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Originally posted by Resnick flyboy said that it was 100% to pick irrational, i said no its not, its only a limit

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.

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,

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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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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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.

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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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.

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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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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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.

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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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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Originally posted by 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?

It's true, you cannot use infinity as a number in an equation.

It is pretty strange that things with 0% probability of occurring can still occur... lots of things are strange about infinities. Who ever would think that there are "countable" and "uncountable" infinities before learning them?

We had to take stats for Computer Engineering, but I doubt you'll learn about this specifically. I'd suggest learning about measure theory, or some other related branches of probability.

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Originally posted by 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?

yes you cannot:

inf/ inf is INDETERMINATE, i.e. can be anything and nothing, which relates to the 'size' of infinity. I remember a friend of mine always yelling "some infinities are bigger than others" which is true. in order to find out what inf/ inf is, you need to know the rate at which either functions approach infinity. And if you relate the integers to the irrational numbers, you will see for each integer step you add an INFINITE amount of irrational numbers, therefore the size of the irrational infinity grows much faster than the size of the integer infinity, therefore it must be much bigger. By how much? by infinity of course.

consider:

(3 * (all integers)) / (4 * (all integers)) = 3/4 only because the two infinities are the same.

now someone, please correct me and make me look like a dumb fool

edit spelling

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Originally posted by drizzt81 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?

That's correct. This is not why it's true for picking a rational out of the reals, however. That has to do with the concept of uncountable infinities being of a higher-order cardinality than the countable ones.

lim x->(infinity) 1/x = 0


As X APPROACHES infinity, 1/x APPROACHES 0. What 1/x really equals when x is EQUAL to infinity is irrelevant.

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Originally posted by drizzt81 yes you cannot: inf/ inf is INDETERMINATE, i.e. can be anything and nothing, which relates to the 'size' of infinity. I remember a friend of min always yelling "some infinities are bigger than others" which is true. in order to find out what inf/ inf is, you need to know the rate at which either functions approach infinity. And if you relate the integers to the irrational numbers, you will see for each integer step you add an INFINITE amount of irrational numbers, therefore the size of the irrational infinity grows much bigger than the size of the integer infinity, therefore it must be much bigger. By how much? by infinity of course. consider: (3 * (all integers)) / (4 * (all integers)) = 3/4 only because the two infinities are the same. now someone, please correct me and make me look like a dumb fool edit spelling

You're wrong, you idiot... it's "friend of mine" not "friend of min"

;-)

Yes, you're right.

k now im confused, let me ask this example again

u have 1 red cup, and infinity blue cups

what is the probability of choosing a red cup?
is the answer dependant on a limit?

and im not trying to argue anything anymore, just wanna see if this has anything to do with the countable/uncountable inifinities (which i havent learned)

everyone feel free to comment

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Originally posted by Flyboy217 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.

woah, now this is really confusing (btw sry i just read this)

my whole argument is that its a limit and ppl are saying its not, and now your saying it IS a limit?

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Originally posted by Resnick k now im confused, let me ask this example again u have 1 red cup, and infinity blue cups what is the probability of choosing a red cup? is the answer dependant on a limit? and im not trying to argue anything anymore, just wanna see if this has anything to do with the countable/uncountable inifinities (which i havent learned) everyone feel free to comment

The problem is that there is no such thing as "infinity" blue cups. However, we can represent this problem by using the set of integers, where the red cup represents some integer (say, zero), and the blue cups are all the other integers.

The probability of picking red is (exactly) zero. It does come from our knowledge of limits. If we had 1 red cup and x blue cups, then the probability would be 1/x. As you know, lim x->inf 1/x = 0. As has been pointed out, it doesn't make sense to say "but what is it when x = infinity" because there is no such number. However, knowing that the cardinality of the real numbers is "countably infinite," we can answer that the probability equals the limit, which is exactly equal to zero.

One problem many people have is when they say that "the limit approaches zero." This is incorrect. The limit doesn't approach zero; the limit IS zero (as the free variable x approaches infinity).

In the cups case, the probability of choosing any item of a finite set from an uncountably infinite one is zero. As noted above, though, this is different from the case of picking a rational from the reals. In this case, the probability of picking an item from a countably infinite-sized one from an uncountably-sized one is zero.

This last point has to do with the "measure" of a set. The measure of a particular interval is equal to the difference of its max and min elements. The probability of picking an integer from [1,2] out of [1,3] is

P(Picking something from [0,1] out of [0,2]) =
Measure [0,1] / Measure [0,2] = 1-0/2-0 = 1/2

The measure of any particular point, x, is x-x = 0. The measure of a countably infinite set (such as the integers or rationals) is actually zero.

P(Picking a rational from the real interval [0,2]) =
Measure (rationals) / Measure [0,2] = 0/(2-0) = 0

To make things weirder, there are sets of uncountably infinite size that also have measure zero (read up on the Cantor set).

Hope this helped.

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Originally posted by Flyboy217 5) You walk into a candy store in a foreign land that has only 2 types of coins. The coins are worth A and B, where A and B are relatively prime (share no common factors other than 1). Assuming you have an infinite number of each coin, what is the price of the most expensive candy that you cannot buy with exact change? That is, find the largest price such that you will require change to purchase it exactly. [20 min.]

solution approach: you know that you CAN buy any candy of cost C, where C = xA+yB [x,y are integers >= 0]

Let's assume that B < A. This will not loose any generality, since it doesn't matter which one is bigger, but one HAS to be, seeing that they do not share common factors, hence are not the same number.

Therefore anything that costs C, C element of (B, thelesserof(2B, A) cannot be bought.

reconsidering:

B/A = q (a number), we can buy anyting that costs
C=B(y + x/q), we know that q < 1, because A > B

q is the 'step' in our function.
-scrap all that-
ok.. i looked at some examples:

The largest price that you canot pay in correct change is:
the product of the first prime number larger than the big number times the first prime number smaller than the smaller number.

why? Because that number will not be a linear combination of the two numbers that you are using...

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Originally posted by drizzt81 solution approach: you know that you CAN buy any candy of cost C, where C = xA+yB [x,y are integers >= 0] Let's assume that B < A. This will not loose any generality, since it doesn't matter which one is bigger, but one HAS to be, seeing that they do not share common factors, hence are not the same number. Therefore anything that costs C, C element of (B, thelesserof(2B, A) cannot be bought. reconsidering: B/A = q (a number), we can buy anyting that costs C=B(y + x/q), we know that q < 1, because A > B q is the 'step' in our function. -scrap all that- ok.. i looked at some examples: The largest price that you canot pay in correct change is: the product of the first prime number larger than the big number times the first prime number smaller than the smaller number. why? Because that number will not be a linear combination of the two numbers that you are using...

A = 5, B = 3. Then you get 7*2 = 14. But 14 = 3*3+5*1. Try again . Think modular arithmetic.

k cool, thanks for the help flyboy and diginut

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Originally posted by Flyboy217 A = 5, B = 3. Then you get 7*2 = 14. But 14 = 3*3+5*1. Try again . Think modular arithmetic.

My knowledge of modular arithmetic is extremely limited and I don't even know the correct terminology, so maybe you could help me along here with my assumptions... unless you think it's totally giving it away.

If we have two primes A and B, then our mod base ought to be A+B. That is to say, we *know* that you can make exact change C with coins A and B if you can make exact change for (C % (A+B)). However, the reverse isn't automatically true, so I'm not sure where to go from there... let's say you take 5 and 7, which is mod 12, then you can't make 11, you can't make 23, but you can make 35...

I think that at the very least, the answer must be some function of A+B, and the largest prime less than (A+B-1)? Or maybe I'm on the wrong track... I don't have time to read all about modular arithmetic.

sry i didnt read other answers to this question

but is it (A*B) - (A+B)??

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Originally posted by Resnick sry i didnt read other answers to this question but is it (A*B) - (A+B)??

Yes, that's exactly correct . It's easy to find by just experimenting with small numbers. Now the trick is proving it.

DigiNut, I think it's actually easier to think of it without specifically using modular arithmetic. Basically, think about a value after which you know ALL values can be achieved, and work your way back. If you have some value x*A + y*B, then what does taking away one A and adding a B do to your value? You can obviously achieve all prices P where P%A = 0 and P%B = 0. How do you achieve the rest?

To be honest, my proof is inelegant... but it's still neat to figure out.

I'll explain more later. I'm currently making a mix

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Originally posted by Flyboy217 DigiNut, I think it's actually easier to think of it without specifically using modular arithmetic. Basically, think about a value after which you know ALL values can be achieved, and work your way back. If you have some value x*A + y*B, then what does taking away one A and adding a B do to your value? You can obviously achieve all prices P where P%A = 0 and P%B = 0. How do you achieve the rest?

Ok, here's one way of looking at it:

If you take A*B, you know that you have a number that's divisible by BOTH A and B (that is, P%A=0 and P%B=0 like you said). If A and B are relatively prime, you know the following things:
- (A+B)%A = B, and (A+B)%B = A. Therefore (A+B)%A > 0, and (A+B)%B > 0. (1)
- If you subtract A from (A*B), you cause the result to be indivisible by B (with a remainder of A). (2)
- If you again subtract B from this total, you cause it to be indivisible by A (with a remainder of B). (3)
- You know that the result after step (2) will not suddenly become divisible by B again after step (3), because of lemma (1).
- (A*B) - A - B = (A*B) - (A+B)
- Therefore, this number is guaranteed to be indivisible by both A and B.

Now, I know that (A*B) - (A+B) can be written as (A-1)(B-1) - 1. So the first value where you CAN make ANY price is (A-1)(B-1). I'm sure there's a way to prove that when you have (B-1) A's and (A-1) B's already, that you can generate prices in increments of exactly 1, but I haven't worked out exactly how yet.

Actually, that last part seems intuitive, but I'm just not sure how to spell it out...

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Originally posted by DigiNut Ok, here's one way of looking at it: If you take A*B, you know that you have a number that's divisible by BOTH A and B (that is, P%A=0 and P%B=0 like you said). If A and B are relatively prime, you know the following things: - (A+B)%A = B, and (A+B)%B = A. Therefore (A+B)%A > 0, and (A+B)%B > 0. (1) - If you subtract A from (A*B), you cause the result to be indivisible by B (with a remainder of A). (2) - If you again subtract B from this total, you cause it to be indivisible by A (with a remainder of B). (3) - You know that the result after step (2) will not suddenly become divisible by B again after step (3), because of lemma (1). - (A*B) - A - B = (A*B) - (A+B) - Therefore, this number is guaranteed to be indivisible by both A and B. Now, I know that (A*B) - (A+B) can be written as (A-1)(B-1) - 1. So the first value where you CAN make ANY price is (A-1)(B-1). I'm sure there's a way to prove that when you have (B-1) A's and (A-1) B's already, that you can generate prices in increments of exactly 1, but I haven't worked out exactly how yet. Actually, that last part seems intuitive, but I'm just not sure how to spell it out...

First, some notation: A%x=0 is properly written as A = 0 (mod x), read "A is congruent to zero, modulo x". (Actually, the congruency sign properly has three bars, not two...)

It's true that AB-A-B is indivisible by both A and B. This can be proven by noting that AB=0 mod A and mod B, and therefore only numbers different from AB by a multiple of A are congruent to 0 mod A (and similarly for B).

What's important is that AB-A-B is not a positive linear combination of A and B though. I'll write up an explanation when I get a chance. You're thinking along the right lines though...

Okay. This is the ugliest, dirtiest proof I've ever written. If you don't follow it, props to you for not being f*ed in the head...

------
Call a price "attainable" if you can purchase an item of that price with exact change.

Using 0 of coin B and as many of coin A as we need, We can attain all prices C where C = 0 (mod A) (i.e., all multiples of A) starting with C = 0*B.

Using 1 of coin B and as many of coin A as we need, we can attain all prices C = xA + B. I.e., we can attain all prices C where C = B (mod A) starting with C = 1*B.

Using at most y of coin B, we can attain all prices C where C = {0*B, 1*B, ..., y*B} mod A.

Note that n*B mod A are distinct for all 0 <= n <= A-1. Thus, we can attain all prices greater than or equal to C = (A-1)B that have the same congruency as (A-1)B mod A.

Hence, (A-1)(B)-A is not attainable (since (A-1)B = (A-1)B-A (mod A) and (A-1)B is the smallest price congruent to (A-1)B mod A that is attainable). However, (A-1)B-A+1 is attainable, because it is congruent to a value other than (A-1) mod A, all other congruencies have been satisfied, and it is greater than (A-2)B (which was the last satsified congruency).

Therefore, (A-1)B - A + 1 = (A-1)(B-1) is the smallest value for which all prices greater than or equal to it can be satisfied.
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