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Re: math q
twelwe wrote:It's like Blink, but you end up drowning.
If it doesn't fit anywhere else, it belongs here. Also, come here if you just need to get hammered.
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twelwe wrote:It's like Blink, but you end up drowning.
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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?
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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
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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).
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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
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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
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
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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:
[b]But that' not considered a valid number, real or unreal, rational or irrational. Yet it still seems to exist.[\b]
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:
[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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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:
[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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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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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.
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twelwe wrote:it felt like i made this thread in june or july, not august
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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.
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n1000 wrote:what number is the closest element to infinity though?
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Arrhythmia wrote:n1000 wrote:what number is the closest element to infinity though?
there isn't in any number system i know of!
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n1000 wrote:what number is the closest element to infinity though?
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papilio wrote:Arrhythmia wrote:n1000 wrote:what number is the closest element to infinity though?
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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papilio wrote:That depends on what metric function you use in your topological space.
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Arrhythmia wrote:n1000 wrote:what number is the closest element to infinity though?
there isn't in any number system i know of!
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TrumpTrain wrote:Imagine a hotel with a infinite number of rooms numbered (1,2,3,4,...)
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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?
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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?Spoiler: show
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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).
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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.
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