20181207, 23:58  #100  
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{3}·53 Posts 
Quote:
If I have an infinite number of $1 bills and you have an infinite number of $100 bills, who has more money? Are there more decimal numbers between 0 and 1 or between 0 and 100? Alternatively, plot y = x. Which has more points, the line or the rest of the coordinate grid? I’m pretty sure that I understand why the answer to both is that they are equal (amounts of money/points). But I’m not sure how to explain to my friends, who think that the amounts in both questions are unequal... Quote:


20181208, 09:30  #101  
Dec 2012
The Netherlands
10110001011_{2} Posts 
Quote:
the elements of A with distinct elements of B, with none left over on either side. In this case, the function f(x)=100x associates each number between 0 and 1 with a distinct number between 0 and 100 and none are left over. 

20181208, 15:11  #102 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
110101000_{2} Posts 
We finally went over the chain rule, which means that I was able to describe the quotient rule in terms of it and the product rule. We start off by using the quotient rule to have something to compare to:\[\frac{d}{dx}\dfrac{7x^24}{3x^2+2}=\dfrac{14x(3x^2+2)6x(7x^24)}{(3x^2+2)^2}=\dfrac{42x^3+28x42x^3+24x}{9x^4+12x^2+4}=\dfrac{52x}{9x^4+12x^2+4}\]Setting this up with the other two rules:\[\frac{d}{dx}(7x^24)(3x^2+2)^{1}+\frac{d}{dx}(3x^2+2)^{1}(7x^24)=\dfrac{14x}{3x^2+2}+\dfrac{1*6x(7x^24)}{(3x^2+2)^2}\]Gotta have the same denominators for both terms...\[\dfrac{14x(3x^2+2)}{9x^4+12x^2+4}+\dfrac{42x^3+24x}{9x^4+12x^2+4}=\dfrac{42x^3+28x42x^3+24x}{9x^4+12x^2+4}=\dfrac{52x}{9x^4+12x^2+4}\]Cool! I’m not sure how someone came up with the quotient rule, which less complicated to use than combining the product and chain rules.

20181209, 10:00  #103  
Dec 2012
The Netherlands
2613_{8} Posts 
Quote:
Take any differentiable functions f and g, and let \(h(x)=\frac{1}{g(x)}\) and \(F(x)=\frac{f(x)}{g(x)}\). Then \(h(x)=g(x)^{1}\) so, by the chain rule, \[ h'(x)=(1)g(x)^{2}g'(x)=\frac{g'(x)}{g(x)^2}\] and \(F(x)=f(x)g(x)^{1}=f(x)h(x)\) so, by the product rule (and using the above expression for \(h'(x)\)), \[ F'(x)=f'(x)h(x)+f(x)h'(x)=\frac{f'(x)}{g(x)}\frac{f(x)g'(x)}{g(x)^2} =\frac{f'(x)g(x)f(x)g'(x)}{g(x)^2}. \] 

20181209, 13:17  #104  
Feb 2017
Nowhere
3,319 Posts 
Quote:
As a historical note, Galileo wrote Two New Sciences after being put under house arrest by the Inquisition. He had, many years previously, done a number of experiments, not yet published. He got out his old notes, and wrote them up. Quote:
Last fiddled with by Dr Sardonicus on 20181209 at 13:21 

20181209, 15:24  #105  
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{3}×53 Posts 
Quote:
Quote:
A side note: on this day in 1968 (thanks Wikipedia!), Douglas Engelbart provided a 90 minute public demonstration of computer components that are essential to modern personal computing: Quote:
It seems that many of the presented technologies had already been developed several years before, but Engelbart was the first to put them all together in a public demonstration. At the time, computers were only used for calculations and other number crunching tasks, but Engelbart had an idea that computers could also be used to augment people's minds. He wanted to create a machine that could be used interactively to “augment their intelligence,” by which I assume he means be more productive or efficient. The article mentions, among other things, that Engelbart's team custom made two modems running at 1200 baud (a unit of data transfer speed? Something to do with dial up?) and that they had a live video feed of Engelbart working at Menlo Park, 30 miles away. How did they do live video back then? I’m somewhat sure that it couldn’t have been over a telephone line... 

20181210, 09:31  #106  
Dec 2012
The Netherlands
1419_{10} Posts 
Quote:
If the line has only 2 possible states then this is the same as the number of bits per second, but some communication channels have more than 2 states, and then the number of bits per second is higher than the baud rate. 

20181210, 14:43  #107  
"William"
May 2003
New Haven
2·3^{2}·131 Posts 
Quote:
Given the list, this constructed number is a well defined number between 0 and 1 that is not on the list  a contradiction. You could make a new list with this number, but then you could construct a new number not on the new list. The list cannot exist  so there are more real numbers between 0 and 1 than there are integers. Some infinities are larger than others, and not just in the "trivial" sense that they are supersets (like the integers contain all the even integers, but the two sets are the same size). 

20181210, 14:46  #108  
Feb 2017
Nowhere
3319_{10} Posts 
Quote:
A general result about sets shows that there is always a "bigger" set than a given set S. Let S be a set, and 2^{S} (sometimes, Pow(S)) the set of all subsets of S, or "power set" of S. (This set is postulated to exist for any "given" set S.) The power set is sometimes formulated as the set of all functions Each function in this set corresponds to a unique subset of S, namely A standard, easytoprove result is Theorem: Let S be a set. For each function from S to 2^{S}, there is at least one element of 2^{S} which is not in the range of f. Proof: Let be a function with domain in S, and range in 2^{S}. (The function is such a function, which is even onetoone with domain all of S, showing that 2^{S} is "at least as big" as S itself.) Now let If A(f) = f(a) for some element a in S, then, is the element a in A(f) = f(a)? Well, if it is, is isn't, and if it isn't, it is. Uhoh, trouble! Conclusion: A(f) is not in the range of f, and the proof is complete. It follows from this result that 2^{S} is "strictly bigger" than S itself. Now, let S be the set of positive integers. Define a function from 2^{S} as follows. If A is a subset of S, let Thus, f(A) is a real number between 0 and 1. Unfortunately, f isn't quite a onetoone correspondence. The finite, nonempty subsets and the complements of finite, nonempty subsets represent the same numbers, namely the set of rational fractions f, 0 < f < 1, with denominator a power of 2. One representation ends in all 0's and the other in all 1's. Luckily, this set is "only as big" as the set of positive integers, so the conclusion still follows, that the set of real numbers in [0,1] is "strictly bigger" than the set of positive integers. Aside: The cardinality of the set of positive integers is often indicated by , "Alephnull." This may defined as the cardinality of the set of all finite ordinal numbers(*), and as such is the first infinite cardinal number. The next cardinal number, is the cardinality of the set of ordinal numbers of sets which are either finite or of cardinality . The hypothesis that is called the continuum hypothesis, and (if memory serves) has been shown to be logically independent of the usual axioms governing these things. (*) An ordinal type is set S with a linear ordering (like the usual ordering of the integers, the rational numbers, or the real numbers). An ordinal number is an ordinal type in which the linear ordering is also a wellordering; that is, in which each nonempty subset has a least element under the ordering. The usual ordering of the positive integers is a wellordering, but the usual linear orderings of the set of all integers, the rational numbers, and the real numbers, (or of the positive rational or positive real numbers) are not wellorderings. Last fiddled with by Dr Sardonicus on 20181210 at 14:51 

20181211, 00:51  #109 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
424_{10} Posts 
I’m kinda confused now, with this whole “orders of infinity” thing. wblipp's response was slightly understandable, but I’m completely lost on subsets/power sets/rational fractions
The main hindrance to my understanding of wblipp's post is the constructed number. Can you elaborate on what you mean by all of the decimal positions, and what would be on the list in the first place? I don’t even know what’s confusing about Dr. Sardonicus's post, it’s just way too complex... I tried using Nick's infinite coin analogy as an explanation to my friend, but we got hung up on the first case, where the coins aren’t numbered. I didn’t have much time to present it, so there may have been a misunderstanding or something Often I find that some things just make sense; Nick's coins immediately clocked for me. When I try to explain things like that, and it doesn’t immediately make sense for others, I have no clue how to explain it. For a really basic example, how would you explain negative numbers to someone who has never thought of them? Someone from several hundred years ago? I don’t remember when negative numbers became widely accepted, although I do remember that Euler pioneered imaginary numbers (or did he invent the notation?), maybe that’s a better example 
20181211, 05:56  #110 
Romulan Interpreter
Jun 2011
Thailand
10000110111011_{2} Posts 
He means, assume you can "count" all real numbers in (0,1) interval (i.e. put them in a correspondence with the natural numbers).
Then you have them in some arbitrary order: 1 : 0.104752003005..... 2 : 0.379805307156..... 3 : 0.952057123459..... 4 : 0.195305712345..... 5 : 0.619527542131..... etc Then you construct the following number (see the underlined positions): if the nth decimal in nth number in the series is not 2, your number has a 2 there, but if it is 2, then your number has a 3 there: 0.22323..... Now, you can prove easy that this number is not in your list, because it is different of any number in the list (they differ at position n, at least  it is not equal to the first, because the first position in the first number is not 2, but in your number it is 2, it is not equal to the third because the third has a 2 on the third position, but your has a 3, etc). This idea is kind of "the same" as when you prove there are infinite prime numbers, just assume you put all in a set and construct one which is not is the set. The method is called "diagonalization" and it should worth mentioning that it is not accepted by all mathematicians (some say the method is not valid, because things get strange when you go to infinities!  fortunately for this particular task, there are other ways to prove it). You can search the web for Cantor sets, alefzero, etc. By the way, the set of rational numbers (Q) has the same cardinality as the set of natural numbers (N), i.e. you can have a bijection from all naturals to all fractions, i.e. there are as many natural numbers as there are fractions with natural/integers numerators and denominators. When I was in grade 11, the teacher shown us the counting method, I had a hard time to grasp this  think about it: all pairs of integers are countable, then all triples are countable (just make a cube and start from a corner, instead a square)  all quadruples are countable... all nuples are countable, but some guy comes and tells me that 2^n is not... Grrrr... . Last fiddled with by LaurV on 20181212 at 07:19 Reason: spaciation... fonts... grrr.... 
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