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[WalrusMcFishSr]

The Axiom of Choice is MURDER!!!

[SwitchedFromStarcraft]

1 - 1 + 1 - 1 + ...

The algebraists and geometers sometimes get up to something a bit like this.

Is the Eilenberg swindle what made Heisenberg uncertain?

[Kirian]

I get the feeling that those uncomfortable with the axiom of choice are really just uncomfortable with infinities. It sometimes goes against "intuition" but really, what is intuition? It's mostly learned. What's intuitive to us is very different than what was intuitive to a mathematician 500 years ago and is different than what will be intuitive for a mathematician 500 years from now.

Also, to really be uncomfortable with the axiom of choice, don't you have to be equally uncomfortable with all of the equivalent statements? Are you really also so uncomfortable with assumption that the Cartesian product of any family of nonempty sets is nonempty? 

Any chance of a layman's explanation of the Axiom of Choice?  I choose to ask this...

[Witherweaver]

There are a lot of formulations, and a lot of things to which it is equivalent, but I think of it as:

If you have any collection of nonempty sets, you can form a new set by taking one element from each set in the collection.  It seems kind of straightforward, but it's a bit confusing because "you can form a new set" makes us think "Okay I'll take one from this one, then one from this other one, then one from this other other one, then...".  Certainly okay if the collection is finite, but confusing when the collection becomes infinite, especially uncountably infinite. 

The wikipedia page gives some layman explanations, I think.

[Polk5440]

If you have any collection of nonempty sets, you can form a new set by taking one element from each set in the collection.

Yep.

[qmech]

Any chance of a layman's explanation of the Axiom of Choice?  I choose to ask this...

You have a drawer full of pairs of socks.  Can you choose one sock from each pair?  Yes: just take the pairs out one by one and choose a sock arbitrarily. 

You have an infinite cupboard full of pairs of shoes.  Can you choose one shoe from each pair?  Yes: just choose the left shoe of each pair.

You have an infinite drawer full of pairs of socks.  Can you choose one sock from each pair?  Yes: just take the pairs out one by one and—

Wait—what does it mean to take the infinitely many pairs of socks out "one by one"?    In what order?  Might there even be too many socks to put them in any order?

The axiom of choice says that you don't have to worry about the details: you have infinitely many choices to make, and you can just make them.  There are lots of statements that are equivalent to the axiom of choice, though not always obviously.  One says that every set can be put in a very nice order, one where at every stage (even if you've just taken infinitely many steps) there's guaranteed to be a well-defined "next" element in the order.  (If you're wondering how this could possibly fail, imagine asking for the next integer after minus infinity, or the next real number after zero.)

People sometimes get hung up on consequences of the axiom of choice like the Banach–Tarski "paradox", which says that you can break two spheres into 5 "pieces" and reassemble them to get two spheres.  But usually the problem is that you were thinking about something in the wrong way.

For example, the naive objection to Banach–Tarski is that you seem to be getting "more stuff" out of nowhere.  But it turns out that Banach–Tarski is really just a cheat.  You start by doing something essentially equivalent to saying that an infinite binary tree is basically two copies of itself glued together, then somehow embed this picture into 3D space.  But the way the embedding is done means that all of the "pieces" are such a horribly mixed up collection of dust that we can't even sensibly talk about their size, so it's not clear that you have any "stuff" at all, never mind extra stuff.

[DStu]

I get the feeling that those uncomfortable with the axiom of choice are really just uncomfortable with infinities. It sometimes goes against "intuition" but really, what is intuition? It's mostly learned. What's intuitive to us is very different than what was intuitive to a mathematician 500 years ago and is different than what will be intuitive for a mathematician 500 years from now.

Also, to really be uncomfortable with the axiom of choice, don't you have to be equally uncomfortable with all of the equivalent statements? Are you really also so uncomfortable with assumption that the Cartesian product of any family of nonempty sets is nonempty?

yepp, also came to  my mind.

For me it's best illustrated (or acts as an reminder for me) at the non-lebessque-measureable sets.  Is it really necessary that each operation we are allowed to do on sets also has to preserve the well-definiedness of something like a volume?

:e ninjad of Banac-Tarski.  Anyway, I think the paradox with the finite deaths for infinite hats is quite similar to Banach-Tarski. In the end, our intuition says that should be impossible, because the probability of guessing incorrect only finite times should be 0.  But what you do is operating with non-measurable sets, so you can't apply probability theorey any more.  And no matter how bad the human intutions on probabilities are, they get worse if the whole theory is not applicable.

[scott_pilgrim]

I checked the logic puzzles thread first this time, and I don't think this one has been done yet (though there was something similar).  Two warm-up questions first.

1. You have three coins, which all look identical.  One is a counterfeit, which weighs more than the others (but you don't know by how much); the other two weigh the same.  You have a balance, so you can put any number of coins on each side and see which group weighs more (or if they weigh the same).  How many weighings do you need to do to find the counterfeit?

2. Same as #1, except now you have nine coins with one counterfeit.

3. Now you have nine coins with one counterfeit, and three balances, one of which is demon-possessed (you don't know which one is demon-possessed).  The demon-possessed balance will always give you a result to try to trick you, so it may give you the correct result, or an incorrect result.  How many weighings are needed in the worst case scenario to find the counterfeit?

[Titandrake]

I checked the logic puzzles thread first this time, and I don't think this one has been done yet (though there was something similar).  Two warm-up questions first.
3. Now you have nine coins with one counterfeit, and three balances, one of which is demon-possessed (you don't know which one is demon-possessed).  The demon-possessed balance will always give you a result to try to trick you, so it may give you the correct result, or an incorrect result.  How many weighings are needed in the worst case scenario to find the counterfeit?

What counts as a weighing in this puzzle? Do you only get to use one balance per weighing?

[Jack Rudd]

Question 1: one weighing. Weigh two coins against each other. If they weigh different amounts, the heavier one is the counterfeit. If they weigh the same amount, the third coin is the counterfeit.

Question 2: two weighings. Put three coins on each side of the balance. If they weigh different amounts, the counterfeit is on the heavier side. If they weigh the same amount, the counterfeit is not on the balance. You now have a set of three coins containing a counterfeit, and the problem has been reduced to the previously solved one.

[pacovf]

3. Now you have nine coins with one counterfeit, and three balances, one of which is demon-possessed (you don't know which one is demon-possessed).  The demon-possessed balance will always give you a result to try to trick you, so it may give you the correct result, or an incorrect result.  How many weighings are needed in the worst case scenario to find the counterfeit?

Is it one plus the number of letters of the word you expected to find in this spoiler?

[Ozle]

Question 1: one weighing. Weigh two coins against each other. If they weigh different amounts, the heavier one is the counterfeit. If they weigh the same amount, the third coin is the counterfeit.

Question 2: two weighings. Put three coins on each side of the balance. If they weigh different amounts, the counterfeit is on the heavier side. If they weigh the same amount, the counterfeit is not on the balance. You now have a set of three coins containing a counterfeit, and the problem has been reduced to the previously solved one.

Am I allowed to chew them?

[scott_pilgrim]

3. Now you have nine coins with one counterfeit, and three balances, one of which is demon-possessed (you don't know which one is demon-possessed).  The demon-possessed balance will always give you a result to try to trick you, so it may give you the correct result, or an incorrect result.  How many weighings are needed in the worst case scenario to find the counterfeit?

Is it one plus the number of letters of the word you expected to find in this spoiler?

No, but it would have been if Ozle had posted that.

I checked the logic puzzles thread first this time, and I don't think this one has been done yet (though there was something similar).  Two warm-up questions first.
3. Now you have nine coins with one counterfeit, and three balances, one of which is demon-possessed (you don't know which one is demon-possessed).  The demon-possessed balance will always give you a result to try to trick you, so it may give you the correct result, or an incorrect result.  How many weighings are needed in the worst case scenario to find the counterfeit?

What counts as a weighing in this puzzle? Do you only get to use one balance per weighing?

Yes, one weighing means putting some number of coins on each side of one balance and then checking the result.

[pacovf]

Is it one plus the number of letters of the word you expected to find in this spoiler?

No, but it would have been if Ozle had posted that.

Huh? So is it or is it not five? What did you expect to find in the spoiler?

[scott_pilgrim]

Is it one plus the number of letters of the word you expected to find in this spoiler?

No, but it would have been if Ozle had posted that.

Huh? So is it or is it not five? What did you expect to find in the spoiler?

No, it is five.  I just didn't think you usually made moat jokes.  Well really, I just didn't think about it until I actually read the spoiler.

[pacovf]

I'm full of surprises!

But to be fair, had you expected the real answer, you would still have come up with the same thing. You see, I was planning for every eventuality.

[heron]

Alright, here's an interesting problem:

What is the sum from n = 1 to infinity of s(n)/((n)(n+1))?
Note that s(n) is the sum of the digits of n.

[ConMan]

Alright, here's an interesting problem:

What is the sum from n = 1 to infinity of s(n)/((n)(n+1))?
Note that s(n) is the sum of the digits of n.

That is quite interesting. Things involving digit sums tend to be a bit weird, because they're dependent on the base being used. That said, there are a few things that are quick to note:

In general, s(n) would be of roughly O(log(n)), since an m-digit number will have a digit sum less than or equal to 9m. Hence, the sum is of terms of order log(n)/n^2, which sits somewhere between 1/n and 1/n^2, so at a guess it will converge but very, very slowly.

Also, since as I already said, the digit sum depends on the base, so presumably this sum will also be some function of b (where for decimal, b = 10).

In looking for some trick to help work out the value of the sum, I stumbled across the answer (but not the proof because it seems to be in a paywalled journal that I don't have access to). I was basically right, and that's pretty cool, and now I want to see the proof - I can kind of see what the outline of the proof would be for binary, but the general proof looks like it will be a bit fiddly.

[Polk5440]

Alright, here's an interesting problem:

What is the sum from n = 1 to infinity of s(n)/((n)(n+1))?
Note that s(n) is the sum of the digits of n.

Nice one. Don't reveal the answer yet. Still thinking......

[sudgy]

So, my brother and I are trying to figure out a question for a practice exam, and are having a hard time with it.  The question is, "Use the method of cylindrical shells to find the volume of the solid of revolution obtained by revolving the region between the graph of the function y = x^-1 and the x-axis over the interval [1, 2]."

The problem is that this isn't starting at 0 so you can't simply use the value of the function at a point as the height of the cylinder, and it's also hard because it's rotating around the x-axis, not the y-axis.  We know how to do it through other methods, but the question specifically asks for cylindrical shells.  Any help?

[scott_pilgrim]

Assuming you mean as x (rather than y) goes from 1 to 2, then yes, shells is a really awkward method to use for that problem.  The height will be a piecewise function in terms of y (and it needs to be in terms of y, since it's shells and it's going around the x-axis), so you'll need to do two separate integrals.

Going from y=0 to y=1/2, the height will just be a constant 1 (2 minus 1).  Then going from y=1/2 to y=1, the height will be 1/y (the x-value) minus 1, because the total distance from the graph to the line x=1 is the same as the distance from the graph to the y-axis (which is the x-value, 1/y), minus the distance from x=1 to the y-axis (which is just 1).  The radius for both parts will just be y.

Therefore, you want to do (2pi)*[(integral from 0 to 0.5 of y)+(integral from 0.5 to 1 of y*((1/y)-1))].

[Witherweaver]

The best thing to do to answer all these problems is to:

1) Draw the shape
2) Draw a representative cross-section
3) Label the cross-section in terms of the relevant geometry to find its volumes. 

For "shells" about the y-axis, the "width" of a shell is like a "dx", the radius is along the x-axis, and the height is along the y-axis.  For "shells" about the x-axis, it's flipped (width dy, radius y, height x). 

For discs/washers/whatever you call them, the height/width is like a dx or dy (dx about x-axis and dy about y-axis).  The radius is along the x-direction if you rotate about the y-axis and the y-direction if you rotate about the x-axis. 

4) Write out the volume in terms of these variables.  This is for a small cross section, so it's like a "dV".
5) Add up all the "dV"s over the relevant range.  This become an integral in the limit as everything is small.
6) Perform the integration.

I would recommend never ever solving one of these problems without drawing the picture, including the "dV" cross section.  In my opinion, that's the essence of the problem.  Your Calc book should have a lot of such pictures. 

[heron]

So, should I give my solution to my problem then? It's been awhile.

[Polk5440]

So, should I give my solution to my problem then? It's been awhile.

I got stuck, so, yeah, I would appreciate it.

[sudgy]

Assuming you mean as x (rather than y) goes from 1 to 2, then yes, shells is a really awkward method to use for that problem.  The height will be a piecewise function in terms of y (and it needs to be in terms of y, since it's shells and it's going around the x-axis), so you'll need to do two separate integrals.

Going from y=0 to y=1/2, the height will just be a constant 1 (2 minus 1).  Then going from y=1/2 to y=1, the height will be 1/y (the x-value) minus 1, because the total distance from the graph to the line x=1 is the same as the distance from the graph to the y-axis (which is the x-value, 1/y), minus the distance from x=1 to the y-axis (which is just 1).  The radius for both parts will just be y.

Therefore, you want to do (2pi)*[(integral from 0 to 0.5 of y)+(integral from 0.5 to 1 of y*((1/y)-1))].

We pretty much got this, but were stupid and thought a few things were wrong but somehow got the right answer (also, doing it piecewise just seems weird).

Also, Witherweaver, we drew pictures like crazy trying to figure things out.