bonanova

You're right of course. My statement should have read [still falling short of a proof] Each individual rational number corresponds to an uncountable infinity of real numbers. A countable infinity of countable infinities is still countable. My intent was to show how much more dense the reals are than the rationals. The reals do not have a vanishing granularity. Your presentation of Cantor's argument provides the proof.

Decimal representation is one way to see that the reals are more dense than the rationals. First, note that finite decimals [1/2 = 0.5] are a special case of infinite repeating decimals: 0.5000... Then the decimal real/rational distinction is just this: Real numbers are represented by infinite decimals. Rational numbers are represented by infinite decimals that eventually repeat. 2 = 2.000... [repeating: rational] 1/2 = 0.5000... [repeating: rational] 1/3 = 0.333... [repeating: rational] pi = 3.1415926536... [not repeating: irrational] e = 2.71828459045... [not repeating: irrational] Why does this difference make the reals [rationals + irrationals] more dense than the rationals? For every rational number there is a value of N such that the decimal digits after the Nth place, since they repeat, are determined. For each N, there is a "next" rational number. There are holes between them; they are discrete. Now since N is arbitrarily large, the "holes" are arbitrarily small. So between any two rational numbers there are an infinity of other rational numbers. Still, the rationals are countable but the reals are not. Why not? Consider the rationals: Specify a number N. There are a countable number of rational numbers with the first N decimal digits freely specified, but the remaining infinite digits fixed [since they repeat.] Precisely, there are 10N such rationals. As N increases without bound, the number of rationals also increases without bound. But the rationals remain countable as this happens. At any granularity level [value of N] the rationals are countable, and they remain countable, even as this granularity becomes arbitrarily fine [N becomes arbitrarily large]. Now consider the reals: Again specify a number N, and consider the 10N values represented by a string of N decimal digits. Recall that for each of these strings there is a unique associated rational number. In contrast, since for reals there is no constraint on the remaining digits, associated with a string of N decimal digits there are an infinite number of reals. This shows that you can't even begin to count [map a single counting number to] the reals. Each individual rational number corresponds to an infinity of real numbers. The rationals are infinitely dense, but the reals in some sense are twice-infinitely dense. In any finite interval [a, b] there are an infinite number of rationals, but there are an infinite number of infinite numbers of reals. To express this difference, infinities are sorted into cardinality classes, the first two of which are Aleph0 [the countable cardinality of the integers and the rationals] and Aleph1 [the uncountable cardinality of the reals]. Aleph is the first letter of the Hebrew alphabet.

Well I think you have stated the idea already, but let me put it into these words. There are just two cases. 1, 2, 3, 4, ... ends at some number N, beyond which one cannot count. There is no such number N. To show that 1, 2, 3, 4, ... does not reach to infinity, one must show that N exists.

Well that was fun. By the way, if the original words maintain their association with their sorted versions, what might this worm-infected dictionary be useful for?

Yes. If p were 4736 then q = 7643, r = 3467 and s = 4176.

I thought of a four-digit number, p = abcd; where each of a, b, c, d are in the set [0, 9]. Boy, was that fun. But how interesting can a four-digit number be? Soon I tired of that number; I wanted a new one. So I made q = klmn, where k, l, m, n are the down-ordering of a, b, c, d. But that looked too familiar, having the same digits and all, and so did r = nmlk. So I subtracted the two: the smaller from the larger, and ended up with s = |q-r| = uvwx. Great, I thought, and I started thinking about s. But s looked even more familiar than q and r did. No wonder: s turned out to be exactly equal to my first number, p. If I give you the hint - if you think you need one - that b < c, can you tell me my original number p?

A vicious worm invaded the [fictitious] online dictionary Sordid Words last night. The effect was all the word entries were scrambled so that the letters in each word were alphabetized. brainden became abdeinnr, puzzle was changed into elpuzz, and so on. And then the scrambled words were listed in their new alphabetical order. Interesting, you thought. Me too. Without thinking about it too hard, what would be the first few and the last few entries in the newly alphabetized word list? And, what's likely the first new word starting with B? Have fun.

When an old riddle [hats on death row] is "solved" again, it gets bumped to the top. In that case, would it be appropriate for the owner or a Mod to convert it to a poll? Agree not to poll every solved puzzle; but if it gets a "new"solution, that bumps it. Converting those solved puzzles to polls might be good idea. - bn

Correct. No one said that. The solution accounts for any distribution.

OK all you programmers, do the exhaustive search now.

This is my first riddle.

Post images: Open a new Reply and type in any text you want in the text entry box. Look below the text box for the Browse button and click it. Locate the image you want to upload and select it. Click Upload. the image file name should appear in the Manage Current Attachments window. Place your cursor in the text entry box where you want the image to appear. Click on the image file name in the Manage Current Attachments window. The code to display the image should appear in the text box. Scroll down and click Preview Post. Edit the text entry box content and repeat Preview until it's what you want. Scroll down and click Add Reply. Hope this helps. - bn

How about the term after that?

Which prisoner will say Red?

As the President looked over the evening mail he was heard to exclaim: Eighteen? Are you kidding me? Dick Cheney gave me 18! The best he got was 1 2 3 4 5 6 7 8 9 10 20 30 40 50 60 70 80 90. No one can do better?

Recasting an old MG favorite: In the U.S. if you wanted to produce $.99, you'd need at least 8 coins; $.49 would require at least 7 coins, and so on. Yesterday our President declared that the current coin denominations of 50, 25, 10, 5 and 1 do not constitute Change you can believe in, and ordered a new set of coins to be minted. I believe that any citizen of this country should be able to produce any amount of money less than one dollar using at most two coins, he said. And he challenged his greatest group of thinkers, the Brain Denizens, to establish a set of coins that accomplishs this feat. How many coin denominations will required?

Spoiler for Suppose the edge of the solid is a centimeters.: The Volume is some constant [say k3] times a3 cm3 and The Area is some other constant [say k2] times a2 cm2. The constants are the same regardless of the units. We could have used meters, inches or miles. When V and A have the same value, then k3 a3 = k2 a2 Dividing both sides by k3 a2 gives the desired value of edge length: a = (k2 / k3) Checking: V = k3 x (k2 / k3)3 = k23 / k32 A = k2 x (k2 / k3)2 = k23 / k32 So the puzzle basically asks to find k2 and k3. The easiest solution is to Google it - they are complicated to calculate. Like if you have no life, it would be a neat way to spend an hour.

Nice result. There is a known proof for 2+2n that would verify your first answer. I don't know the proof for 1+2n. I imagine working out the answer for say n=5 would suggest the proof, and the odd answer would follow. Just a hunch.

Close - retired research engineer and editor.

Now that Bushindo has introduced us to the mysteries of the dodecahedron's dihedral angle, can we determine the length of the dodecahedron edge for which the volume and surface area have the same value?

Stunned, if not a bit chagrinned at bit at the rapidity of the reply, Alex nevertheless tips is cap to Glycereine and mutters, alright then; I suppose a proof is a little much to ask for, so let's move on to the next issue: who are the winning players?

I thought my last puzzle was more challenging than that, said Alex after his first swallow of ale at Morty's last night. You lads have been boning up, it seems. But that's alright. This one will take a bit longer, I expect, although it's a child's game. Here; let me show you. At this, he was joined by Jamie, Ian and Davie at his table. Look. Each of us starts with a marble. The person who starts passes a marble to his left. That person then passes two marbles to his left. That person then passes one marble to his left, and so on. The number of marbles passed alternates between one and two. At any point, if a player has no marbles in front of him he drops out of the game. And depending on the number of starting players, one person may end up with all the marbles to himself. To demonstrate, Alex passed his marble to Jamie, who passed two marbles to Ian, who passed one marble to Davey, who passed two marbles to Ian. [Alex and Jamie had been eliminated after their first pass.] At that point Ian had all four of the marbles. Now it might be interesting to figure out which player it will be that gets all the marbles when it happens, Alex mused, but we'll leave that thought for another night. Tonight's question is: for an n-player game, what are the values of n that will result in one player finally getting all the marbles? If you understand the problem, you'll find that n=2, 3, 4, 5 and 6 all terminate with a winner, but the next terminating number is n=9. What are the other values of n?