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Smart? (pg. 9)
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| drizzt81 |
| quote: | Originally posted by Flyboy217
It in fact fails for n=3. Not sure where you're going with the GCF of 2 and 4, but since I myself never solved this one, I can't comment :). I'm told the solution is non-trivial. I'll get on it soon :). |
it fails for 3, yet works for 4? that blows my mind. |
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| drizzt81 |
| quote: | Originally posted by Resnick
heres the quote from the proof you gave:
Now, that doesn't mean choosing a rational number is impossible, but it just means it's VERY unlikely if you're truly choosing at random. |
yes but that is what he said.. and in the limit it approaches 1 (i.e. 100%) it seems quite clear to me :) |
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| Resnick |
| quote: | Originally posted by drizzt81
yes but that is what he said.. and in the limit it approaches 1 (i.e. 100%) it seems quite clear to me :) |
nonononon
heres how this discussion started:
flyboy said that it was 100% to pick irrational, i said no its not, its only a limit, then he said no its 100% everytime no matter what, and diginut and i argued this for a while,
so now im saying yes it is a limit and flyboy and diginut dont know what theyre talking about.
Furthermore, i would like to see someone actually prove this limit (im not saying that its not true, just saying that if ur so confident in ur answer, you should be able to mathematically prove this) |
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| Resnick |
| quote: | Originally posted by Flyboy217
Not really sure what you mean. The actual value of the probability of choosing an irrational number is 1. Not just close to 1, but 1. You will, with 100% probability, pick an irrational number. But you still may pick a rational one. (This sounds strange, but read on).
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^^^ thats his exact quote in reply to my limit answer originally |
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| Flyboy217 |
| quote: | Originally posted by Resnick
flyboy said that it was 100% to pick irrational, i said no its not, its only a limit
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Yes, I did say that. And it's correct. Perhaps you didn't read the article I gave you. Keep in mind that "1" and "100%" are the same thing:
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So if you choose a random number from the real line, the probability of choosing an irrational number is 1.
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Now this makes no sense:
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, then he said no its 100% everytime no matter what, and diginut and i argued this for a while,
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Sigh... I said the probability is 100%. "100% everytime no matter what" has no meaning, and I challenge you to find where I said such a thing. Read again and you'll see where I explained that you may pick a rational number.
This point is, and never was, in contention; you simply don't understand it. As I said before, please seek a math professor or other trusted source to learn about measure theory. While I enjoy helping, I can only do so much online.
I don't intend to flame, but please be advised that any further contention of this point will be ignored. |
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| Resnick |
| quote: | Originally posted by Flyboy217
Perhaps you didn't read the article I gave you. Keep in mind that "1" and "100%" are the same thing:
Now this makes no sense:
Indeed I said it's 100%. "100% everytime no matter what" has no meaning, and I challenge you to find where I said such a thing. Read again and you'll see where I explained that you may pick a rational number.
This point is not in contention; you simply don't understand it. As I said before, please seek a math professor or other trusted source to learn about measure theory. While I enjoy helping, I can only do so much online.
I don't intend to flame, but please be advised that any further contention of this point will be ignored. |
no it is you who doesnt understand the concept of a limit.
the author of the article was just trying to tone it down for the person who posed the question, he didnt imply that its actually 1, he meant the limit is 1.
Furthermore i would be imbarassed to ask such a stupid question from one of my profs.
But just answer this one last question before you leave, what you are saying is that it is actually 100% to pick an irrational # and not a limit in any way? if your saying that than it CANT be possible to pick a rational number. for instance if u have 1 blue cup and infinite red cups, what is the probability that you will pick the blue cup? NO mathematician will ever say its zero..NONE. because thats pure BS. |
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| Flyboy217 |
| quote: | Originally posted by Resnick
no it is you who doesnt understand the concept of a limit.
the author of the article was just trying to tone it down for the person who posed the question, he didnt imply that its actually 1, he meant the limit is 1.
Furthermore i would be imbarassed to ask such a stupid question from one of my profs.
But just answer this one last question before you leave, what you are saying is that it is actually 100% to pick an irrational # and not a limit in any way? if your saying that than it CANT be possible to pick a rational number. for instance if u have 1 blue cup and infinite red cups, what is the probability that you will pick the blue cup? NO mathematician will ever say its zero..NONE. because thats pure BS. |
I'm going to break my promise to ignore this. You asked me to answer this question before I left. Yes, it is exactly 100%. The author of the article was not trying to tone anything down, and he was not "implying" anything. As he said, it is 1. As a final reference, I give you:
http://www.greylabyrinth.com/Puzzles/answer006.htm
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100% of all integers contain at least one three.
What?!? How can this be? The solution is so surprising, it is difficult, if not impossible to believe that 100% of integers contain the digit three at least once. The simple fact that the number 8, for example, has exactly zero threes in it seems to dispute this.
This seeming paradox illustrates one of the many "problems" associated with trying to apply concepts (like percentages) used for regular sets on the infinite. This puzzle, to the best of my knowledge, was originally posed by Clifford Pickover, the author and mathematician.
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The point to understand here is that, just because something has a 0% probability of occurring (selecting a rational number), that doesn't mean it will not occur.
Instead of being embarrassed to ask a professor, see it as a way to learn. Ask him what the *exact* probability of choosing a rational amongst a nonzero real number interval is. If he answers anything other than "100%", I shall eat my hat. |
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| Resnick |
| quote: | Originally posted by Flyboy217
The point to understand here is that, just because something has a 0% probability of occurring (selecting a rational number), that doesn't mean it will not occur.
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yet again u dont understand the concept of a limit.
first of all that quote makes no sense.
second that link you gave me proves that all digits have 3 using a limit.
YOU go ask your prof this question exact question and explicitly ask him if it is really 1 or a limit.
oh and diginut, plz reply to this, i wanna know whose right, even tho u said i was wrong before, maybe u thought i was saying its not true, but in fact all im saying is that its a limit. |
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| DigiNut |
| quote: | Originally posted by Resnick
yet again u dont understand the concept of a limit.
first of all that quote makes no sense.
second that link you gave me proves that all digits have 3 using a limit.
YOU go ask your prof this question exact question and explicitly ask him if it is really 1 or a limit.
oh and diginut, plz reply to this, i wanna know whose right, even tho u said i was wrong before, maybe u thought i was saying its not true, but in fact all im saying is that its a limit. |
Flyboy is right.
This has nothing to do with limits. It's not a calculus problem. Res, I'm not trying to put you down but it seems like you just haven't learned stats yet, so you don't have the knowledge to be arguing this issue.
Think of it this way: if a random integer function is uniformly distributed over the interval [1,n], then the probability of it being a particular value "x" is exactly 1/n. Yes it's true, that this function has a "limit" of zero as n -> infinity. However, that has nothing to do with it. If your random integer function is distributed over the ENTIRE INTEGER SET, which means that the value of "x" is unbounded and therefore n IS infinity, then the probability of picking any individual integer "x" from that last is EXACTLY zero.
So to correlate this to picking a rational number on the set of real numbers (which includes irrational numbers), refer to Flyboy's post on different types of infinite. |
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| Flyboy217 |
| quote: | Originally posted by Resnick
yet again u dont understand the concept of a limit.
first of all that quote makes no sense.
second that link you gave me proves that all digits have 3 using a limit.
YOU go ask your prof this question exact question and explicitly ask him if it is really 1 or a limit.
oh and diginut, plz reply to this, i wanna know whose right, even tho u said i was wrong before, maybe u thought i was saying its not true, but in fact all im saying is that its a limit. |
Probability(contains a three) = lim x -> inf (1-.9^x) = 1
Yes, it's a limit, and that limit is *exactly equal* to 1.
Yes, outcomes with 0% probability of occurring can occur when dealing with infinities.
As for asking my professors, I have asked this and much more. In fact, I went on to author several publications in Quantum Computing with them by age 18, just before graduating from college. Google my name if you're interested in reading them:
http://www.google.com/search?hl=en&lr=&ie=UTF-8&oe=UTF-8&q=%22Aditya+K+Prasad%22&btnG=Google+Search
Now go back to solving those problems. ;) |
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| Resnick |
ok ok one last point and i promise ill shut up, your right, i havent taken stats
all im arguin is that you cant say something like:
inf/inf = 1
0/inf = 0
0/0= 0
n/inf=0
and stuff like that
one last thing, is stats an engineering course? what faculty is it from and what year course is it? |
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| drizzt81 |
| quote: | Originally posted by DigiNut
Think of it this way: if a random integer function is uniformly distributed over the interval [1,n], then the probability of it being a particular value "x" is exactly 1/n. Yes it's true, that this function has a "limit" of zero as n -> infinity. However, that has nothing to do with it. If your random integer function is distributed over the ENTIRE INTEGER SET, which means that the value of "x" is unbounded and therefore n IS infinity, then the probability of picking any individual integer "x" from that last is EXACTLY zero. |
OK, sorry for jumping in on this, but i am a curious mind -sometimes and right now everything is better than more 8051 code ;) - so:
you say that by definition 1/(infinity) = 0, and that is the reason why the probability of picking any single number is zero, right?
the only reason we can say that 1/(infinity) is zero is using the limit concept on 1/n as n -> (infinity) ?
Do I understand this correctly? |
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