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3dknight
2007-Sep-17, 02:33 AM
The game is a math game, we start out with a simple math question then whoever answers it has to come up with a slightly harder question. Ok what is 12x12?

bluebanana
2007-Sep-17, 04:06 AM
144

ok
(17x4)+37

sciencefair
2007-Sep-17, 05:48 PM
Okay, 105.

Now, what is the square root of 0.1?

DanishDynamite
2007-Sep-17, 09:01 PM
0.316....

What is the last digit of the number 2007^2007?

JMV
2007-Sep-17, 09:55 PM
3?

DanishDynamite
2007-Sep-17, 10:00 PM
3?
Correctomunto!

You now have the honor of asking the next question, my friend. :)

JMV
2007-Sep-18, 12:20 AM
Thank you.

What's √123456789ABCDEFEDCBA987654321 in base-16 (hexadecimal)?

pghnative
2007-Sep-18, 03:08 AM
111,111,111,111,111

presuming I'm correct, what is the first set of "10s" (e.g. 20 through 29, or 140 through 149) for which there are no prime numbers.

The_Radiation_Specialist
2007-Sep-18, 03:37 AM
200-209

what is the smallest number which can be written as the sum of two (positive) cubes in two different ways?
except 2

The_Radiation_Specialist
2007-Sep-18, 01:46 PM
I'll post the answer in 12 hours, for now I'll need to ask for the number to get a taxicab.

The_Radiation_Specialist
2007-Sep-19, 02:47 AM
http://en.wikipedia.org/wiki/1729_%28number%29
1729 = 1^3 + 12^3 = 9^3 + 10^3

Post a question anyone...

DanishDynamite
2007-Sep-19, 07:44 PM
A rectangular floor is covered by square tiles. The tiles along the edge of the floor are red. The rest are white. There is an equal number of red and white tiles.

How many tiles are there?

The_Radiation_Specialist
2007-Sep-20, 07:31 AM
8x6

Mellow
2007-Sep-20, 08:40 AM
Nice, you need to post a slightly harder question now, TRS.

The_Radiation_Specialist
2007-Sep-20, 09:00 AM
Using only three "9's" and any mathematical sign (+ - X / sqrt ^ ! etc) make it so that it can equal digits 1 to 10.

eg. for 1 you could write (9/9)^9

note there must be three "9's"

:)

pghnative
2007-Sep-20, 08:59 PM
1: (9/9)^9
2: (9+9)/9
3: sqrt (9*9/9)
4: sqrt(9) + 9/9
5: sqrt(9)!-9/9
6: (sqrt (9*9/9))!
7: sqrt(9) + 9/9
8: 9 - (9/9)
9: 9*9/9

edit: Thanks to Eroica for pointing out the missing factorial:
7: sqrt(9)! + 9/9

Coincidentally (re TRS's comment), I gave some thought to being a smart aleck and using 0.999... as the answer to "1", but since it used a "0", it technically would have been wrong.

Mister Earl
2007-Sep-20, 09:13 PM
*Reads the thread, then slowly backs away*

Aha! My math-fu is too weak...

The_Radiation_Specialist
2007-Sep-21, 03:01 AM
7: sqrt(9) + 9/9

?

more like 9-sqrt(9) + 0.9(recurring) :D

No Discussions about the validity of 0.999... = 1 ! I'm serious. :D

Eroica
2007-Sep-21, 07:12 AM
7: sqrt(9) + 9/9
Just a typo, I think. He forgot the factorial sign:

(sqrt(9))! + 9/9 = 7

pghnative
2007-Sep-21, 01:30 PM
I'll post the answer in 12 hours, for now I'll need to ask for the number to get a taxicab.I've been meaning to ask: how was that a hint for 1729?

Anyway: What's the smallest number that, when squared, produces a four digit number comprised of two 2-digit primes.

e.g. 37^2 = 1369 -- thirteen is prime, but 69 isn't, so 37 isn't the answer.

tdvance
2007-Sep-21, 02:33 PM
"69", dude!

Find the largest 51-digit number divisible by 11. (probably takes fewer keystrokes to describe the number than to type it out!)

Todd

The_Radiation_Specialist
2007-Sep-21, 04:28 PM
I've been meaning to ask: how was that a hint for 1729?

Taxicab numbers (http://en.wikipedia.org/wiki/Taxicab_numbers) ;)

Find the largest 51-digit number divisible by 11.

999,999,999,999,999,999,999,999,999,999,999,999,99 9,999,999,999,990

tdvance
2007-Sep-21, 05:04 PM
I didn't count the 9s, but unless you miscounted, it's right.

Todd

The_Radiation_Specialist
2007-Sep-21, 05:33 PM
How much dirt is there in a pentagonal prism-shaped hole with each side of the pentagon 1 meter and the depth 4 meters? Approximate to the nearest meter cube.

pghnative
2007-Sep-21, 05:34 PM
none

The_Radiation_Specialist
2007-Sep-21, 05:38 PM
:)

pghnative
2007-Sep-21, 05:59 PM
121 is an example of a palindromic perfect square. Of course, it's square root (11) is also a palindrome.

What is the smallest palindromic perfect square whose square root is not a palindrome.

DanishDynamite
2007-Sep-22, 12:11 AM
8x6Very good. But is this the only possible answer? (Hint: No)

mr obvious
2007-Sep-22, 12:21 AM
121 is an example of a palindromic perfect square. Of course, it's square root (11) is also a palindrome.

What is the smallest palindromic perfect square whose square root is not a palindrome.

26^2 =676, assuming standard conditions (base 10 and all that).

DanishDynamite
2007-Sep-22, 01:56 AM
A fisherman catches a number of fish. The 3 heaviest together make up 35% of the total weight of the catch. He immediatelly sells these fish. At this point the 3 lightest fish make up 5/13 of the weight of the remaining fish.

How many fish did the fisherman catch?

Sarawak
2007-Sep-22, 02:11 AM
Very good. But is this the only possible answer? (Hint: No)

8x6
12x5

l assume you do not consider 6x8 and 5x12 to be distinct from the above.

Do you also allow 0x0?

mr obvious
2007-Sep-22, 02:14 AM
A fisherman catches a number of fish. The 3 heaviest together make up 35% of the total weight of the catch. He immediatelly sells these fish. At this point the 3 lightest fish make up 5/13 of the weight of the remaining fish.

How many fish did the fisherman catch?

Ten, not counting those he threw back.

mr obvious
2007-Sep-22, 02:20 AM
8x6
12x5

l assume you do not consider 6x8 and 5x12 to be distinct from the above.

Do you also allow 0x0?

In general, where a=4(b-2)/(b-4) is a natural number for natural number b, the solution axb (or bxa) works.

Sarawak
2007-Sep-22, 02:27 AM
In general, where a=4(b-2)/(b-4) is a natural number for natural number b, the solution axb (or bxa) works.

I got them all, though, right?

a=4(b-2)/(b-4)

so

a=4+8/(b-4)

So 8 has to be divisible by (b-4). So that means b=0 or b=2 (which are "paired" solutions, and leave a floor with an area of zero), or b=5, b=6, b=8, or b=12.

DanishDynamite
2007-Sep-22, 02:58 AM
8x6
12x5

l assume you do not consider 6x8 and 5x12 to be distinct from the above.

Do you also allow 0x0?
8X6 and 12X5 does it for me. Well done!

DanishDynamite
2007-Sep-22, 03:06 AM
Ten, not counting those he threw back.
Congratulations. :)

Ten it is.

Next question:

Given that n is a whole number and that n^2 has 7 as its last but one digit, what is the last digit in n^2?

Sarawak
2007-Sep-22, 03:20 AM
6, and there are four different combinations of the last two digits in n that get it.

DanishDynamite
2007-Sep-22, 03:43 AM
6, and there are four different combinations of the last two digits in n that get it.
Correct!

You have the honor of asking the next question.

Sarawak
2007-Sep-22, 03:54 AM
Correct!

You have the honor of asking the next question.

Uh oh, now I must think of a clever question. I displayed remarkable lack of foresight in choosing to answer the last question.

Is it dishonorable or insulting in BAUT culture to give away my turn to anyone who likes to take it? I am still working on the airplane circumnavigation problem.

Eroica
2007-Sep-22, 10:11 AM
Is it dishonorable or insulting in BAUT culture to give away my turn to anyone who likes to take it?Not at all. I'll go next.

My attempt, I remember, to remember ? finished in disaster, mate!

What does ? represent?

3dknight
2007-Sep-22, 11:01 AM
I don't understand the question?

The_Radiation_Specialist
2007-Sep-22, 11:03 AM
I understand the question!

no wait...I don't.

tdvance
2007-Sep-22, 06:09 PM
"I" is the only English word that could possibly fit ("a" doesn't quite work, neither does the contraction "o" ).

Todd

mr obvious
2007-Sep-22, 10:39 PM
Does the ? need to be only a single letter?

Eroica
2007-Sep-23, 08:43 AM
Does the ? need to be only a single letter?That's for me to know and for you to find out! It's a puzzle - but a mathematical one.

mr obvious
2007-Sep-23, 06:33 PM
Not sure what you mean. The number of letters in each word correspond to the digits of e, insofar as your sentence is a finite representation of the first few digits of e.

So, the ? should be a single letter long. So, I guess tdvance was correct as far as the number of letters is concerned. But ?=e is probably better, except that that leads to a paradox since you clearly remember the mnemonic for e, you can determine quite a few digits and that would hardly constitute a disaster. So what's the rub?

Eroica
2007-Sep-24, 06:39 AM
Not sure what you mean. The number of letters in each word correspond to the digits of e, insofar as your sentence is a finite representation of the first few digits of e.
:clap: e is the correct answer.

... But ?=e is probably better, except that that leads to a paradox since you clearly remember the mnemonic for e, you can determine quite a few digits and that would hardly constitute a disaster. So what's the rub?
Unfortunately success has only 7 letters! I was only a kid when I devised this mnemonic, so the illogic escaped me at the time. But it has served me well over the years.

Over to you, mrobvious.

mr obvious
2007-Sep-24, 11:04 AM
Take a chessboard (8 by 8 grid) and remove one set of opposite corners (A1 and H8, for example, for those that know chess notation). There are 62 squares left. Take a single knight and place it anywhere on the board.
(For those that don't know how knights move, see Diagram 2 here:
http://en.wikipedia.org/wiki/Knight_(chess) )

Your job is to describe a sequence of moves such that the knight visits every square exactly once (the square you start on counts as visited immediately), or explain why this is not possible.

Eroica
2007-Sep-24, 03:07 PM
It should be impossible, since you removed two black squares, leaving 30 black and 32 white squares. When a knight moves, the squares it lands on are alternately black and white. So after visiting 30 black squares and 31 white squares, the knight will run out of black squares.

The_Radiation_Specialist
2007-Sep-24, 05:52 PM
I assume that is the correct answer.

Next?

DanishDynamite
2007-Sep-24, 09:44 PM
There are exactly 2 proofs regarding number theory that I remember from my high school days. I hesitate to ask for either one as I'm sure they will be easily available using Google. Still, they are the only mathematical questions I can think of at this moment, so here's one of them:

Prove that sqrt(2) is an irrational number.

mr obvious
2007-Sep-24, 11:22 PM
Let a/b = the rational expression of sqrt(2). Further, assume that a and b have gcd=1, or else reduce them so that their gcd =1.

a=sqrt(2) * b
a*a = 2*b*b (1)
Thus, a must be even (if a is odd, then the '2' on the right can't divide a).
let a = 2*n
Substituting in equation 1,
(2n)*(2n)=2*b*b
4*n*n=2*b*b
Dividing both sides by 2,
2*n*n=b*b
By the same reasoning as previously applied, b must be even. However, now both a and b are even, contradicting the initial condition that a and b must have gcd=1. Therefore, such a and b do not exist, therefore by definition sqrt(2) must be irrational.

DanishDynamite
2007-Sep-24, 11:43 PM
Let a/b = the rational expression of sqrt(2). Further, assume that a and b have gcd=1, or else reduce them so that their gcd =1.

a=sqrt(2) * b
a*a = 2*b*b (1)
Thus, a must be even (if a is odd, then the '2' on the right can't divide a).
let a = 2*n
Substituting in equation 1,
(2n)*(2n)=2*b*b
4*n*n=2*b*b
Dividing both sides by 2,
2*n*n=b*b
By the same reasoning as previously applied, b must be even. However, now both a and b are even, contradicting the initial condition that a and b must have gcd=1. Therefore, such a and b do not exist, therefore by definition sqrt(2) must be irrational.
Congrats!

Next question, please?

mr obvious
2007-Sep-25, 12:43 AM
For whatever reason, a king has decreed that he wishes to increase the female population in his kingdom. Without advanced medicine, he orders all families to stop having children once they have a son. His reasoning is that even though the chances of male vs. female in a given birth is 50-50, after a while, there will be families with one girl, two girls, three girls, etc. whereas there will be no families with more than one boy (and of course, there may be families with no boys - the people are not forced to have children until they have a boy; they simply can't have more once they do have a boy).

Assuming this rule is extended to infinite time, that families have many children (e.g., no self-imposed abstinence), that natural resources are not a concern (this is a math problem, after all), what is the asymptotic ratio of females to males?

DanishDynamite
2007-Sep-25, 12:53 AM
For whatever reason, a king has decreed that he wishes to increase the female population in his kingdom. Without advanced medicine, he orders all families to stop having children once they have a son. His reasoning is that even though the chances of male vs. female in a given birth is 50-50, after a while, there will be families with one girl, two girls, three girls, etc. whereas there will be no families with more than one boy (and of course, there may be families with no boys - the people are not forced to have children until they have a boy; they simply can't have more once they do have a boy).

Assuming this rule is extended to infinite time, that families have many children (e.g., no self-imposed abstinence), that natural resources are not a concern (this is a math problem, after all), what is the asymptotic ratio of females to males?
Can a man have more than 1 wife (i.e. do surplus girls simply die as old spinsters or are they also impregnated?)

DanishDynamite
2007-Sep-25, 01:06 AM
Nevermind. It seems to me that the asymptotic ratio would still be 1.

mr obvious
2007-Sep-25, 01:37 AM
Correct, but an explanation would be nice :)

Sarawak
2007-Sep-25, 02:35 AM
If the probability of a boy is 1/2, then 1/2 of all families will have no girls, 1/4 of all families will have one girl, 1/8 of all families will have two girls, and so on. The expected number of girls is:

0/(2^1)+1/(2^2)+2/(2^3)+3/(2^4)+...

This series converges to one, so each family has children until it has one boy, but it also has one girl on average.

It is simpler than that, though, if the probability of a boy is 1/2, then on average 1/2 of all children are boys, regardless of what rule families apply in when to stop.

mr obvious
2007-Sep-25, 02:51 AM
Fair enough. One of you two gets to ask a question. Please go ahead.

Sarawak
2007-Sep-25, 03:06 AM
DanishDynamite gave a correct answer, so probably he did everything correctly even though he did not explain. So I think it is his turn :)

victor003
2007-Sep-25, 06:19 AM
Cool. Really interesting game

DanishDynamite
2007-Sep-25, 08:51 PM
DanishDynamite gave a correct answer, so probably he did everything correctly even though he did not explain. So I think it is his turn :)Such trust! :)

All I can think of at the moment is the other proof from high school number theory that I remember. It is almost too banal to post, but I suppose it might be new for one or two readers, so here goes:

Prove that there are an infinite number of primes.

The_Radiation_Specialist
2007-Sep-26, 07:36 AM
My unprofessional proof.
Suppose there is a number P(q) which is the biggest prime.
Then suppose M = P1 * P2 * .... P(q) Where P(n) is the nth prime.
M+1 would not factor any of the other primes because any division by a prime would result in remainder 1.
So M+1 would be the biggest prime.

There is a contradiction.

Sarawak
2007-Sep-26, 01:32 PM
My unprofessional proof.
Suppose there is a number P(q) which is the biggest prime.
Then suppose M = P1 * P2 * .... P(q) Where P(n) is the nth prime.
M+1 would not factor any of the other primes because any division by a prime would result in remainder 1.
So M+1 would be the biggest prime.

There is a contradiction.

It is a contradiction, but there are two possibilities. Either M+1 is prime, or M+1 can be divided by some prime larger than P(n). In either situation, there is a prime larger than P(n).

The_Radiation_Specialist
2007-Sep-26, 01:59 PM
Either M+1 is prime,

In which case it would be the largest prime and larger than p(q) we assumed in the beginning. So there is a contradiction to the biggest prime. There cannot be a biggest prime.

or M+1 can be divided by some prime larger than P(n).

The primorials (product of all primes smaller or equal to a number) also included the largest prime we assumed in the beginning. So, no, M+1 cannon be divided without remainder by any prime based on our assumption in the beginning.

In either situation, there is a prime larger than P(n).

:)

P(n) is not a prime by itself. It means simply the nth prime.
We assumed there p(q) is the biggest prime if primes were a finite number.

Sarawak
2007-Sep-26, 02:04 PM
The primorials (product of all primes smaller or equal to a number) also included the largest prime we assumed in the beginning. So, no, M+1 cannon be divided without remainder by any prime based on our assumption in the beginning.

You are right, I did not apply the assumption that P(n) is the largest prime at this point. It is a different proof than the proof I know, but it is valid :)

The_Radiation_Specialist
2007-Sep-26, 02:41 PM
What's the answer to life, the universe and everything?

DanishDynamite
2007-Sep-26, 05:02 PM
Sorry, TRS, but I agree with Homo Bibiens. The proof I was looking for, and the one you attempted, was Euclid's proof (given in Euclid's own words here (http://en.wikipedia.org/wiki/Prime_number)).

You cannot say "M+1 would be the biggest prime" because you don't know if it is a prime.

You can only say that either M+1 is a prime, because none of the primes in our list of primes can divide it, or there exists some other prime number not in our list which can divide it. But in either case, a prime number larger than p(q) exists, hence contradicting the premise.

DanishDynamite
2007-Sep-26, 05:05 PM
What's the answer to life, the universe and everything?
42. :)

The_Radiation_Specialist
2007-Sep-26, 05:19 PM
You cannot say "M+1 would be the biggest prime" because you don't know if it is a prime.

We assumed p(q) was to be the largest prime if there were finite primes. Then took M as the product of all primes from 2 to p(q). M+1 would not have any factors of all the primes we had in M. Therefore it would be a bigger prime.

So based on the assumption that there are finite primes (and there is a biggest prime) we arrive at a contradiction.

This is a bit different with Euclids proof as it assumes a biggest prime in the beginning.

DanishDynamite
2007-Sep-26, 05:32 PM
We assumed p(q) was to be the largest prime if there were finite primes.
Yes, as does Euclid.

Then took M as the product of all primes from 2 to p(q). M+1 would not have any factors of all the primes we had in M. Therefore it would be a bigger prime.
But you don't know that it is a prime. It could very well be the product of 2 numbers larger than p(q). If so, it would not be a prime number.

So based on the assumption that there are finite primes (and there is a biggest prime) we arrive at a contradiction.

This is a bit different with Euclids proof as it assumes a biggest prime in the beginning.
Sorry, I don't see it.

The_Radiation_Specialist
2007-Sep-26, 05:51 PM
But you don't know that it is a prime. It could very well be the product of 2 numbers larger than p(q). If so, it would not be a prime number.

Based on our assumption that p(q) is the biggest prime, M+1 must be a prime also.

I have to go to bed now. Will make more clarifications tomorrow if needed.

DanishDynamite
2007-Sep-26, 06:51 PM
Based on our assumption that p(q) is the biggest prime, M+1 must be a prime also.
Yes, but as you say, that is based on your assumption. The point of this exercise is to show your assumption is wrong. In order to show that your assumption is wrong, you must then show that M+1 is in fact not a prime.

DanishDynamite
2007-Sep-26, 07:01 PM
I've found another exercise (stolen from a math problem page I just found) which is the following:

Assume a cube. Assume that at one corner of the cube a fly sits, perfectly still. Assume that at the opposite corner of the cube (the corner furhest away from the fly corner) a blind spider is released. Assume that the spider can only move along the edges of the cube and does so randomly.

What is the expected number of edges that the spider will traverse before it hits upon the fly corner?

Sarawak
2007-Sep-26, 10:56 PM
I've found another exercise (stolen from a math problem page I just found) which is the following:

Assume a cube. Assume that at one corner of the cube a fly sits, perfectly still. Assume that at the opposite corner of the cube (the corner furhest away from the fly corner) a blind spider is released. Assume that the spider can only move along the edges of the cube and does so randomly.

What is the expected number of edges that the spider will traverse before it hits upon the fly corner?

Ten.

Every corner is 0, 1, 2, or 3 distance (measured by shortest path) from the fly corner. We can call x0 the expected number of edges to reach the fly corner from a distance of zero, x1 the expected number of edges to reach the fly corner from a distance of one, and so on.

x0=0 because the spider is already there.

From a distance of one, there is one path to the fly, and two paths to corners with a distance of two. So x1=1+(1/3)*x0+(2/3)*x2.

From a distance of two, there are two paths to corners with a distance of one, and one path to a corner with distance of three. So x2=1+(2/3)*x1+(1/3)*x3.

From a distance of three, all three paths lead to a distance of two, so x3=1+x2.

The equations are:

a) x0=0
b) x1=1+(1/3)*x0+(2/3)*x2
c) x2=1+(2/3)*x1+(1/3)*x3
d) x3=1+x2

Solution is:

x1=0
x1=7
x2=9
x3=10

Sarawak
2007-Sep-26, 10:58 PM
Yes, but as you say, that is based on your assumption. The point of this exercise is to show your assumption is wrong. In order to show that your assumption is wrong, you must then show that M+1 is in fact not a prime.

I think TRS has a correct proof by contradiction.

Assume p(n) is largest prime, and define M=p(1)*...*p(n).

Then M+1 is not divisible by p(1) through p(n) by construction. By assumption, there is no larger prime, so M+1 cannot be divisible by any number. But then, M+1 must be prime, so this is a contradiction. It is different from the proof you know (which is the same proof I know), but it is still valid.

DanishDynamite
2007-Sep-27, 12:57 AM
Ten.

Every corner is 0, 1, 2, or 3 distance (measured by shortest path) from the fly corner. We can call x0 the expected number of edges to reach the fly corner from a distance of zero, x1 the expected number of edges to reach the fly corner from a distance of one, and so on.

x0=0 because the spider is already there.

From a distance of one, there is one path to the fly, and two paths to corners with a distance of two. So x1=1+(1/3)*x0+(2/3)*x2.

From a distance of two, there are two paths to corners with a distance of one, and one path to a corner with distance of three. So x2=1+(2/3)*x1+(1/3)*x3.

From a distance of three, all three paths lead to a distance of two, so x3=1+x2.

The equations are:

a) x0=0
b) x1=1+(1/3)*x0+(2/3)*x2
c) x2=1+(2/3)*x1+(1/3)*x3
d) x3=1+x2

Solution is:

x1=0
x1=7
x2=9
x3=10
Congrats!

10 is the answer!

DanishDynamite
2007-Sep-27, 12:59 AM
I think TRS has a correct proof by contradiction.

Assume p(n) is largest prime, and define M=p(1)*...*p(n).

Then M+1 is not divisible by p(1) through p(n) by construction. By assumption, there is no larger prime, so M+1 cannot be divisible by any number. But then, M+1 must be prime, so this is a contradiction. It is different from the proof you know (which is the same proof I know), but it is still valid.
No, I'm afraid it isn't. See answer above. :)

Sarawak
2007-Sep-27, 01:49 AM
No, I'm afraid it isn't. See answer above. :)

I did see above, but the solution of TRS looks acceptable to me. I said before it had a problem because I read it carelessly, and only noted that it was different from the usual proof. But I think both are correct.

Proof by contradiction is to proof something is both true and false. If this is possible, then the axioms contain some inconsistency, and some axiom must be rejected. If we assume there is some largest prime and call it p(n), and define M as the product of all primes, then we have two statements:

A: There is some prime larger than p(n) but smaller than M+1 that divides M+1.
B: M+1 is prime.

The traditional proof shows that the statement (A or B) is false (by assumption that p(n) is largest prime), and that (A or B) is also true (by laws of arithmetic). So there is a contradiction.

The proof of TRS finds the contradiction in a different place. It shows that both A and B are false, by assumption that p(n) is largest prime. But by the laws of arithmetic, (A or B) is true. Since A is false, B is true. Therefore it proves that B is false and that B is true, so there is a contradiction.

Sarawak
2007-Sep-27, 01:49 AM
Congrats!

10 is the answer!

I will take a little time to try to think of a good question :)

mr obvious
2007-Sep-27, 11:20 AM
If we assume there is some largest prime and call it p(n), and define M as the product of all primes, then we have two statements:

A: There is some prime larger than p(n) but smaller than M+1 that divides M+1.
B: M+1 is prime.

Even though I'm quoting HB, I'm not saying HB is wrong, just making a quick point.

While it is true that proof by contradiction can be satisfied by showing either A or B must be true, in most cases, people leave out A. This is because you wish to show that if your assumptions are all correct, then you can derive a result contradictory to those assumptions. There is no need to further state that it's possible that your assumptions were incorrect to begin with, since that is what you are trying to show.

Thus, if you start out with a list p1, p2, p3... pq of primes for which pq>= all primes, there is no purpose in saying that M+1, where M is the product of all primes, might factor into some prime you left off the list. Your list must be comprehensive, and it can be since there are only finitely many primes if you assume there is a fixed largest prime.

After all, if you are allowed to have an incomplete list, you then also need to show what happens if M+1 factors into some prime pr such that pr<pq, that wasn't on the list. Then what will you do? If you say 'well, pr should've been in the product seeing as it's a prime,' then the same applies to the mysterious prime pw with M+1>pw>pq and M+1 factoring into pw. Why wasn't it on the list to begin with?

There's nothing wrong with including it, but it doesn't add anything to the proof, Euclid's version (admittedly, I haven't read it) notwithstanding.

The_Radiation_Specialist
2007-Sep-28, 04:43 AM
. . .

;)

DanishDynamite
2007-Sep-28, 06:35 PM
I did see above, but the solution of TRS looks acceptable to me. I said before it had a problem because I read it carelessly, and only noted that it was different from the usual proof. But I think both are correct.

Proof by contradiction is to proof something is both true and false. If this is possible, then the axioms contain some inconsistency, and some axiom must be rejected. If we assume there is some largest prime and call it p(n), and define M as the product of all primes, then we have two statements:

A: There is some prime larger than p(n) but smaller than M+1 that divides M+1.
B: M+1 is prime.

The traditional proof shows that the statement (A or B) is false (by assumption that p(n) is largest prime), and that (A or B) is also true (by laws of arithmetic). So there is a contradiction.

The proof of TRS finds the contradiction in a different place. It shows that both A and B are false, by assumption that p(n) is largest prime. But by the laws of arithmetic, (A or B) is true. Since A is false, B is true. Therefore it proves that B is false and that B is true, so there is a contradiction.
TRS makes no mention of A at all in his proof. This is the problem. He simply states that M + 1 is not only a prime, but the biggest prime. The first part of that statement, that M + 1 is a prime, is simply not true and a wrong conclusion to draw. Once you have the product M + 1 and determine that this number cannot be divided (without remainder) by any of the known primes, the only conclusion one can draw at this point is that either A is true or B is true. Drawing the conclusion that B is true only, is simply wrong math.

An example of TRS's proof with a small introduction:

A prime number is a whole number which has no other factors than 1 and itself. Without giving the proof here, this means that any whole number is either a prime or a product of primes. Now on to TRS's proof....

Suppose the sixth prime p(6) = 13 is the biggest prime. Let M = p(1)*p(2)*...*p(6). Then the number M + 1 cannot be divided by any of the known primes and M + 1 is therefore a prime, and since it is bigger than p(6) it is also the largest prime. Since we assumed p(6) was the largest prime, we have a contradiction and p(6) cannot be the largest prime.

The part in bold is a wrong conclusion to draw as shown below:

M + 1 = p(1)*p(2)*p(3)*p(4)*p(5)*p(6) + 1
= 2*3*5*7*11*13 + 1
= 30031
= 59*509

Since the part in bold is wrong, everything that follows (in italics) is also wrong.

The_Radiation_Specialist
2007-Sep-28, 07:19 PM
You could also argue that including that there is a prime pw, pk<pw<M+1 would contradict the first assumption which is saying that assume pk is the biggest prime.

The definition of a prime is as you stated:

A prime number is a whole number which has no other factors than 1 and itself.

So if it was a composite number (not a prime) it would be composed of a product of several primes. But in our assumption that p(k) was the biggest prime. No prime that was in our assumption (and we assumed these are all the primes that exist) would factor M+1. So based on our assumption and definition of a prime we could say M+1 would be another prime.

M + 1 = p(1)*p(2)*p(3)*p(4)*p(5)*p(6) + 1
= 2*3*5*7*11*13 + 1
= 30031
= 59*509

While true that primorial primes (http://en.wikipedia.org/wiki/Primorial_prime) are not always a prime and can have factors that are larger than the largest prime factored, as you just showed, my example was based on an assumption that there weren't any primes bigger than p(k). When we establish that and make a primorial of p(k) then that + 1 would not factor all the primes we assumed there to exist. So it would be a new prime.

DanishDynamite
2007-Sep-28, 08:16 PM
You could also argue that including that there is a prime pw, pk<pw<M+1 would contradict the first assumption which is saying that assume pk is the biggest prime.Yes, I could argue that, and that is exactly what Euclid argued. But you cannot simply omit this step in a proof.

So if it was a composite number (not a prime) it would be composed of a product of several primes.
This is true, and the proof is simple, but it is a statement that needs to be proved.

But in our assumption that p(k) was the biggest prime. No prime that was in our assumption (and we assumed these are all the primes that exist) would factor M+1. So based on our assumption and definition of a prime we could say M+1 would be another prime.
Yes, we could say it would need to be a prime because although it doesn't need to be, the only alternative is that a number >p(q) divides M + 1 and this would also go against our assumption. BUT, the point is that we need to say this in our proof, otherwise the proof is not a proof.

While true that primorial primes (http://en.wikipedia.org/wiki/Primorial_prime) are not always a prime and can have factors that are larger than the largest prime factored, as you just showed, my example was based on an assumption that there weren't any primes bigger than p(k). When we establish that and make a primorial of p(k) then that + 1 would not factor all the primes we assumed there to exist. So it would be a new prime.
Yes, but you need to say this. A proof is a step by step affair. Your proof is Euclid's proof, you just forgot to explicitly mention a step in the logical process.

No worries.

The_Radiation_Specialist
2007-Sep-28, 09:07 PM
Yes, but you need to say this. A proof is a step by step affair. Your proof is Euclid's proof, you just forgot to explicitly mention a step in the logical process.

I did mention rather "unprofessional proof" ;)

Well speaking of infinite prime proving there is another one from Euler which is elegant.

First consider the sequence:
1+1/4+1/9+1/16+... = Infinite Sum (n=1) [1/n^2]

This is bigger than zero, obviously and converges (in another proof which I won't include here).

Now according to the fundamental theorem of arithmetic every number has a unique combination of product of primes. Which is elegantly included in:

(1 + 1/2 + 1/2^n + 1/2^2n + 2^3n +...) * (1 + 1/3 + 1/3^n + 1/3^2n +.....) * (1 + 1/5 ...)

This would produce the sum of all inverse square numbers for n=2, sum of all inverse numbers (does not converge) for n=1 and so on.

(1 + 1/2 + 1/2^n + 1/2^2n + 1/2^3n +...) can be written as 1/(1-(1/2^n))

as it is what it converges to in an infinite summation.

and generally it would be for any number (m) , 1/(1-(1/m^n))

And using the fundamental theorem of mathematics and this result we can have an infinite product of all the primes this way.

So it would be
Infinite Sum (n=1) [1/n^q] = Infinite product (n=1) [1/(1-(1/p^q))

where n is all integers and p is all primes. Therefore there has to be infinite primes as we know that the replacing q by 2 would converge and be greater than zero.

mr obvious
2007-Sep-29, 12:19 AM
Yes, but you need to say this. A proof is a step by step affair. Your proof is Euclid's proof, you just forgot to explicitly mention a step in the logical process.

Depending on the phrasing of the assumptions in the original proof, this does not need to be said. The basic core of the proof is thus:
We hypothesize there are finitely many primes of which pq is the largest.
If M=the product of all primes, then M+1 is itself a prime. Since M+1>pq, we have a contradiction to disprove the hypothesis.

That's it. There is no need to say, maybe M+1 is composite and can factor into some prime pr that wasn't in the product somewhere. This is because of the word 'all' in the antecedent. Once you've made an assumption, you cannot challenge its validity later in the same proof.

Your own example (of factoring a short product of a few primes) does not have significance in this proof because you are showing the inverse of the statement. Specifically, you show:
If M=the product of some primes, then M+1 can be composite.

That does not mean anything since the original antecedent states that M must be the product of all primes, not some small finite set of primes.

To extend this to another example, in the proof that sqrt(2) is irrational, you must then also agree that my proof was incomplete. I must say, that either (a,b,)>1, which creates a contradiction, or that a/b was not sqrt(2) to begin with. This latter part would then need to be accounted for somehow, and I have never seen that in any proof.

mr obvious
2007-Sep-29, 12:29 AM
Once you've made an assumption, you cannot challenge its validity later in the same proof.

Just to clarify, you cannot challenge its validity in the same proof by pulling an exception out of the air. A proof by contradiction has at its root a demonstration of inconsistency that challenges the assumption, but said challenge must be derived.

So if you say 'I have a finite set of all the primes,' I can say, 'you can't, because from whatever finite set you have, I can generate a new one that isn't in that set.' I cannot say 'Maybe there is one that you missed,' as a mathematical proof.

If you wish to think about this some more, why don't you assume that M+1 is composite (where M is the product of all primes). Now prove to me that some prime factor pw that divides M+1 must be greater than pq, which was the largest prime postulated in the hypothesis. Remember, you have relaxed the condition in the antecedent so that when I said 'all primes,' obviously there are some primes less than M+1 that haven't been included. Your job is to demonstrate why a prime factor of M+1 must be greater than pq, without introducing new terms into the proof (e.g., you can't assume consecutiveness or anything that wasn't already explicitly stated in your proof).

DanishDynamite
2007-Sep-29, 12:53 AM
Depending on the phrasing of the assumptions in the original proof, this does not need to be said. The basic core of the proof is thus:
We hypothesize there are finitely many primes of which pq is the largest.
If M=the product of all primes, then M+1 is itself a prime. Since M+1>pq, we have a contradiction to disprove the hypothesis.

That's it. There is no need to say, maybe M+1 is composite and can factor into some prime pr that wasn't in the product somewhere. This is because of the word 'all' in the antecedent. Once you've made an assumption, you cannot challenge its validity later in the same proof.

Your own example (of factoring a short product of a few primes) does not have significance in this proof because you are showing the inverse of the statement. Specifically, you show:
If M=the product of some primes, then M+1 can be composite.

That does not mean anything since the original antecedent states that M must be the product of all primes, not some small finite set of primes.

To extend this to another example, in the proof that sqrt(2) is irrational, you must then also agree that my proof was incomplete. I must say, that either (a,b,)>1, which creates a contradiction, or that a/b was not sqrt(2) to begin with. This latter part would then need to be accounted for somehow, and I have never seen that in any proof.
I have already explained where TRS went wrong. If you still feel his proof is valid and different from Euclid's proof, I suggest you send in your thoughts to the current Mathematical body in charge of these things. I'm sure they will be utterly amazed at this new revolutionary way of proving this simple concept.

Let us know about the feedback you receive.

mr obvious
2007-Sep-29, 03:17 AM
To DanishDynamite:

I am not saying Euclid's proof is incorrect. I am saying your claim that TRS's proof is incomplete is incorrect. I have looked at the Wiki page you linked to. There, Euclid simply states that the product of some finite number of primes plus one is either prime or must factor a prime that wasn't in the product. He made no claims that the finite set of primes was exhaustive. Since TRS claimed (or rather strongly implied) at the very beginning of his proof that he is enumerating all primes P1, P2, etc until Pq as the largest prime, he can skip the extra step.

So, yes, TRS's proof is complete and does not require any additional statements, even if it is not fluently stated. For example, see here:
http://mathforum.org/library/drmath/view/55855.html

He says there are two possibilities, that M+1 is prime, or that M+1 divides the existing, enumerated primes. Note that he does not include a random prime that divides M+1 and was not enumerated. Or see here:
http://www.jcu.edu/math/vignettes/primes.htm

You see, they also do not postulate the existence of some mysterious prime pw > pq. In all proofs that I see where the primes are listed exhaustively and there is a maximum designated, the only comment is that if M+1 is not prime, it must factor into an existing, enumerated prime, but it can't because there would be a remainder of one.

Again, here:
http://www.math.utah.edu/~pa/math/q2.html

You see in the upper box of the proof again, they do not list the condition where M+1 can factor into some mysterious unlisted prime.

In all of these cases, the argument is as I have outlined (although in most cases, the argument is flipped - but still equivalent). There is no need to postulate that there may be some Pw, where Pw wasn't on the list of the primes being multiplied, that divides M+1.

Since I have provided many sources and have explained why your understanding of the Wiki section is incorrect, I hope you'll take back your unnecessarily sarcastic and rude comments. I also note that you did not rise to the challenges posed by me in the preceding posts, which were not addressed in any of your previous explanations.

The_Radiation_Specialist
2007-Sep-29, 05:41 AM
This seems to be cleared. Thanks mr. obvious.

Now who's posting the next question? o_O

hhEb09'1
2007-Sep-29, 02:12 PM
So, yes, TRS's proof is complete and does not require any additional statements, even if it is not fluently stated. For example, see here:
http://mathforum.org/library/drmath/view/55855.htmlHa! Check it out, even the statement "or it is prime itself" can be removed from that proof! :)

Now who's posting the next question? o_OHomo Bibiens, but he hasn't posted since:

I will take a little time to try to think of a good question :)

The_Radiation_Specialist
2007-Sep-29, 04:58 PM
Homo Bibiens hasn't been online since he posted that so I assume he is away or very busy...

Moving swiftly on...

Prove by induction

11^n - 1 is divisible by 10 for all positive integer values of n.

hhEb09'1
2007-Sep-29, 05:16 PM
Prove by induction

11^n - 1 is divisible by 10 for all positive integer values of n.Well, it's certainly true for n=1 (or 0, or 2, or 3... :) )

So, assume for some n that (11^n - 1) is divisible by 10, that is (11^n - 1) = 10A for some integer A.

Then (11^(n+1) - 1) = 11(11^n - 1) + 10 = 11(10A) + 10 = 10(11A + 1), and since (11A+1) is also an integer, 11^(n+1)-1 is also divisible by 10.

QED :)

My question? Solve this (http://www.xkcd.com/287/) :) (You_must_order_two_of_some_item)

mr obvious
2007-Sep-29, 09:33 PM
Seven mixed fruits will work, I think. I'm assuming tax/tip, etc. are included or not part of the \$15.05 total.

hhEb09'1
2007-Sep-29, 10:19 PM
There's at least one more answer, I'll hold out for it.

The_Radiation_Specialist
2007-Sep-30, 02:58 PM
There's at least one more answer, I'll hold out for it.

Please do, this is an interesting question and one which might only be solved with an exhaustive search program.

hhEb09'1
2007-Sep-30, 03:20 PM
The old aphorism, many a truth is shared in jest, certainly seems appropriate in this case! The references to the knapsack and traveling salesman problems are not irrelevant. :)

pghnative
2007-Sep-30, 04:28 PM
one mixed fruit: \$2.15
two hot wings: \$7.10
one sampler: \$5.80

solving hh's puzzle: priceless!

It's a shame that the answer isn't unique. I wonder if the answer would be unique if the fruit were \$2.25 and the sampler \$5.70? (And no, that's not my next question!)

The_Radiation_Specialist
2007-Sep-30, 05:16 PM
Cool, how did you do it? :)

hhEb09'1
2007-Sep-30, 05:29 PM
It's a shame that the answer isn't unique. Seems even more like a cruel joke on a waitperson, right? :)

I even sent the artist an email asking about that. But surely some other geeks have attacked it.

You and mr obvious may share the next honors (or, how about, next one of you to post a question gets to post the next question?)

PS:
I wonder if the answer would be unique if the fruit were \$2.25 and the sampler \$5.70? (And no, that's not my next question!)one mixed fruit: \$2.25
one french fries: \$2.75
three side salads: \$10.05

The_Radiation_Specialist
2007-Sep-30, 05:40 PM
Well the answer can be obtained by a simple Google search (http://www.google.com.my/search?hl=en&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=gd2&q=knapsack+problem+xkcd+answer&btnG=Search&meta=). I'm interested if Pghnative found it by some algorithm or some exhaustive search.

:)

hhEb09'1
2007-Sep-30, 06:13 PM
Well the answer can be obtained by a simple Google search (http://www.google.com.my/search?hl=en&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=gd2&q=knapsack+problem+xkcd+answer&btnG=Search&meta=). Ah, the author has commented in his blag (sic, another xkcd joke) that he thought the problem had one solution. "Long story short, it claimed 2.15*7 did not equal 15.05, and so it missed the second answer in the search, though it found the first just fine"

Floating point math was the culprit :)

mr obvious
2007-Sep-30, 07:52 PM
pghnative may have the honors for asking a question since I have already done so in this thread and would like some variety.

pghnative
2007-Oct-01, 12:30 PM
Well the answer can be obtained by a simple Google search (http://www.google.com.my/search?hl=en&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=gd2&q=knapsack+problem+xkcd+answer&btnG=Search&meta=). I'm interested if Pghnative found it by some algorithm or some exhaustive search.

:)
By a semi-exhaustive search. Used a spreadsheet to let me test permutations quickly. Since there had to be an odd number of the first 4 items, that cut down on the possible permutations.

OK, so there's this bar where you can get a free beer if you know a secret code. You go into it, and watch for a while. Guy # 1 comes in, the bartender says "6" and the guy answers "3" and gets a free beer. Guy # 2 comes in, the bartender says "12" and the guy answers "6" and gets a free beer. Finally, guy # 3 comes in, the bartender says "14" and the guy says "8", and gets a free beer.

The bartender looks at you and says "22" --- what do you say?

tdvance
2007-Oct-01, 01:36 PM
9 (10 if you count the dash).

Todd

hhEb09'1
2007-Oct-01, 01:41 PM
Weirdly enough, seventh graders in the class I was teaching on Friday were talking about this function. They claimed that every number became four, if you continued iterating the function. And, of course, that's where it stops.

pghnative
2007-Oct-01, 01:45 PM
9 (10 if you count the dash).

Todd
correct --- your turn

UFOvsUSO
2007-Oct-01, 02:27 PM
144 of course

hhEb09'1
2007-Oct-01, 02:46 PM
144 of courseWelcome to BAUT!

21.

Which, if anybody hasn't already noticed, is exactly half of 42.

tdvance
2007-Oct-01, 04:43 PM
Ok....

Two cars play "chicken" on a perfectly straight road, starting one mile apart. These "magical" cars accelerate instantly to 30 miles per hour and drive toward each other, ultimately crashing. Exactly halfway between the cars is a "magical" fly. This fly starts instantly at 5 miles per hour toward one of the cars. As soon as he contacts that car, he reverses course instantly with no deceleration and flies 5 miles per hour toward the other car. Upon contacting that car, it again instantly reverses direction and returns to the first car, and so on and so on. This continues until the two cars crash with the fly flattened in between them.

How far does the fly ultimately travel, adding all the distance it flies in both directions all the time from when the cars start a mile apart to the time they collide?

(Hint, it can be done without resorting to geometric series).

hhEb09'1
2007-Oct-01, 05:07 PM
Two cars play "chicken" on a perfectly straight road, starting one mile apart. These "magical" cars accelerate instantly to 30 miles per hour and drive toward each other, ultimately crashing. Exactly halfway between the cars is a "magical" fly. This fly starts instantly at 5 miles per hour toward one of the cars. As soon as he contacts that car, he reverses course instantly with no deceleration and flies 5 miles per hour toward the other car. Upon contacting that car, it again instantly reverses direction and returns to the first car, and so on and so on. This continues until the two cars crash with the fly flattened in between them.Shouldn't the fly's speed be greater than the car's speed? Otherwise, after it contacts the first car, the car will pull away from it and the fly will never be able to catch up to get caught between them when they crash.

Maybe you meant, 5 mph relative to the following car? Which would be a speed of 35 mph, for the fly?

The_Radiation_Specialist
2007-Oct-01, 05:09 PM
oops.
I think you need to check your numbers.

The_Radiation_Specialist
2007-Oct-01, 05:40 PM
The numbers may not be right but if they were, the answer would be the speed of the fly multiplied by the time it takes the cars to crash, assuming the speed of the fly is faster than the speed of the car (relative to a point on the road).

mike alexander
2007-Oct-01, 06:04 PM
220 feet.

mike alexander
2007-Oct-01, 06:08 PM
Which is TRS's answer with a calculator. Assuming 5MPH.

hhEb09'1
2007-Oct-01, 06:50 PM
220 feet.That's half of what I get. The cars meet in one minute, right?

And where will the fly be when they crash? :)

DanishDynamite
2007-Oct-01, 07:24 PM
To DanishDynamite:

I am not saying Euclid's proof is incorrect. I am saying your claim that TRS's proof is incomplete is incorrect. I have looked at the Wiki page you linked to. There, Euclid simply states that the product of some finite number of primes plus one is either prime or must factor a prime that wasn't in the product. He made no claims that the finite set of primes was exhaustive. Since TRS claimed (or rather strongly implied) at the very beginning of his proof that he is enumerating all primes P1, P2, etc until Pq as the largest prime, he can skip the extra step.

So, yes, TRS's proof is complete and does not require any additional statements, even if it is not fluently stated. For example, see here:
http://mathforum.org/library/drmath/view/55855.html

He says there are two possibilities, that M+1 is prime, or that M+1 divides the existing, enumerated primes. Note that he does not include a random prime that divides M+1 and was not enumerated. Or see here:
http://www.jcu.edu/math/vignettes/primes.htm

You see, they also do not postulate the existence of some mysterious prime pw > pq. In all proofs that I see where the primes are listed exhaustively and there is a maximum designated, the only comment is that if M+1 is not prime, it must factor into an existing, enumerated prime, but it can't because there would be a remainder of one.

Again, here:
http://www.math.utah.edu/~pa/math/q2.html

You see in the upper box of the proof again, they do not list the condition where M+1 can factor into some mysterious unlisted prime.

In all of these cases, the argument is as I have outlined (although in most cases, the argument is flipped - but still equivalent). There is no need to postulate that there may be some Pw, where Pw wasn't on the list of the primes being multiplied, that divides M+1.

Since I have provided many sources and have explained why your understanding of the Wiki section is incorrect, I hope you'll take back your unnecessarily sarcastic and rude comments. I also note that you did not rise to the challenges posed by me in the preceding posts, which were not addressed in any of your previous explanations.My comments were not sarcastic. I don't see any new version of Euclid's proof here, I just see a missing step in the proof. You apparently feel this is a new version and hence I encouraged you, and still encourage you, to send this in to the appropriate Mathematical body in charge of rendering verdict on these things.

If it truely is a new version, you could get your name in the history books.

DanishDynamite
2007-Oct-01, 07:28 PM
That's half of what I get. The cars meet in one minute, right?

And where will the fly be when they crash? :)
Stuck on the fender of the car it meets first. :)

tdvance
2007-Oct-01, 10:58 PM
Oops--I guess I did misremember the problem wrong. But yes, the solution is to find out how long (in hours) the cars take to crash, and multiply that by 5 miles per hour, to get the distance in miles the fly flew (or convert to more reasonable units!). So, it takes one minute for the cars to collide. In one minute, the fly can go 1/12 of a mile, or 440 feet. So, HeebeeJeebee99 or whatever that is, wins for two reasons--being first to point out my screw up..I mean "Lesson Learned" (at work, a manager suggested I never say "screw up" in front of budget people, but "lessons learned" instead), and for getting the correct answer to the flawed problem! (that would be the real answer, since the fly would be stuck to the one bumper flipping back and forth an infinite number of times per second till the crash).

Todd

DanishDynamite
2007-Oct-02, 12:18 AM
I'm not sure I have the privelidge, but never the less I'll post an oldy, but goody:

Suppose you had a rope around the Earth. It fit snugly.

Suppose you now inserted a stick, 1 meter long, between the string and the Earth. How long would the stretched string be now?

hhEb09'1
2007-Oct-02, 12:28 AM
Go ahead and play through, DanishDynamite, but you probably should be spending the time on those prime proofs. :)

DanishDynamite
2007-Oct-02, 12:44 AM
Go ahead and play through, DanishDynamite, but you probably should be spending the time on those prime proofs. :)
Not understood my friend. Could you clarify?

pghnative
2007-Oct-02, 01:17 AM
I'm not sure I have the privelidge, but never the less I'll post an oldy, but goody:

Suppose you had a rope around the Earth. It fit snugly.

Suppose you now inserted a stick, 1 meter long, between the string and the Earth. How long would the stretched string be now?3.14 meters (approximately).

The_Radiation_Specialist
2007-Oct-02, 05:46 AM
Not understood my friend. Could you clarify?

...So, HeebeeJeebee99 or whatever that is, wins ...(snip)

Todd

HeebeeJeebbbe99 = hhEb09'1

Sarawak
2007-Oct-03, 12:56 AM
I am sorry, I left this thread for some days without posting my question. But someone else went in my place, this is reasonable, the game should not be interrupted by my absence.

If I understand the current question correctly, then I think the answer is:

2 [(pi-theta)*r+sqrt(2*r+1)]

where

theta=ArcSec[1+1/r]

and r is the radius of the earth in meters. If r is 40,000,000 divided by (2*pi), then I get that the string is stretched by 0.747 millimeters.

If this is the correct answer, then anyone who likes can ask the next question. If it is a wrong answer, then maybe I will ask for some clarifications.

hhEb09'1
2007-Oct-03, 02:26 AM
2 [(pi-theta)*r+sqrt(2*r+1)]That's kinda the answer that I got, but without the pi, I think.

Sarawak
2007-Oct-03, 03:14 AM
That's kinda the answer that I got, but without the pi, I think.

My formula is for the total length of the string. If you calculate only the incremental distance from the stretching, then you would subtract 2*pi*r from my answer. Did you calculate it in this way? Maybe we have the same answer then.

hhEb09'1
2007-Oct-03, 03:17 AM
I didn't actually substitute the values in, but the formula is the same, without the pi.

Sarawak
2007-Oct-03, 03:34 AM
Then I think we have the same answer in substance. My answer gives the total length of the stretched string, your answer gives the additional length due to stretching.

Or maybe it is possible that we use theta for different angles, so that your theta is equal to pi minus my theta.

DanishDynamite
2007-Oct-03, 07:55 PM
Sorry, I should have provided a value for the radius of the Earth. Let's suppose it is 6367 km exactly.

Sarawak
2007-Oct-04, 10:59 PM
I find 40,005,040.8515597 meters. Without the stick, it is 40,005,040.8508124 meters.

Is my answer correct?

The_Radiation_Specialist
2007-Oct-07, 06:18 PM
BUMP! :)

Try this:

Simplify,

[1000]
Summation (n-1) /n!
[n=2]

= 1/2! + 2/3! + 3/4! + 4/5! + .... + 999/1000!

The_Radiation_Specialist
2007-Oct-09, 04:34 AM
Any takers?

Sock puppet
2007-Oct-09, 01:39 PM
Tried it, but i'm busy so i didn't have much time. Got nowhere so far...

pghnative
2007-Oct-09, 06:21 PM
BUMP! :)

Try this:

Simplify,

[1000]
Summation (n-1) /n!
[n=2]

= 1/2! + 2/3! + 3/4! + 4/5! + .... + 999/1000!1

hhEb09'1
2007-Oct-09, 06:24 PM
1Almost. 1 - 1/1000!

The_Radiation_Specialist
2007-Oct-09, 06:27 PM
1

That is close but not exact.

It would approach 1 as n goes to infinity but here I'm looking at the exact answer when there are 1000 terms.

The_Radiation_Specialist
2007-Oct-09, 06:28 PM
Almost. 1 - 1/1000!

Correct. :)

This is one of my favorite problems. It was included in the Polya Student problems of the Mathematics Challenge for young Australians program, two years ago and this was one of the problems I got right. My love for maths started as soon I stared at the finished proof of this problem. :)

tdvance
2007-Oct-09, 07:20 PM
Correct. :)

This is one of my favorite problems. It was included in the Polya Student problems of the Mathematics Challenge for young Australians program, two years ago and this was one of the problems I got right. My love for maths started as soon I stared at the finished proof of this problem. :)

Is there a simple proof?

Todd

Sarawak
2007-Oct-09, 07:36 PM
Is there a simple proof?

Todd

Define s(n) is sum up to (n-1)/(n!) term. Then proof is s(n)=1-1/(n!) by induction.

For n=2, this is by simple verification: 1/(2!)=1/2=1-1/(2!)

Now assume it is true up to some n. Then:

s(n+1)=s(n)+(n)/(n+1)!=1-1/(n!)+(n)/(n+1)!=1-(n+1)/(n+1)!+(n)/(n+1)!=1-1/(n+1)!

So it is proved.

I did not find the solution myself, but only found the proof after seeing the solution here.

hhEb09'1
2007-Oct-09, 07:43 PM
I did not find the solution myself, but only found the proof after seeing the solution here.Still, I think that you are next, since you answered DanishDynamite's problem. I left a message for them over at the other thread.

The_Radiation_Specialist
2007-Oct-09, 08:04 PM
Is there a simple proof?

Todd

This is the way I did it. I'm sure there would be a shorter proof using mathematical induction to first prove the general formula and then to show for this answer.

= [ (3*4*5*...*1000) + (2*4*5*...*1000) + (3*5*6*...*1000) + ... + 999 ] /1000!

Temporarily add 1 to the numerator, it will be subtracted. And we are only working on the numerator.

= [ (3*4*5*...*1000) + (2*4*5*...*1000) + (3*5*6*...*1000) + ... + 1000 ] -1

Factor by 1000.

= [ 1000*[(3*4*5*...*999) + (2*4*5*...*999) + (3*5*6*...*999) + ... + 998 + 1] ] -1

= [ 1000*[(3*4*5*...*999) + (2*4*5*...*999) + (3*5*6*...*999) + ... + 999] ] -1

Now Factor by 999.

= [ 1000*999[(3*4*5*...*998) + (2*4*5*...*998) + (3*5*6*...*998) + ... + 997 +1 ] ] -1

And continue the same cycle by factoring by 998, 997 ... , until 4.

= [ 1000*999*...*4[(3) + 2 + 1 ] ] -1

= [ 1000*999*...*4[(3) + 3 ] ] -1

And finally factor by 3.

= [ 1000*999*...*4*3[ 2] ] -1

= 1000! - 1

putting it into the fraction

= 1000! - 1 /1000!

=1- 1/1000!

And generally it can be shown:

m
Sum (n-1) / n! = 1 - 1/m!
n=1

toejam
2007-Oct-09, 09:47 PM
Well, that's simple enough... :)

tdvance
2007-Oct-10, 03:42 AM
ah, good. Yeah, induction is often simple after you know what you are trying to prove. But the Radiation proof is what's needed to come up with it in the first place.

Todd

Sarawak
2007-Oct-10, 01:33 PM
ah, good. Yeah, induction is often simple after you know what you are trying to prove. But the Radiation proof is what's needed to come up with it in the first place.

Todd

I am certain that if I added the first few terms, I would recognize the pattern, and then prove the result by induction in a minute or two. But since I did not think to try this, I wasted much time with other approaches, still without finding the answer :(

So it is my turn? I hope my answer to the stick and string question is correct.

But here is one. I think it is easy, but maybe people do not work with probability so much, then it might be more difficult.

X and Y are uncorrelated random variables. If the distribution of X is normal (or Gaussian) and the distribution of Y is also normal, are X and Y independent? Prove so or give counter-example.

DanishDynamite
2007-Oct-12, 06:26 PM
I find 40,005,040.8515597 meters. Without the stick, it is 40,005,040.8508124 meters.

Is my answer correct?
Yes, it is correct. Well done. :)

But I now realize I made a mistake in the question I asked. I should not have asked how much longer the rope becomes, mainly because the answer is so unsurprising. It hardly gets any longer at all, does it? I mean, we are talking milimeters here!

The question I should have asked and which gives a surprising answer, is: How long is the part of the rope no longer touching the Earth? And the answer to this is approximately 7.1 kilometers!

That was surprising, at least to me.

Sorry for amswering late and asking wrong.

Sarawak
2007-Oct-15, 11:58 AM
The question I should have asked and which gives a surprising answer, is: How long is the part of the rope no longer touching the Earth? And the answer to this is approximately 7.1 kilometers!

I calculated this as a step to the calculation of the final answer. I did not save my work, but this number sounds familiar.

X and Y are uncorrelated random variables. If the distribution of X is normal (or Gaussian) and the distribution of Y is also normal, are X and Y independent? Prove so or give counter-example.

It is several days now with no replies. Is this question too difficult?

hhEb09'1
2007-Oct-15, 12:54 PM
Too much like homework :)

DanishDynamite
2007-Oct-15, 06:10 PM
It is several days now with no replies. Is this question too difficult?
Not having done any serious math for over 15 years (and by serious I just mean university level math), it is a difficult question to solve from first principles. If I could just find my statistics books from back then (I suspect they are in a box in the attic) it would be a very easy question. I have definitely met this question before.

Give us a few days more.

tdvance
2007-Oct-15, 07:29 PM
Let X be uniformly distributed on [-1,1]. Let Y=X^2.

Then, they are not independent:
probability that X>0.9 and Y<0.5 is 0, but prob X>0.9 is 0.1, prob Y<0.5 is something close to .7 or so, product is .07 != 0.

However, they are uncorrelated (i.e. have 0 covariance):

COV(X,Y) = E((X - E(X))(Y-E(Y)) = E((X - 0)(X^2 - 1/2))

= E(X^3 - 1/2X) = 0.

Todd

hhEb09'1
2007-Oct-15, 08:12 PM
Let X be uniformly distributed In the problem, both X and Y are normally distributed

tdvance
2007-Oct-15, 08:36 PM
oops--I forgot about that.

Todd

The_Radiation_Specialist
2007-Oct-20, 08:04 PM
Did this game die?

Sarawak
2007-Oct-20, 08:29 PM
I think I killed it :(

Maybe someone else should ask a new question.

The_Radiation_Specialist
2007-Oct-20, 08:44 PM
Since hhEb09'1 answered my question correctly, I think he may go now.

hhEb09'1
2007-Oct-20, 09:59 PM
An oldie but a goodie. And the eighth graders were working on it last month :)

Take a sphere, and drill an eight inch long (or 8 cm long, if you prefer) hole through it. What is the volume of the material that is left in the sphere?

The_Radiation_Specialist
2007-Oct-20, 10:18 PM
I remember this...

Ah yes, the mighty Google gives me the thread...

http://www.bautforum.com/off-topic-babbling/40327-core-blimey-puzzle-really-intelligent-people.html

So that is 4*4*4*4/3*pi = (85 + 1/3)*pi cubic inch (or cm) :)

hhEb09'1
2007-Oct-21, 09:12 AM
the thread...

http://www.bautforum.com/off-topic-babbling/40327-core-blimey-puzzle-really-intelligent-people.htmlI was sick that week :)

Yep, that is the answer (and the discussion), you're up!

The_Radiation_Specialist
2007-Oct-21, 09:24 AM
Prove that

9a4-8a3b+b4 ≥ 0

holds for all real numbers a and b.

Hint: use the AM GM Inequality

Sarawak
2007-Oct-21, 12:23 PM
Probably there is a more elegant solution, but:

9a4-8a3b+b4 = (b-a/sqrt(2)(i-sqrt(1-4isqrt(2))))*(b-a/sqrt(2)(i+sqrt(1-4isqrt(2))))*(b-a/sqrt(2)(-i-sqrt(1+4isqrt(2))))*(b-a/sqrt(2)(-i+sqrt(1+4isqrt(2))))

For a=0, the function is ≥0 because it is b4. For real a<>0, the function is continuous and all its zeros (in b) have non-zero imaginary part, so the function cannot have both positive and negative values. And for b=0, it is 9a4, so for real a<>0 and real b it is always positive.

The_Radiation_Specialist
2007-Oct-21, 01:02 PM
Probably there is a more elegant solution

The other solution is we divide it into 4 parts ... 9a^4 + b^4 = 4a^4 + 4a^4 + a^4 + b^4
and take the AM of LHS and GM of RHS and according to the inequality we can put the sign ... (9a^4 + b^4)/4 ≥ 4th root(4a^4 * 4a^4 * a^4 * b^4 )
so ... 9a^4-8a^4b+b^4 ≥ 0

Your go. :)

Sarawak
2007-Oct-21, 04:31 PM
My last turn was not a big success, I will try to do better this time. First I will give a solution to my earlier question.

Choose X as a normal random variable, and Z with values +1 or -1 with equal probability, and Z is independent of X. Then Y=Z*X is a normal random variable, it is uncorrelated with X, but it is not independent of X.

Now here is my new question.

In a gambling game, a gambler wins 0.5 for each 1.0 of the bet if a coin toss is "heads," and loses 0.4 if the coin toss is "tails." So if he bets five Australian dollars, then he wins 2.5 if the coin is heads, and loses 2.0 if it is tails. Now, suppose someone plays this game repeatedly with the following strategy. He begins with a gamble of one Australian dollar. Whatever he has after this gamble (1.5 or 0.6), he bets the entire amount on a second coin toss. After this, he can have 2.25, 0.9, or 0.36, but he then bets the full amount on a third coin toss. And he does this repeatedly until N coin tosses are complete.

What is his average winning (including the one Australian dollar he began with) after N coin tosses? As N goes to infinity, what is his average winning?

What is his median winning (including the one Australian dollar he began with) after N coin tosses? As N goes to infinity, what is his median winning?

hhEb09'1
2007-Oct-21, 04:51 PM
In a gambling game, a gambler wins 0.5 for each 1.0 of the bet if a coin toss is "heads," and loses 0.4 if the coin toss is "tails." So if he bets five Australian dollars, then he wins 2.5 if the coin is heads, and loses 2.0 if it is tails. There is a different way of representing the game. I would have said he starts with a stake of five dollars, bets 40% of it (2 dollars) and has good odds of winning (he wins \$2.50 half of the time on a \$2 bet.)

Sarawak
2007-Oct-21, 10:55 PM
There is a different way of representing the game. I would have said he starts with a stake of five dollars, bets 40% of it (2 dollars) and has good odds of winning (he wins \$2.50 half of the time on a \$2 bet.)

Yes, this is an equivalent formulation. So if anyone prefers to think in this way, he bets 40% of his money on each coin toss. If he loses, all the amount bet is lost, but if he wins, he receives back 1.25 times the bet, plus the original bet. So if he has \$10, he bets \$4. If he loses, the \$4 is gone, so he has \$6 remaining, and his next bet will be \$2.40. If he wins, he receives back his \$4 plus \$5, so he has \$15 in total, and the next bet will be \$6.

Also, for the median, it is simpler to think only of the case where N is even. The case where N is odd has some slight complication, and does not add anything interesting to the problem.

Sarawak
2007-Oct-24, 10:21 AM
I hope I have not killed this thread again :(

Someone please try, it is not so difficult. If no one can answer, I will give some hints. Here is the first hint. For the median case, it can be helpful to consider the logarithm of is winnings instead of winning. When N is even, the median of the logarithm is the logarithm of the median.

hhEb09'1
2007-Oct-24, 10:54 AM
Then Y=Z*X is a normal random variable, it is uncorrelated with X, but it is not independent of X.What do you mean by Z*X?

Sarawak
2007-Oct-24, 12:34 PM
What do you mean by Z*X?

It is just ordinary multiplication, not convolution or anything strange. Z has values +1 or -1 with equal probability, and X has a random value with the normal distribution. If X and Z are independent, then the product of the two random variables is also a normal distribution, and it is uncorrelated with X. But it is not independent.

The_Radiation_Specialist
2007-Oct-31, 06:19 PM
I got some help from a friend, will that count? :)

notice that if the gambler wins with a bet x, his total amount of money becomes 1.5x. If he loses, then it becomes 0.6x. So if there are N tosses, the money he gets, y(x), is so that y(x) = x*1.5^a * 0.6^b where a + b = N. So a = 0,1,2, 3 ... N and b = N, N-1, N-2...0. So there are N + 1 ways of having a + b = N. That said, we look at 1.5^0*0.6^N + 1.5^1*0.6^N-1 + ... + 1.5^N*0.6^0. This is a geometric sum of N +1 terms with the first term equal to 0.6^N and the argument being 1.5/0.6. So that sum is equal to 0.6^N * ((1.5/0.6)^N+1 - 1)/(1.5/0.6 -1). This really simplifies to 5/3 * 1.5^N - 2/3 *0.6^N. Taking the average, we get 5/3 * 1.5^N - 2/3 *0.6^N / N + 1. From this it's clear that the average as N goes to infinity is infinity: 0.6^N / N + 1 obviously goes to 0, and 1.5^N / (N + 1) goes to infinity since 1.5 > 1 and the order of magnitude of e^x is greater than x + 1.

The_Radiation_Specialist
2007-Oct-31, 06:22 PM
There is something wrong with this thing, it won't let me post the entire proof in one post. :eh:

Now considering the median: from the geometric sum above, we have an increasing sequence of N + 1 terms. So if N is even, the median is 0.6^N * 1.5^(N/2 + 1) = . If N is odd, the median is the average of 0.6^N * 1.5^(N+1 /2) and 0.6^N * 1.5^(N+1 / 2 + 1). But in all cases the we get a development that yields numbers lesser than 1 to unbounded powers. So the median goes to 0.

Sarawak
2007-Oct-31, 10:55 PM
The Radiation Specialist, I think you have the right idea, but there are some details. There are N+1 ways to have a+b=N, but they do not all have equal probability. To have a=b=N/2 is much more probable than to have a=0 and b=N. The probability of a=M and b=N-M is {(N)!/[(M)!(N-M)!]}/(0.5^N), so to find the mean you need to weight by these probabilities, not give equal weight to all values of M as you did.

But I think an easier way to find the mean is to call X(i) the multiplier of gamble i, so X(i) is 1.5 if he wins, and X(i) is 0.6 if he loses. Then he has X(1)*X(2)*...*X(N-1)*X(N). The problem is to find the mean of this quantity. But if random quantities are independent, then the mean of their product is the product of their means. And the mean of X(i) is the same for all values of i, so the mean of his winnings after N gambles is some number raised to the power of N.

For the median, I think you have the right idea but just made a small typo. The median outcome for even N has N/2 wins and N/2 losses, so the median amount of winnings is 0.6^(N/2)*1.5^(N/2).

You are very close, I think you can find an exact expression for the mean if you do it in this way :)

The_Radiation_Specialist
2007-Nov-03, 08:06 PM
Prove that the fraction,

21n+4
-------
14n+3

is irreducible for every natural number n

tdvance
2007-Nov-04, 03:00 AM
Prove that the fraction,

21n+4
-------
14n+3

is irreducible for every natural number n

Let (.,.) represent GCD. Because n is a natural number, the arguments to GCD below are all integers, and, just writing out Euclid's algorithm by hand:

(21n+4,14n+3) = (21n+4 - (14n+3),14n+3) = (7n+1,14n+3)
= (7n+1,14n+3-2*(7n+1)) = (7n+1,1) = 1, so the numerator and denominator have no common factor (other than 1 or -1) and the fraction is irreducible.

The_Radiation_Specialist
2007-Nov-04, 10:41 AM
Yes, your turn. :)

Sarawak
2007-Nov-04, 11:22 PM
I want to say a few things about the solution to my question. The mean is 1.05N, and the median is 0.9N/2. It is true also for any percentile that it approaches zero as N goes to infinity. So the gambler following this strategy becomes rich on average, but becomes nearly broke (you can define "nearly broke" to be as close to zero as you like) with probability that approaches one.

tdvance
2007-Nov-06, 04:13 PM
For n a nonnegative integer, what (in terms of n) is the sum of the n-th powers of the roots of the polynomial x^6 - 1 ?

The_Radiation_Specialist
2007-Nov-07, 04:09 AM
1+(-1)^n

Sarawak
2007-Nov-07, 12:09 PM
If you mean to include all complex roots in addition to real roots, then the sum is 6 if n is divisible by 6, and 0 otherwise. But if this is the correct answer, then I give my turn to The Radiation Specialist, because my questions have not been very successful.

tdvance
2007-Nov-07, 12:59 PM
Homo Bibiens has the correct answer, but has given his turn to The Radiation Specialist.

Todd

The_Radiation_Specialist
2007-Nov-07, 03:39 PM
Show that there are infinitely many pairs of positive integers (m, n) such that

(m + 1) / n + (n + 1) / m

is a positive integer.

I'll be away for a few days so if anyone thinks they got it right they may go ahead with the next question.

The_Radiation_Specialist
2007-Nov-12, 01:25 PM
Any updates?

Did I do what Homo Bibiens dreaded? :doh:

hhEb09'1
2007-Nov-13, 08:42 PM
Any updates?Wait, don't tell me :)

These things are supposed to be getting harder, so they should take more time :)
The game is a math game, we start out with a simple math question then whoever answers it has to come up with a slightly harder question. Ok what is 12x12?

The_Radiation_Specialist
2007-Nov-17, 09:54 PM
Clue: Try going case by case, such as case m = n, m = n + 1, m > n + 1. With the last case you may arrive at an alternative identity which may need to be proved by itself.

The_Radiation_Specialist
2007-Nov-18, 11:50 PM
Do you want more time or should I give the answer? I'm starting to think the interest in this waning...

Sock puppet
2007-Nov-19, 01:51 PM
I'd like to take a crack at this, but I've been too busy lately. Would you mind giving me until Thursday evening?

hhEb09'1
2007-Nov-27, 07:04 AM
Do you want more time or should I give the answer? I'm starting to think the interest in this waning...Here's one way to go. It's pretty easy to find a few pairs for (n,m), like (1,1), (1,2), (2,3), or (2,2). Take any m and n, such that m is greater than or equal to n and A(n,m) = (n+1)/m + (m+1)/n is an integer.

Then, it is easy to show that n must divide (m2+m) so M=(m2+m)/n is an integer, and it is greater than m. Also, A(m,M) = A(n,m), so the larger pair (m, M) also produces an integer--in fact, the same integer. Apply, rinse, repeat. :)

MY QUESTION: How many different values can A(n,m) take on? Prove it.

hhEb09'1
2007-Dec-04, 12:06 PM
Been a week, water low, haven't seen anything on the horizon, no contrails, where is everybody?? BAUT bunnies are starting to taste funny, and everything smells like molasses. I may change my name to Legend.

Interesting observation: If you take consecutive odd members of the Fibonacci sequence and add one to each, the resulting pair of numbers satisfies the (m + 1) / n + (n + 1) / m integer condition. Another way (sorta!) of producing an infinite number of solutions. I heard that from my imaginary friend Vladeta Jovovic.

The_Radiation_Specialist
2007-Dec-05, 08:03 AM
I'm here, well I was busy last week and right now I'm not clear minded enough to think about it (my brain needs more caffeine) . Hold on there with the bunnies!

If you take consecutive odd members of the Fibonacci sequence and add one to each, the resulting pair of numbers satisfies the (m + 1) / n + (n + 1) / m integer condition.
What you mean 1,1,2,3,5,8,13 so I take 5 and 13 then it is 6 and 14. 7/14+15/6= 1/2+3/2 = 2 Oh cool! Must. Get. Proof. :)

hhEb09'1
2007-Dec-05, 11:21 AM
Hold on there with the bunnies!The bunnies are safe for now

What you mean 1,1,2,3,5,8,13 so I take 5 and 13 then it is 6 and 14. 7/14+15/6= 1/2+3/2 = 2 Oh cool! Must. Get. Proof. :)7/14+15/6= 1/2+5/2=3 :)

hhEb09'1
2007-Dec-06, 05:09 PM
OK it's been over 24 hours. The bunnies are looking pretty delectable...
...lectable
...table
...ble

Sean Clayden
2007-Dec-07, 01:40 PM
If the radius of a circle is 5m what is the area of that circle in m ?

hhEb09'1
2007-Dec-24, 02:55 AM
Ba-da-bump
If the radius of a circle is 5m what is the area of that circle in m ?There's a question already on the table. Besides, that one seems more philosophical than mathematical. :)

Sarawak
2007-Dec-24, 03:03 AM
A clarification please - do you mean how many integer values can A(n,m) take? Otherwise, I think the answer is, infinitely many :)

hhEb09'1
2007-Dec-24, 03:14 AM
A clarification please - do you mean how many integer values can A(n,m) take? Yes, A(n,m) is associated with The_Radiation_Specialist's question, and it is only defined for A, n, m, positive integers. :)

Sarawak
2007-Dec-24, 03:19 AM
Yes, A(n,m) is associated with The_Radiation_Specialist's question, and it is only defined for A, n, m, positive integers. :)

But A(n,m) must also be an integer, correct? Otherwise, we can choose infinitely many values, just by setting m=1.

hhEb09'1
2007-Dec-24, 03:28 AM
I don't think the board software would allow infinitely recursive threads, so I think I can safely say yes. :)

Sarawak
2007-Dec-24, 04:33 PM
Use the procedure of hhEb09'1 in reverse.

If A(m,n) is an integer, then M=(m2+m)/n is an integer also, whether or not m is larger than n, and A(M,m)=A(n,m). So assume with no loss of generality that m is less than or equal to n, then define M=(m2+m)/n. Repeat this procedure (reversing m and n if m > n) until the M produced is not smaller than n.

But M<n if (m2+m)<n2. If m is equal to n, this is impossible for m>0. But (m2+m)<(m+1)2, so this is always true if m<n. So this procedure terminates when m and n are equal.

So for any integer m>0 and n>0 where A(m,n) is an integer, there is some integer k>0 so that A(k,k)=A(m,n). But A(k,k)=2+2/k. This is only an integer for k=1 and k=2, and A(1,1)=4 and A(2,2)=3.

So A(m,n) for integer m>0 and n>0 can only take two integer values, 3 and 4.

hhEb09'1
2007-Dec-24, 05:16 PM
So for any integer m>0 and n>0 where A(m,n) is an integer, there is some integer k>0 so that A(k,k)=A(m,n). But A(k,k)=2+2/k. This is only an integer for k=1 and k=2, and A(1,1)=4 and A(2,2)=3.

So A(m,n) for integer m>0 and n>0 can only take two integer values, 3 and 4.That's it! You're next.

By building the sequence an = (an-12-an-1)/an-2, and starting with (1,1) and (2,2) you get two different sequences such that consecutive pairs of terms satisfy The_Radiation_Specialist's criteria. The second sequence is just the odd terms of the Fibonacci sequence each plus one, so an alternative definition is just an = 3an-1 - an-2 - 1. An alternative definition for the first one is an = 4an-1 - an-2 - 1

Sarawak
2008-Jan-01, 02:33 PM
So, I will have to think of a good question now. I will spend some time on it. But my questions have not been very successful, so if anyone thinks of a good question, that person is free to take my turn.

Sean Clayden
2008-Jan-22, 12:28 PM
Ba-da-bumpThere's a question already on the table. Besides, that one seems more philosophical than mathematical. :)

Juat worry about the bunnies

hhEb09'1
2008-Jan-22, 12:36 PM
I can now answer your question: a buck two fifty five

hhEb09'1
2008-Feb-02, 08:35 AM
:think: (http://www.math.niu.edu/~rusin/problems-math/MP1975.html) Take the group of all ordered pairs of integers generated by component-wise addition by the three elements (3,8), (4,-1), and (5,4). There is another set of generators (1,b) and (0,a), with a, b integers, a>0. Find a.

tdvance
2008-Feb-02, 05:37 PM
#2 - #1 gives (1,-9) (#4)
#3 - #2 gives (1,5) (#5)

#4 + #5 gives (2,4) (#6)
#1 - #6 gives (1,4) (#7)

#6 - #7 gives (1,0) (#8)

4 * #8 - #2 gives (0,1)

Thus, (1,0) and (0,1) are in the same group. Since they clearly generate #1, #2, and #3, they are another set of generators. I.e. you have the entire integer lattice.

hhEb09'1
2008-Feb-02, 10:22 PM
#2 - #1 gives (1,-9) (#4)
#3 - #2 gives (1,5) (#5)

#4 + #5 gives (2,4) (#6)#4 + #5 gives (2,-4) (#9) :)

tdvance
2008-Feb-03, 03:37 AM
Darn...

Let a=(3,8)
Let b=(4,-1)
Let c=(5,4)

d=2a+b+2c=(0,7)
e=c-b=(1,5)

and we have

3e-d = a
4e-3d = b
5e-3d = c

So, a=7, b=5.

hhEb09'1
2008-Feb-03, 05:55 AM
Of course, you can assume that the original three are generated by those two. :)

Your turn!

The_Radiation_Specialist
2008-Feb-04, 06:11 PM
Hey all, What's happening? I've been busy with college life, and haven't been online in a while. This thread is the only thread I can let myself waste time on (sorry).

Solve this,
x^2 + 1/x^2 = 7

x + 1/x = ???

find "???"

Cheers, and apologies to whoever's turn I just poked my nose into.

tdvance
2008-Feb-04, 06:24 PM
(x+1/x)^2 = x^2 + 2x*1/x+1/x^2 = 7+2=9, so x+1/x is +/- 3.

I still need to come up with a math problem.

Sarawak
2008-Feb-04, 06:27 PM
:( I was not fast enough :(

tdvance
2008-Feb-06, 06:07 PM
take four "4"s and five possible binary operations: +, -, *, /, and concatenation (e.g. 4 concatenated with 4 is 44), and you can make a lot of numbers: e.g. 15 = 4+(44/4) (parentheses can be placed wherever you want).

I want 5 numbers:

there are exactly 2 numbers from 0 to 20 that cannot be created this way. One of them is 13. What is the other? ____

There are exactly 5 numbers having 4 digits that can be created this way. One of them is 1776. What are the other 4? ____ ____ ____ ____

edit: by "numbers" I mean integers, and division is exact (i.e. no rounding allowed).

edit #2--use exactly four 4s, not 1 2 or 3

Sarawak
2008-Feb-06, 07:23 PM
Is it required to use all four "4"s? There is one number between 0 and 20 I can create with three "4"s, but I cannot create it with four. If it is acceptable to use fewer than the full allotment of 4s, then I can solve both problems.

Sarawak
2008-Feb-06, 07:46 PM
there are exactly 2 numbers from 0 to 20 that cannot be created this way. One of them is 13. What is the other? ____

If we need to use exactly four 4s, then I cannot find a way to generate three numbers, including 13.

There are exactly 5 numbers having 4 digits that can be created this way. One of them is 1776. What are the other 4? ____ ____ ____ ____

This one I could find in less than 30 seconds.

tdvance
2008-Feb-06, 10:46 PM
I'm teaching myself "scheme" and wrote a scheme program to list all possibilities. I found and corrected a bug-- It is true, there are 3 that can't be done from 0 to 20. And there are actually 9, not 5, 4 digit numbers; all from the same bug.

go ahead and post your answers--it sounds like you got it.

Todd

Sarawak
2008-Feb-06, 11:58 PM
I'm teaching myself "scheme" and wrote a scheme program to list all possibilities. I found and corrected a bug-- It is true, there are 3 that can't be done from 0 to 20. And there are actually 9, not 5, 4 digit numbers; all from the same bug.

go ahead and post your answers--it sounds like you got it.

Todd

For the second question, I stopped looking when I found five :) But now I have found ten of the nine possibilities.

For the first question, 13, 14, and 19.

For the second question, 1616, 1644, 1664, 1764, 1776, 1936, 4164, 4176, 4416, and 4444.

tdvance
2008-Feb-07, 12:06 AM
13, 14, 19 are correct, as well as

1616, 1644, 1764, 1776, 1936, 4164, 4176, 4416, and 4444.

How did you get 1664?

Sarawak
2008-Feb-07, 12:20 AM
How did you get 1664?

i) Multiply 4 by 4 = 16.

ii) Concatenate 4 with 16 = 416.

iii) Multiply 4 by 416 = 1664.

tdvance
2008-Feb-07, 06:33 PM
ah ok---

of course, you're the winner and it's your turn.

I checked the output again--1664 is in there after all--I had scanned it by eye (visual grep) and extracted the four-digit solutions--and clearly I missed that one.

Todd

Sarawak
2008-Feb-08, 07:47 AM
ah ok---

of course, you're the winner and it's your turn.

I checked the output again--1664 is in there after all--I had scanned it by eye (visual grep) and extracted the four-digit solutions--and clearly I missed that one.

Todd

I do not know what method you used, but if there are redundancies in the list, it is easy to mistake 1644 and 1664 as the same number :(

But here is a new problem.

Assume the earth is a perfect sphere with a radius of 6,378 km. What percentage of the world's surface is more than 9,000 km (measured by distance along the surface, not through the earth) from the Komsomolskaya-Koltsevaya Metro station in Moscow?

hhEb09'1
2008-Feb-08, 02:44 PM
Assume the earth is a perfect sphere with a radius of 6,378 km. What percentage of the world's surface is more than 9,000 km (measured by distance along the surface, not through the earth) from the Komsomolskaya-Koltsevaya Metro station in Moscow?(1 + cos(9000/6378))/2 times 100%

58%

Sarawak
2008-Feb-08, 06:36 PM
Looks right to me. Of course, the particular choice of metro station does not affect the answer.

Your turn!

hhEb09'1
2008-Feb-09, 04:53 AM
It is a not well-enough appreciated fact that the probability of two randomly aligned axes being within theta of each other is sin theta. I just had to tranlate it into cosine, divide by two, and add in the other hemisphere.

Next problem: take the vertices of an acute triangle, in flat space, and show that there are two more points such that no three of the five are collinear, and the line through any two is perpendicular to the plane determined by the other three.

Sarawak
2008-Feb-09, 07:39 AM
It is a not well-enough appreciated fact that the probability of two randomly aligned axes being within theta of each other is sin theta. I just had to tranlate it into cosine, divide by two, and add in the other hemisphere.

I found the answer the hard way :) But your formula is valid even when the included portion is smaller than a hemisphere.

Next problem: take the vertices of an acute triangle, in flat space, and show that there are two more points such that no three of the five are collinear, and the line through any two is perpendicular to the plane determined by the other three.

The assumption that the triangle is acute is unnecessary; it is only required that it is not a right triangle. Use an (x,y,z) coordinate system, and choose one vertex to be (x1,y1,z1)=(0,0,0). By rotation of axes and scaling, choose another vertex to be (x2,y2,z2)=(1,0,0). By further rotation of the y and z axes, choose the third vertex to be (x3,y3,z3) with z3=0. For the three vertices not to be collinear, y3 is not zero.

Then choose (x4,y4,z4)=(x3,(1-x3)x3/y3,z4) with z4 not equal to zero, and (x5,y5,z5)=(x3,(1-x3)x3/y3,(1-(1-x3)x3/y32)(1-x3)x3/z4).

Then the vector between any two points is perpendicular to the vector between any other two points, which can be verified by taking the cross-product of each pair of vectors, equal to zero.

hhEb09'1
2008-Mar-11, 07:12 PM
Then the vector between any two points is perpendicular to the vector between any other two points, which can be verified by taking the cross-product of each pair of vectors, equal to zero.I haven't actually done this algebra, but it looks right to me! :)

For a triangle lying in the plane, the other two points would have to be on a plane perpendicular to each side that goes through the third vertex. That means that the additional two points have to be on the line normal to the plane, through the orthocenter of the triangle--that's your first two coordinates.

You chose an arbitrary point on that line, and the next point is determined by the intersection of the perpendicular planes.

You're up.

The_Radiation_Specialist
2008-Nov-08, 07:20 PM
Bump!

Suppose you have a bag with 50 black marbles and 49 white marbles. You take two out marbles at a time. If they are the same you put back a black marble (from an infinite supply of black and white marbles next to you) and if they are different you put back a white marble.

Justify what will be the last marble remaining.

hhEb09'1
2008-Nov-08, 07:45 PM
Justify what will be the last marble remaining.Has to be white. The rules require you to take the white marbles out two at a time. Since there are an odd number of white marbles, the last one has to be white, if there ever is a last one.

The_Radiation_Specialist
2008-Nov-08, 07:47 PM
cool, your turn.

hhEb09'1
2008-Nov-08, 08:01 PM
For a convex polygon, possibly irregular, of n sides, for each point of the interior, compute a function F that represents the sum of the distances from the point to each side (the distance calculated perpendicular to the side, as if the side were infinitely extended). For each polygon, F is a polynomial whose degree D is a function of n. What is the function D(n)?

Sarawak
2008-Nov-15, 06:21 PM
For a convex polygon, possibly irregular, of n sides, for each point of the interior, compute a function F that represents the sum of the distances from the point to each side (the distance calculated perpendicular to the side, as if the side were infinitely extended). For each polygon, F is a polynomial whose degree D is a function of n. What is the function D(n)?

I think maybe some clarification will be helpful. It is a polynomial in what variable? Or maybe multiple variables? Is it the x and y coordinates of the point that is chosen, if we place the polynomial in a two-dimensional Cartesian coordinate system?

hhEb09'1
2008-Nov-15, 09:28 PM
I think maybe some clarification will be helpful. It is a polynomial in what variable? Or maybe multiple variables? Is it the x and y coordinates of the point that is chosen, if we place the polynomial in a two-dimensional Cartesian coordinate system?Thank you for that clarification!

Sarawak
2008-Nov-15, 11:48 PM
Thank you for that clarification!

It was a question, not a clarification :)

But then, I am getting D(n)=1. But I think this is based on an incorrect interpretation of your question.

tdvance
2008-Nov-15, 11:53 PM
Ok, an affine change of variables, i.e. functions like x' = ax+b, y'=ay+b (linear polynomials) can map any line to any other line in the plane. If the problem is not flawed, F is a polynomial. One can choose an invertible affine transformation mapping any side to, say, the x axis, and it will not change the degree of F. Thus, for each side in turn, transform the point (x,y) and F to make the side be the x-axis, and then the distance is just the y coordinate--a linear function! Add each of these up for each side to get the sum of distances. This is a sum of linear functions, so is a linear function, and it came from composed affine transformations on the variables so the original F is linear.

Thus, D(n)=1 for all n>0.

(Either that, or the original F is not a polynomial).

hhEb09'1
2008-Nov-16, 12:00 AM
Thus, D(n)=1 for all n>0.Remember, n > 2 :)

Hint: what's the function F for the interior points of a square (n=4)?

Sarawak
2008-Nov-16, 12:18 AM
I have a feeling the answer is D(n)=1 for odd n, and D(n)=0 for even n, but I have no proof yet.

Sarawak
2008-Nov-16, 12:39 AM
I have a feeling the answer is D(n)=1 for odd n, and D(n)=0 for even n, but I have no proof yet.

No, I do not think this is correct either, with a square it is 0, but with a trapezoid it is not.

If the answer is not D(n)=1, with some possibility of 0 instead depending on particular characteristics of the polygon besides n, then I do not think I understand the question properly.

tdvance
2008-Nov-16, 01:50 AM
ah gee--it's not 0 is it--a constant function?

Yeah, that's right for regular polygons. And if the problem is sound, should be the same formula for any (convex) polygon.

hhEb09'1
2008-Nov-16, 01:52 AM
ah gee--it's not 0 is it--a constant function?

Yeah, that's right for regular polygons. And if the problem is sound, should be the same formula for any (convex) polygon.Do you feel like justifying that conjecture? :)

Otherwise, we have a winner!

hhEb09'1
2008-Nov-16, 03:01 AM
No, I do not think this is correct either, with a square it is 0, but with a trapezoid it is not.I just sat down and tried the trapezoid, and either I'm too tired, or I screwed up the wording of the problem. I'll get back to you as soon as I figure out which.

tdvance
2008-Nov-16, 04:03 AM
That's right--the sum for a trapezoid is a linear polynomial in the distance from the point to the smaller (of the parallel) side. Putting the smaller of the parallel sides on the x axis makes it a linear polynomial in the variable y.

hhEb09'1
2008-Nov-16, 09:06 AM
I guess I can change the problem to when does D=0?

I don't know what I was thinking. I'm going to have my blood pressure checked as soon as it is light. In the meantime, do not trust anything I say, that's an order. Even if I say "cancel that order, that's an order," don't believe me.

tdvance
2008-Nov-16, 03:07 PM
A sufficient condition for D=0 would be: the polygon has an even number of sides, and opposite sides are parallel.

Sarawak
2008-Nov-19, 12:18 AM
For an equilateral triangle also, it is D=0.

Sarawak
2008-Nov-23, 12:38 AM
An equivalent formulation is to say, take for each side a unit vector that is perpendicular to the side and points to the inside of the polygon. D=0 if these vectors all add up to zero.

For triangles, it can only be equilateral. For quadrilaterals, it can only be a parallelogram. I do not have a general result at this time.

tdvance
2008-Nov-23, 10:50 PM
I think that's a pretty good answer--necessary and sufficient, and gets all the cases we've noted so far, e.g. even number of sides and regular, or more generally, opposite sides are parallel. Also, it shows D=0 for any regular polygon, since the unit normals must add to 0 for the same reason that the nth roots of unity add to 0.

Sarawak
2008-Nov-24, 12:10 AM
I am just realizing this point, why do we use the word "triangle" for three but "quadrilateral" for four? Why not "trilateral" or "quadrangle"? They are both words for other things :) But it would be consistent to use always "angle" or "lateral," but we don't.

tdvance
2008-Nov-24, 07:17 PM
In "projective geometry" a quadrangle is four points, no three collinear, and a quadrilateral is four lines, no three concurrent. (no parallel lines in projective geometry, so any four lines, no three concurrent, have four intersection points, actually six). So, it really depends on what you emphasize.

I don't know why in ordinary Euclidean geometry we choose to emphasize the vertices of a triangle but the sides of a quadrilateral.

hhEb09'1
2008-Nov-26, 10:58 PM
And, it's not pentangle/pentalateral nor hexangle/hexalateral neither! :)
I think that's a pretty good answer--necessary and sufficient, and gets all the cases we've noted so far, e.g. even number of sides and regular, or more generally, opposite sides are parallel. I'm not sure it's sufficient. What if three of the normals are parallel? Say, by adding a square to one side of a regular septagon? Of course, that's no longer convex.

connor240287
2008-Nov-27, 10:40 PM
Ok no 1 is posting questions/answers so I'll post 1.

98 to the power of 7 = 8.68 × 1013
(Hope non of you are useing calculators, took me 10 mins to figure that out)

and my question, 934 to the power of 11

hhEb09'1
2008-Nov-28, 05:29 AM
(Hope non of you are useing calculators,
we all are! :)

The_Radiation_Specialist
2008-Nov-28, 06:31 AM
In which direction is the bus travelling?

left or right?
http://img503.imageshack.us/img503/6826/schoolbuslu7.jpg

connor240287
2008-Nov-28, 01:45 PM
It's travelling right, just a gess cos I'm looking at the wheels and the circles inside are more to the right.