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Old 01-31-2011, 09:12 PM
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Default Calling all Math Geeks

I see that there hasn't been a Math thread here in quite some time, so I'm going to start one.

I think I'll go with one of my all time favorite topics, the Axiom of Choice. But before we dive in, a little background would be nice.

In the late Nineteenth Century, Georg Cantor began working with sets, in what is now called a Naive Set Theory. The idea caught on quickly, as most mathematicians saw how incredibly useful his ideas were. However, just as quickly (around 1900), problems with the theory began to come to light, most notably Russell's Paradox. The issue here is that "set" was not well defined, and any collection that could be conceived was allowed to be a set.

Let us define a "normal" set as any set that does not contain itself, and say that any set that does contain itself is abnormal. Let S be the set of all normal sets. Is S normal or abnormal? If S is normal, then it belongs to S, which makes S abnormal. But if S is abnormal, then it is not in S, which means S must be normal. Hence S is normal if and only if S is abnormal. This is called Russell's Paradox.

There were two major proposals to fix this issue, Russell's Type Theory and Zermelo's axioms of set theory. Later, Zermelo together with Fraenkel proposed what is now called ZF set theory, which is an extension of Zermelo's first Axiomatic Set Theory. During this time, Zermelo also proposed the Axiom of Choice. ZF Set Theory together with the Axiom of Choice is called ZFC Set Theory and is, today, the foundation for all of modern mathematics.

The Axiom of Choice states:

Quote:
Let C be a collection of nonempty sets S. Then there exists a function f defined on C such that for all S in C, f(S) is in S.
In other words, we can choose one element out of each of the sets S in C. So, why do we even need this axiom? Russell explained it this way:

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The Axiom of Choice is necessary to select a set from an infinite number of socks, but not an infinite number of shoes
IOW, for many collections the choice can be obvious. In the case of the shoes just select, say, the left-hand ones. But how do you differentiate between socks?

Seems pretty obvious and noncontroversial, doesn't it? And yet the debate over whether or not it should be accepted raged for over 2 decades, and has been called "The Last Great Controversy in Mathematics".

The main reason it was so controversial is that as obvious as it seems, it leads to a great many results that are just as obviously false. One such result is called "The Well Ordering Principle", which basically states

Quote:
Any nonempty set can be Well Ordered
For a set S to be Well Ordered, every nonempty subset of S must contain a least element. Let's look at the Real Numbers, R. For R to be well ordered, every subset of R must contain a least element. But consider the open interval (0,1). Obviously, there is no least element for this interval, at least not under the usual ordering <. So, for the reals to be well ordered, there must exist some other nonstandard ordering. What this ordering might look like is anybody's guess, as nobody has been able to propose such an ordering.

There's even a math geek joke about it:

Quote:
The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?
These three statements are all, in fact, logically equivalent; if any one is true, so are the other two.

Another well known theorem whose proof is dependent on the Axiom of Choice is the Banach-Tarsky Paradox. This is not a true paradox. Its named that because the result is so incredibly counter-intuitive. It basically says that you can take the unit ball (a solid sphere with radius 1) in 3-dimensional space and carve it up into a finite number of pieces and then, without any stretching or other deformation of the pieces, reassemble them into two balls, each of which is identical to the original.

On the other hand, there are many basic, or even fundamental theorems in a large number of branches of mathematics whose proofs rely on the Axiom. If we reject the Axiom of Choice then huge amounts of mathematics go bye-bye unless or until we can find alternate proofs for these results.

What say you? Do you accept the Axiom of Choice or should we try to do without it?
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  #2  
Old 01-31-2011, 09:44 PM
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Default Re: Calling all Math Geeks

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Any nonempty set can be Well Ordered
For a set S to be Well Ordered, every nonempty subset of S must contain a least element. Let's look at the Real Numbers, R. For R to be well ordered, every subset of R must contain a least element. But consider the open interval (0,1). Obviously, there is no least element for this interval, at least not under the usual ordering <. So, for the reals to be well ordered, there must exist some other nonstandard ordering. What this ordering might look like is anybody's guess, as nobody has been able to propose such an ordering.
Why did you switch from set to subset? Is Well Ordered not a property of sets, but of subsets? And why can, but not must?
(I tend to be very literal when trying to understand math, and this looks confusing to me)
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Old 01-31-2011, 10:08 PM
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Default Re: Calling all Math Geeks

The reason to switch to subsets is to give you the guarantee that, say, for any two items in the set, one is less than the other. It's not just that the set as a whole has a least element, but that you can pick the least element of any subset of it. And "can" vs. "must" is there because "can" be well ordered only means that it's possible, not that you've done it.
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Old 01-31-2011, 10:14 PM
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Default Re: Calling all Math Geeks

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Originally Posted by Chris Porter View Post
Why did you switch from set to subset? Is Well Ordered not a property of sets, but of subsets? And why can, but not must?
(I tend to be very literal when trying to understand math, and this looks confusing to me)
Not a problem. Its much better to be literal or even pedantic here. Otherwise, we get sloppy and things start to go wrong :)

Well-ordered is a property of a Set, but for S to have this property requires all of the nonempty subsets of that set to meet a particular criteria. So, in this case for a particular set S to be well ordered, every subset of S must have a least element.

For example, the Natural Numbers N is the set of all nonnegative integers N = {0, 1, 2, 3, ...}. Some subsets of N include {1, 2, 3}, {1000, 10003, 100003}and the prime numbers. If we take the usual < as our order over N, the we can see that N is well-ordered under <, since every possible subset of N contains a least element. Another example of a well ordering on N is let all odd integers be greater than all even integers, and use the standard ordering with the evens and odds. Then the ordering would look like {0, 2, 4, 6, ... 1, 3, 5, ...}. Here, we still have the property that every subset of N has a least element, and thus N is still well-ordered with this ordering.

There are quite a few definitions and principles involved here that I didn't go into. (what an order is, strict ordering, etc) I felt that it may detract from the overall post, If you like, I can make a post on ordering and such that may make this clearer.

As to can vs must, not all orders on a given set will be well ordered, we are saying that for any given nonempty set, we can find an ordering that will make it well ordered.
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Old 01-31-2011, 10:15 PM
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Default Re: Calling all Math Geeks

Thanks seebs. Your explanation of can vs must is much clearer than mine!
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Old 01-31-2011, 10:17 PM
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Default Re: Calling all Math Geeks

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Originally Posted by seebs View Post
The reason to switch to subsets is to give you the guarantee that, say, for any two items in the set, one is less than the other. It's not just that the set as a whole has a least element, but that you can pick the least element of any subset of it. And "can" vs. "must" is there because "can" be well ordered only means that it's possible, not that you've done it.
I'm fascinated in so much of this stuff, especially as I'm getting into robotics and cybernetics (which is just scifi bullshit if you don't do any math) but my neural chips don't have all the needed circuitry and quickly start to overheat. So I have to use tricks and shortcuts and such. I'm also figuring out better and better ways to overclock, so to speak.

I'm hoping Aqua can be back online and post in this thread soon. That woman can crunch numbers like they are pretzels.
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Old 01-31-2011, 11:14 PM
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Default Re: Calling all Math Geeks

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I'm fascinated in so much of this stuff, especially as I'm getting into robotics and cybernetics (which is just scifi bullshit if you don't do any math) but my neural chips don't have all the needed circuitry and quickly start to overheat.
Will be interesting to see how this changes should my viral therapy succeed in completely eradicating the cancer.

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Old 01-31-2011, 11:30 PM
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Default Re: Calling all Math Geeks

So, if the Universe is just plain dumb, and yet it's described as the complete set of everything contained within, how does the subset "intelligence" fit in? This is in fact how we determine one thing is evident of another, by drawing such correlations isn't it?
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Old 02-01-2011, 03:51 AM
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Default Re: Calling all Math Geeks

Can you have a set of all radixes? Then I'm not sure if it can be well ordered.
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Old 02-01-2011, 04:31 AM
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Default Re: Calling all Math Geeks

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Can you have a set of all radixes? Then I'm not sure if it can be well ordered.
Yes, we can have a set of all radixes. Depending on how we take the scope of allowed bases, we could have something as simple as just the Natural Numbers, all the way to the Complex Numbers.

Either way, according to the Well Ordering Principle, there is some well ordering. Remember, the Principle does not guarantee that we can find the well ordering, only that it exists.
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Old 02-01-2011, 05:48 AM
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Default Re: Calling all Math Geeks

In mathematics, ordering is a formal consideration of the common notion that we can compare some pairs of elements of a set to determine which precedes the other in the set. There are several types of orderings, including preorders, partial orders, and total orders. All orders are binary relations on a set S that fulfill certain properties.

Consider a set S with a binary relation < on S. Then < is a Preorder on S if for all a,b and c in S:

a<a (reflexivity)
if a<b and b<c then a<c (transitivity)

A set S with a preorder < is called a preordered set.

A set S with a binary relation < on S is a Partial Order on S if < is a preorder on S and for all a,b in S if a<b and b<a then a=b. (antisymmetry)

Such a set is called a Partially Ordered Set, or a Poset.

A set S with a binary relation < on S is a Total Order on S if:

if a<b and b<c then a<c (transitivity)
if a<b and b<a then a=b. (antisymmetry)
a<b or b<a (totality)

As might be expected, a Set with a total order is called a Totally Ordered Set.

Note that totality implies reflexivity, so all Total Orders are also Partial Orders.

A Totally Ordered Set S is Well Ordered if and only if every nonempty subset of S contains a least element.

Please note that the binary relations here do not need to be the "less than" we are all familiar with, nor do the sets have to contain numbers. One common example of this is the common alphabet. We can then define our relation to be the standard Dictionary precedence.
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Old 02-01-2011, 01:18 PM
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Default Re: Calling all Math Geeks

It's the notes from my course on Frege's Grundlagen!

Good stuff. Thanks, John. (And seebs.)
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Old 02-01-2011, 04:38 PM
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Default Re: Calling all Math Geeks

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Please note that the binary relations here do not need to be the "less than" we are all familiar with, nor do the sets have to contain numbers. One common example of this is the common alphabet. We can then define our relation to be the standard Dictionary precedence.
Aha! Ok, then yes, I see how radixes could be well ordered. I was thinking "less than", and in that case, it must be you were only talking about numbers.

If this is the case... how about the set of all truths? I don't see how they can be well ordered.

(actually, just kidding around here with the idea of sets and order)
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Old 02-01-2011, 08:16 PM
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Default Re: Calling all Math Geeks

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What say you? Do you accept the Axiom of Choice or should we try to do without it?
You should attach a poll to this thread.
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Old 02-01-2011, 09:07 PM
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Default Re: Calling all Math Geeks

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If this is the case... how about the set of all truths? I don't see how they can be well ordered.
In order of truthiness, of course. :P

But yeah, I think the point is that it's pretty easy to imagine sets for which we have trouble imagining an ordering principle.

Normally lack of imagination isn't very convincing in logic/mathy stuff, but I guess that because we're talking about an axiom here it can be a useful guide, because we want axioms to be straightforward and non-controversial if we're going to build a whole mathematical edifice on them.
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Old 10-02-2015, 02:52 AM
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Question Re: Calling all Math Geeks

So many great maths threads to resurrect. I choose this one! :pikachu:

I was given a challenge to simulate a random 6-sided die using a random coin. I figure an 8-sided die is easy-peasy. Just flip 3 coins and assign 0 or 1 to three bits. For a 6-sided die, is there anything wrong with just throwing out the result when comes out 0 or 7? Or does that some how mess up the mathemagical randomness of it? Like will that fuck up the distribution or anything?

Code:
def roll_die():
  result = 0
  while result == 0 or result == 7:
    result = result << 1    # shift left 1 bit
    result += flip_coin()   # flip_coin() returns 0 or 1
    result = result << 1
    result += flip_coin()
    result = result << 1
    result += flip_coin()
    result = result % 8     # trim result to last 3 bits
  return result
I can, and have, tested it by running it thousands of times and checking the distribution, and it looks fairly even to me. :shrug: But let us not continue to dirty our hands with such things and instead dive into some pure theory. What say you?

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Old 10-02-2015, 07:43 AM
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Default Re: Calling all Math Geeks

I think your method is sound, in that it results in a flat distribution of 1-6. The drawback is that it's possible to end up in a long loop of 0s and 7s. I don't think it's possible to simulate a six-sided die using a random coin with a finite number of coin tosses, for the simple reason that tossing a coin N times gives you 2N possibilities which can never be a multiple of 6.
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Old 10-02-2015, 07:57 AM
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Default Re: Calling all Math Geeks

I think you're right in assuming it won't make a difference to the randomness of the remaining numbers.

I don't have any sort of proof, but if I visualise a string of say 10,000 results from your method before removing the 0s and 7s and then picture the 0s and 7s being lifted out, the remaining numbers still look randomly generated. Ha, that sounds so cheesy when I write it out. :D

But I don't see how they couldn't be, if you know what I mean.

I suppose you could say the positions of the removed 0s and 7s are also random, so it shouldn't make a difference to the remaining numbers' distribution.

For instance, picture that same 10,000 string of 8 randomly distributed numbers and just randomly generate 2500 positions to remove (removing any number in that position), wouldn't the remaining 7500 numbers be randomly distributed? I think they would and I think yours are too.
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Old 10-02-2015, 02:59 PM
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Default Re: Calling all Math Geeks

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Originally Posted by slimshady2357 View Post
...
Ha, that sounds so cheesy when I write it out. :D

But I don't see how they couldn't be, if you know what I mean.
Let's try ...

This looks like a proof by contradiction.

So assume we have a sequence of numbers 0 to 7 of a fixed finite length distributed randomly (from the 3 coin tosses). But also assume that the sub-sequence of numbers <=5 (pause to check that's unambiguous ... yes) is non-random.

(Pause to think what "non-random" means ... um ... we mean, of course, that the process generating this is random, not that we can test a particular output for being "random" - that doesn't make sense. We mean that all possible outcomes, namely sequences of numbers, are equally likely.)

This is equivalent to saying there are sequences of numbers 0-5 that are more likely than others.

The function of stripping out the 6s and 7s is, in fact, a function. (I am resisting the temptation to add lots of notation here...) And the probability of getting a particular 0-5 sequence is the sum of the probabilities of the 0-7 sequences that map to it (... since finite and disjoint events).

Now I think I'm stuck, and I need to go to a meeting. We may need to show that the number of sequences 0-7 that map to a given sequence 0-5 is constant. In fact, that's a necessary condition of the probabilities being equal.
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Old 10-02-2015, 03:41 PM
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Default Re: Calling all Math Geeks

To present another way of looking at it:

Assuming fair coin tosses, at each iteration step the probability of coming up with i (which is an element of {1,2,3,4,5,6}) 1/8, and there is a 1/4 chance that another iteration is needed (for results 0 and 7). Therefore, the total probability of coming up with i is

(probability of i in iteration 1) + (probability of i in iteration 2)*(probability of reaching iteration 2) + (probability of i in iteration 3)*(probability of reaching iteration 3) + ...

The infinite sum is implied rather than specified. This is equal to

1/8 + 1/8 * 1/4 + 1/8 * 1/4 * 1/4 + ...

or

1/8 * ( 1 + 1/4 + (1/4)2 + (1/4)3 + ...)

Since for all |x| < 1, 1 + x + x2 + x3 + ... = 1/(1-x), this is equal to 1/8 * 1/(1 - 1/4) = 1/8 * 1/(3/4) = 1/8 * 4/3 = 1/6

which is the value we are aiming for. QED, checkmate, ball center.
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Old 10-02-2015, 04:07 PM
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Default Re: Calling all Math Geeks

If the random number generator generates numbers in the range 0 to 7 with equal distribution, then any sub-set (up to size 6) of those numbers can be discarded and the remaining numbers must necessarily still be equally distributed. Q.E.D.

In terms of code the mod function is unnecessary (and may be costly) - just generate three random bits by the coin tossing, join them together either by shifting or maybe there is a less costly way (depends on the processor architecture). This will give a binary representation of an integer in the range 0 to 7. We only want 1 to 6 so start over when the result is 0 or 7.

As Pan pointed out there is a slight chance that the function may never terminate, but in the real world that possibility is vanishingly small.

If the coin toss is a costly process for the processor (for example if it stops and waits for a human to toss a real coin and type in the result) then you can use the two 'throw away' values to generate the first bit for the next attempted dice roll - if you're throwing away a 0 then the first bit would be 0 and if you throw away a 7 the first bit would be a 1. Generally the coin toss function would be not very costly in terms of computer time, so there would be no need to employ this technique.
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Old 10-02-2015, 07:20 PM
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Old 10-02-2015, 07:56 PM
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Default Re: Calling all Math Geeks

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Old 10-02-2015, 08:34 PM
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Default Re: Calling all Math Geeks

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:fflove: :squeezle: I love you guys.
∀m(m∈:ff:, m∈TMath Geeks): mES~m
:fixed:
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Old 10-02-2015, 10:01 PM
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Default Re: Calling all Math Geeks

We like to leave errors in for students to spot.
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