It's 1 out of infinity. Yet, it's not quite 0, because a point will be picked, and any given point has a chance of being picked.

No, you can't write this number, you can't represent it just like you can't write the number that i represents, nor can you write the number pi represents for that matter. The closest you can come to writing it is:
  Code:
  _
0.01


[b]But that' not considered a valid number, real or unreal, rational or irrational. Yet it still seems to exist.[\b]



Actually, it's a hyperreal number: https://ncatlab.org/nlab/show/nonstandard+analysis. The details of nonstandard analysis are unfortunately rather technical, but the basic idea is as follows: we replace numbers by sequences of numbers. For instance, the number 1 is replaced by an infinite sequence of 1's (1,1,1,1,...), 0.9999 repeating is replaced by the sequence (0.9,0.99,0.999,...) and their difference is (1-0.9,1-0.99,1-0.999,...) = (0.1,0.01,0.001,...), which informally is "0.0000 repeating, followed by 1." Now there are problems with this scheme - for instance, which number is bigger, (1,0,1,0,1,0,...) or (0,1,0,1,0,1)? - but with the help of something called an ultrafilter you can answer all of these questions in a logically consistent way. Note however that (1) you still can't divide by 0 (although (0.1,0.01,0.001,...) /= 0 and so 1/(0.1,0.01,0.001,...) = (10,100,1000,...) is a sensible, infinite number) and (2) that there are not one but many "infinities" (for instance (10,100,1000,...) \= (1,2,3,4,...) yet both are infinite).
His claim that the probability is "not quite 0" is also wrong in the first place. It's not not quite 0. It is 0. The probability of selecting any given point being 0 doesn't mean that you can't select a point. See almost surely (in this case, almost never).

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Post Sunday, 4th September 2016, 13:11

Re: math q

ManMan wrote:
dowan wrote:An infinitesimally small number would be the difference between 0.333333333333 repeating * 3 vs 1/3*3.
In other words, 1 - 0.9999999 repeating.
It's equivalent to 0, clearly, but equally clearly it's not quite the same. The difference between those two numbers is 0.0000 repeating, followed by 1. That number is "one infinitith" or the result of 1 divided by infinity.

Let's say you have a line extending to infinity in both directions. Clearly there are infinite points on this line (2 * infinite points, if we consider + and - directions). Lets say you pick a point on this line at random. What are your odds of any given point being chosen? It's 1 out of infinity. Yet, it's not quite 0, because a point will be picked, and any given point has a chance of being picked.

No, you can't write this number, you can't represent it just like you can't write the number that i represents, nor can you write the number pi represents for that matter. The closest you can come to writing it is:
  Code:
  _
0.01


[b]But that' not considered a valid number, real or unreal, rational or irrational. Yet it still seems to exist.[\b]



Actually, it's a hyperreal number: https://ncatlab.org/nlab/show/nonstandard+analysis. The details of nonstandard analysis are unfortunately rather technical, but the basic idea is as follows: we replace numbers by sequences of numbers. For instance, the number 1 is replaced by an infinite sequence of 1's (1,1,1,1,...), 0.9999 repeating is replaced by the sequence (0.9,0.99,0.999,...) and their difference is (1-0.9,1-0.99,1-0.999,...) = (0.1,0.01,0.001,...), which informally is "0.0000 repeating, followed by 1." Now there are problems with this scheme - for instance, which number is bigger, (1,0,1,0,1,0,...) or (0,1,0,1,0,1)? - but with the help of something called an ultrafilter you can answer all of these questions in a logically consistent way. Note however that (1) you still can't divide by 0 (although (0.1,0.01,0.001,...) /= 0 and so 1/(0.1,0.01,0.001,...) = (10,100,1000,...) is a sensible, infinite number) and (2) that there are not one but many "infinities" (for instance (10,100,1000,...) \= (1,2,3,4,...) yet both are infinite).


I don't think I entirely agree with this. 0.999... is still equal to 1 in an ultrapower, as it is equal to 1 in the original structure. It's more correct to say that the sequence 0.9,0.99,0.999,.. whose limit is 0.999... (and thus 1) in the reals no longer has a limit in a (non-trivial) ultrapower. The element defined by (0.9,0.99,0.999,...) breaks the limit, but does not replace it, as smaller upper bounds can always be constructed (at least over a countable set. I don't know if something changes if you take an ultrapower over a weird cardinal).
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Post Monday, 5th September 2016, 12:01

Re: math q

Man, this thread just keeps giving.
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Post Monday, 5th September 2016, 16:24

Re: math q

Ok, I've seen this proof countless times.

And like I say it's logically flawed because it requires the a completed list of numerals must be square, which they can' t be.

~~~~

First off you need to understand the numerals are NOT numbers. They are symbols that represent numbers. Numbers are actually ideas of quantity that represent how many individual things are in a collection.

So we aren't working with numbers here at all. We are working with numeral representations of numbers.

So look at the properties of our numeral representations of number:

Well, to begin with we have the numeral system based on ten.

This includes the ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

How many different numbers can we list using a column that is a single digit wide?

Well, we can only list ten different numbers.

0
1
2
3
4
5
6
7
8
9

Notice that this is a completed list of all possible numbers. Notice also that this list is not square. This list is extremely rectangular. It is far taller than it is wide.

Let, apply Cantor's diagonal method to our complete list of numbers that are represented by only one numeral wide.

Let cross off the first number on our list which is zero and replace it with any arbitrary number from 1-9 (i.e. any number that is not zero)


0
1
2
3
4
5
6
7
8
9

Ok we struck out zero and we'll arbitrary choose the numeral 7 to replace it.

Was the numeral 7 already on our previous list? Sure it was. We weren't able to get to it using a diagonal line because the list is far taller than it is wide.

Now you might say, "But who cares? We're going to take this out to infinity!"

But that doesn't help at all.

Why not?

Well what happens when we make the next step? We need to make the list 2 digits wide now.

What happens?

Here is a 2-digit list of all possible numbers represented by 2 numerals.

00
01
02
03
04
05
06
07
08
09
11
12
13
14
15
.
.
.

95
96
97
98
99

What happened? Well, our completed list of possible numerals that is two digits wide has incrusted in vertical height exponentially. This list is now 100 rows tall and only 2 column wide.

Now let's cross off the first two digits of our list and replace them with arbitrary numerals.

0 0
0 1

Ok, for the first zero being stuck off the list, I'll chose to arbitrarily replace that with a 5. For the second digit being struck off the list I'll replace that arbitrarily with a 7.

My new number is 57.

Is 57 already on my completed list? Yes. It's just further down the list where I couldn't possibly reach it by drawing a diagonal line.

Now you might say, "But who cares? We're going to take this out to infinity!"

But duh? We can already see that in a finite situation we are far behind where we need to be, and with every digit we cross off we get exponentially further behind the list.

Taking this process out to infinity would be a total disaster.

You could never claim to have "completed" this process because you can't move down the list fast enough using a diagonal line that crosses off each digit diagonally.

The very nature of our system of numerical representation forbids this. You can't complete this process in a finite situation, and it gets exponentially worse with every digit you add to the width, then you could never claim to have completed this process by claiming to have taken it out to infinity.

"Completed Lists" of numerical representations of numbers are NOT SQUARE.

Yet Cantor claims to be creating a "Completed List" here. It's a bogus proof that fails. Cantor didn't stop to realize that our numerical representations of number do not loan themselves to nice neat square competed lists. And that was the flaw in his logic.

By the way you can't even do this using binary representations of numbers.

In Binary Representation

A completed list of binary numbers 2 digits wide:

00
01
10
11

It's not square. It's twice as tall as it is wide.

Add another digit it gets worse:

000
001
010
011
100
101
110
111

There is no way that a completed list of numbers can be represented numerically in square lists.

Yet Cantor's diagonal argument demands that the list must be square. And he demands that he has created a COMPLETED list.

That's impossible.

Cantor's denationalization proof is bogus.

It should be removed from all math text books and tossed out as being totally logically flawed.

It's a false proof.

Cantor was totally ignorant of how numerical representations of numbers work. He cannot assume that a completed numerical list can be square. Yet his diagonalization proof totally depends on this to be the case.

Otherwise, how can he claim to have a completed list? Think

If he's standing there holding a SQUARE list of numerals how can he claim that he has a completed list?

Yet at what point does his list ever deviate from being square?

It never deviates from being square. It can't because he's using a diagonal line to create it. That forces his list to always be square.

Georg Cantor was an idiot.

He didn't even understand how numerical representations of numbers work.

His so-called "proof" doesn't prove anything. It's totally bogus.

He can't claim to have a "completed list" by the way he is generating his list. Claiming to take this out to infinity doesn't help. With every new digit he creates he falls exponentially behind where he would need to be to create a "Completed List".

Yet that's what he claims to have: A Completed List.

It's a bogus proof, and I'm shocked that no mathematicians have yet recognize this extremely obvious truth.

They keep publishing this proof and teaching it like as is it has merit when in fact it's totally bogus.
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Post Monday, 5th September 2016, 17:15

Re: math q

"It's not a number, it's a numeral!"

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Post Monday, 5th September 2016, 19:15

Re: math q

I've never felt so intellectual while reading copypasta

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Post Monday, 5th September 2016, 19:20

Re: math q

Es tut weh.

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Post Tuesday, 6th September 2016, 05:18

Re: math q

I just wish the diagonal argument cranks would turn the page in their 'introduction to set theory' book and read the proof that no set can be in bijection with its powerset.
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Post Tuesday, 6th September 2016, 16:03

Re: math q

I have to go teach calculus now but I read this thread and now I can't stop crying

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Post Wednesday, 7th September 2016, 01:35

Re: math q

i never learned (surprise, keep reading) calculus
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Post Monday, 9th January 2017, 03:53

Re: math q

The answer to this question is not in baby rudin as it is trivial. I checked it

Consider the sequence a_n = {.9, .99, .999, ...}, and the set A for which every element of the sequence a_n is contained in A.

Then the limit of the sequence, as n approaches infinity, approaches 1.

In other words, lim sup a_n =1. Where sup is the least upper bound of the set.

which is a fancy way of saying that 1 is bigger than all those numbers in the sequence, yet smaller than any other number above that sequence. In other words, 1 is the least upper bound for the sequence a_n.

There are many fancy ways to prove that lim sup a_n=1.. or equivalently we can prove sup A=1. ( which is the same thing just proving it as a set instead of a sequence)

we start with the definition of upper bound: a set in R is bounded above if there exists a number b in R st a<b for all a in A. The number b is called the upper bound for A.* Where R is the complete real number line, a is any element in A, and b =1

Now we can use the Axiom of completeness, Every nonempty set of real numbers that is bounded above has a least upper bound.

So all we have to do now is show that 1 is the least upper bound for A.

We can show this by definition (Abbott)

A real number b is the least upper bound for a set A in R if it meets the following two criteria:
(i) b is an upper bound for A
(ii) if c is any upper bound for A then b <= c, (less than or equal to) ( means that b is the 'smallest' upper bound)

We showed (i).
(ii) is also trivial but can be shown by using Theorem 1.11 (Rudin)

Thus: sup A=1 , which implies that lim sup a_n =1 , which implies .99999.... =1

*
tldr;
since the set A is bounded, by axiom of completeness, and the fact that 1 is the smallest number above .9999999... , we can deduce that .99999...=1
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Post Monday, 9th January 2017, 04:20

Re: math q

TrumpTrain wrote:The answer to this question is not in baby rudin as it is trivial. I checked it

Consider the sequence a_n = {.9, .99, .999, ...}, and the set A for which every element of the sequence a_n is contained in A.

Then the limit of the sequence, as n approaches infinity, approaches 1.

In other words, lim sup a_n =1. Where sup is the least upper bound of the set.

which is a fancy way of saying that 1 is bigger than all those numbers in the sequence, yet smaller than any other number above that sequence. In other words, 1 is the least upper bound for the sequence a_n.

There are many fancy ways to prove that lim sup a_n=1.. or equivalently we can prove sup A=1. ( which is the same thing just proving it as a set instead of a sequence)

However we can use the Axiom of completeness, Every nonempty set of real numbers that is bounded above has a least upper bound. ( Stephan Abbott)

So all we have to do is show that the least upper bound for A is actually 1. Fuck.

By definition of upper bound: a set in R is bounded above if there exists a number b in R st a<b for all a in A. The number b is called the upper bound for A.* Where R is the complete real number line, a is any element in A, and b =1 ( For context )

However we have only shown that 1 is an upper bound for A. We need to show it is the smallest upper bound for A. ie, least upper bound:

We can show this by definition (Abbott)

A real number b is the least upper bound for a set A in R if it meets the following two criteria:
(i) b is an upper bound for A
(ii) if c is any upper bound for A then b <= c, (less than or equal to) ( means that b is the 'smallest' upper bound)

We proved (i).
(ii) is also trivial but can be proved by using Theorem 1.11 (Rudin)
However we don't need to keep proving it. The sup A=1, which implies that lim sup a_n =1 as n approaches infinity, which implies " .99999.... =1 "

*

There are definitely other ways to prove it but I'm not about that life anymore.


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Post Monday, 9th January 2017, 05:23

Re: math q

what the heck

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Post Monday, 9th January 2017, 10:42

Re: math q

Infinity is not a whole number, so the OP's opening statement is actually not true. Infinity is more like a concept. It's not really applied to math the way most people think.

It's like asking someone what's the closest number to 0? Then you say, "take a number and halve it". They say, "ok. is that that the closest number to zero?". Then you say, "Now, halve it again." And, you repeat that process forever.
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Post Monday, 9th January 2017, 11:01

Re: math q

Of all the threads to zombify, I vote this one as the most worthy.
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Post Monday, 9th January 2017, 11:57

Re: math q

TonberryJam wrote:Infinity is not a whole number, so the OP's opening statement is actually not true. Infinity is more like a concept. It's not really applied to math the way most people think.

It's like asking someone what's the closest number to 0? Then you say, "take a number and halve it". They say, "ok. is that that the closest number to zero?". Then you say, "Now, halve it again." And, you repeat that process forever.



I definitely see your point here, however, notice that i use the word approaches infinity. This means I will never actually reach infinity.

I used a limit that 'approaches' infinity. As it is an idea not a number.

My statement would've been false if I had said. N is infinity.
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Post Monday, 9th January 2017, 15:42

Re: math q

it felt like i made this thread in june or july, not august
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Post Monday, 9th January 2017, 23:59

Re: math q

twelwe wrote:it felt like i made this thread in june or july, not august

i thought it was a year old, personally
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Post Tuesday, 10th January 2017, 00:04

Re: math q

TonberryJam wrote:Infinity is not a whole number, so the OP's opening statement is actually not true. Infinity is more like a concept. It's not really applied to math the way most people think.

It's like asking someone what's the closest number to 0? Then you say, "take a number and halve it". They say, "ok. is that that the closest number to zero?". Then you say, "Now, halve it again." And, you repeat that process forever.

Infinity is not a whole number, nor is it a rational or real number, but it is an ordinal, cardinal, and extended real number. What is or isn't a number just depends on what numbers you're working with. Similarly, in the whole, ordinal, and cardinal numbers there is a non zero number closest to zero, but there isn't in the rational, real, or extended reals. There's even two in the integers!
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Post Tuesday, 10th January 2017, 03:04

Re: math q

what number is the closest element to infinity though? :twisted:
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Post Tuesday, 10th January 2017, 03:05

Re: math q

n1000 wrote:what number is the closest element to infinity though? :twisted:


there isn't in any number system i know of!
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Post Wednesday, 11th January 2017, 00:54

Re: math q

Arrhythmia wrote:
n1000 wrote:what number is the closest element to infinity though? :twisted:


there isn't in any number system i know of!


That depends on what metric function you use in your topological space.
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Post Wednesday, 11th January 2017, 03:14

Re: math q

n1000 wrote:what number is the closest element to infinity though? :twisted:


Haha you just opened up a new can of worms lmao. Ill just leave this here

https://en.wikipedia.org/wiki/Cantor's_paradox


tldr;

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Post Wednesday, 11th January 2017, 07:19

Re: math q

papilio wrote:
Arrhythmia wrote:
n1000 wrote:what number is the closest element to infinity though? :twisted:


there isn't in any number system i know of!


That depends on what metric function you use in your topological space.


fair enough but i don't know of any reasonable one.
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Post Wednesday, 11th January 2017, 11:53

Re: math q

papilio wrote:That depends on what metric function you use in your topological space.


Just to be that guy: Not all topological spaces have their topology derived from a distance, and some just can't be metrizable.
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Post Wednesday, 11th January 2017, 13:45

Re: math q

Arrhythmia wrote:
n1000 wrote:what number is the closest element to infinity though? :twisted:


there isn't in any number system i know of!

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Post Wednesday, 11th January 2017, 17:09

Re: math q

I like where this tread is going.

Okay here is a question.

Imagine a hotel with a infinite number of rooms numbered (1,2,3,4,...) and each room is occupied.

would you be able to make room for one more guest?


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Post Wednesday, 11th January 2017, 20:39

Re: math q

TrumpTrain wrote:Imagine a hotel with a infinite number of rooms numbered (1,2,3,4,...)


as a follower of Brouwer I reject anyone's claim to have completed this step

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Post Thursday, 12th January 2017, 00:39

Re: math q

TrumpTrain wrote:Imagine a hotel with a infinite number of rooms numbered (1,2,3,4,...) and each room is occupied.

would you be able to make room for one more guest?

Even an infinite number of more guests.
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Post Thursday, 12th January 2017, 01:46

Re: math q

TrumpTrain wrote:I like where this tread is going.

Okay here is a question.

Imagine a hotel with a infinite number of rooms numbered (1,2,3,4,...) and each room is occupied.

would you be able to make room for one more guest?


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https://en.wikipedia.org/wiki/List_of_paradoxes


better question: the premise is the same, but an infinite (read: aleph_0) number of busses each filled with an infinite (aleph_0) number of guests show up. now can you make more room??
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Post Thursday, 12th January 2017, 06:22

Re: math q

Here's a fun one I proved today in algebraic topology.

Let f be a continuous real function defined on S^2 (thought of as unit vectors in R^3). Then there exist three mutually orthogonal vectors, v_1, v_2, and v_3 \in R^3 such that f(v_1) = f(v_2) = f(v_3).
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Post Friday, 13th January 2017, 01:57

Re: math q

Arrhythmia wrote:Here's a fun one I proved today in algebraic topology.

Let f be a continuous real function defined on S^2 (thought of as unit vectors in R^3). Then there exist three mutually orthogonal vectors, v_1, v_2, and v_3 \in R^3 such that f(v_1) = f(v_2) = f(v_3).



I'd like to see this proved. I have no idea where to start this.
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Post Saturday, 14th January 2017, 09:02

Re: math q

TrumpTrain wrote:
Arrhythmia wrote:Here's a fun one I proved today in algebraic topology.

Let f be a continuous real function defined on S^2 (thought of as unit vectors in R^3). Then there exist three mutually orthogonal vectors, v_1, v_2, and v_3 \in R^3 such that f(v_1) = f(v_2) = f(v_3).



I'd like to see this proved. I have no idea where to start this.


Essentially:

1. Show every continuous function SO(3) \rightarrow R^3 - \{(x, x, x) : x \in R\} is null-homotopic by looking at the fundamental groups (\pi_1(SO(3)) = C_2, \pi_1(R^3 - \{(x, x, x) : x \in R\} = Z (deformation retractable to the circle)) so the only homomorphism between them is the trivial one).

2. Represent orthogonal triples of vectors in S^2 as elements of SO(3).

3. Assume towards contradiction that there is a continuous function f which doesn't satisfy the problem's conditions. Then (its representation in SO(3)) must map to R^3 - \{(x, x, x) : x \in R\} so it's null-homotopic, but you can come up with a path in S^2 which won't be null-homotopic because its image (under the SO(3) representation) will wind around the pole.

I could write it up more formally, if you want.
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