|
|
| Author |
Message |
server
science forum beginner
Joined: 24 Mar 2005
Posts: 26
|
 Posted: Thu Apr 28, 2005 3:19 pm Post subject:
All Integers are Interesting (with Proof)
|
|
|
|
message unavailable |
|
| Back to top |
|
 |
Reef Fish
science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 200
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: All Integers are Interesting (with Proof)
|
|
|
Bob Pease wrote:
| Quote: | "Reef Fish" <Large_Nassau_Grouper@Yahoo.com> wrote in message
news:1114374431.963748.290770@o13g2000cwo.googlegroups.com...
|
<Gigantic snip for brevity>
| Quote: |
Suppose that Dustin Hoffman found nothing interesting about anything
except
1,2,
Someone points out that the number 3 is the first uninteresting
number,
then.
So Dustin admits this and groups all numbers this way.
so if you name any number whatever, it is now interesting to HIM
because of
its residue mod 2.
|
This joke is almost as bad as the joke of Cauchy's dog leaving his
residue at the pole.
| Quote: | ( reducing Dustin's world to binary)
|
Another oldie:
There are 10 kinds of people in the world:
those who understand binary and those who don't.
| Quote: | Even *I* think this post is getting odd!!
|
On even and odd, Goldback's conjecture that all EVEN numbers greater
than 2 can be expressed as the sum of two (ODD of course) primes
is still up for grabs.
The conjecture had been verified to all even numbers up to 100 million,
but has not been proved to be universally true for all even numbers.
A final interesting number for today. Mersennes conjectured that
2^67 - 1 was a prime, after it was found that 2^p -1 were prime for
p = 2,3,5,7 resulted in primes, but failed when p =11.
The conjecture stood for 250 years unchallenged until Frank Nelcon Cole
gave a "talk" at the American Mathematics Society in 1903. Cole, a man
of few words, did not say a single during the entire presentation.
He went to the blackboard. raised 2 to the 67th power and subtracted 1
to get 147,573,952,589,676,412,927.
Without a word, he moved to the clear side of the blackboard and
multiply out long hand,
195,707,721 x 761,838,257,287
which was 147,573,952,589,676,412,927.
That poor sap had more time on his hand than any of us!  But
he certainly made the Mersenne number 147,573,952,589,676,412,927
interesting.
-- Bob. |
|
| Back to top |
|
|
Reef Fish
science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 200
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
$1,000,000 Prize
|
|
|
Reef Fish wrote:
| Quote: |
On even and odd, Goldback's conjecture that all EVEN numbers greater
than 2 can be expressed as the sum of two (ODD of course) primes
is still up for grabs.
|
Faber and Faber offered a $1,000,000 prize prize to anyone who proved
Goldbach's conjecture between March 20, 2000 and March 20, 2002, but
the prize went unclaimed and the conjecture remains open. Not sure if
the prize is still open, but I am sure the PROOF will be more than $1M
to the methematician, on just Nike sponsored ads.
| Quote: |
The conjecture had been verified to all even numbers up to 100
million,
but has not been proved to be universally true for all even numbers.
|
This number is way out of date.
http://www.ieeta.pt/~tos/goldbach.html
By March 20, 2005, all even numbers through 2^17 or
10,000,000,000,000,000
had been double-checked, and still going strong. But this will never
constitute a proof, though a counterexample may DISPROVE the
conjecture.
I had a proof written on the margin of a note book, but I couldn't find
the notebook or remember the proof. :-)
-- Bob. |
|
| Back to top |
|
|
Proginoskes
science forum Guru
Joined: 29 Apr 2005
Posts: 2593
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: $1,000,000 Prize
|
|
|
Reef Fish wrote:
| Quote: | [about Goldbach's conjecture]
I had a proof written on the margin of a note book, but I
couldn't find the notebook or remember the proof.
|
Or remember what language it's in. 8-)
--- Christopher Heckman
Lunatic idea of the week: The Voynich Manuscript is really a proof of
Goldbach's Conjecture. |
|
| Back to top |
|
|
Frank J. Lhota
science forum beginner
Joined: 09 May 2005
Posts: 43
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: All Integers are Interesting (with Proof)
|
|
|
"Reef Fish" <Large_Nassau_Grouper@Yahoo.com> wrote in message
news:1114374431.963748.290770@o13g2000cwo.googlegroups.com...
| Quote: | Actually, that statement can only be PROVED if the number is an
INTEGER.
|
Not necessarily. Using the Axiom of Choice, every set can be well ordered,
that is, every set S has a linear ordering such that every non-empty subset
of S has a minimal member. Your proof can be modified to show that any set
that can be well ordered has no uninteresting members. |
|
| Back to top |
|
|
Reef Fish
science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 200
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: All Integers are Interesting (with Proof)
|
|
|
Mike Terry wrote:
| Quote: | "Frank J. Lhota" <NOSPAM.Lhota.adarose@verizon.net> wrote in message
news:GyRbe.7643$WX.1703@trndny01...
"Reef Fish" <Large_Nassau_Grouper@Yahoo.com> wrote in message
news:1114374431.963748.290770@o13g2000cwo.googlegroups.com...
Actually, that statement can only be PROVED if the number is an
INTEGER.
Not necessarily. Using the Axiom of Choice, every set can be well
ordered,
that is, every set S has a linear ordering such that every
non-empty
subset
of S has a minimal member. Your proof can be modified to show that
any set
that can be well ordered has no uninteresting members.
This only works if the well ordering is a "particular" well ordering
that is
interesting in some way...
|
Even THEN, the set of all intergers, well ordered into its natural
order,
which is rather interesting, has no minimal member.
And if anyone else is going to be pedantic about this, I'll have to
bring my
friend Godel here to lecture you on HIS principle, which pulled the rug
out of 20 years of hard labor by Russell and Whitehead on Principia
Mathematica! :-)
While we are at it, here are still other interesting facts about the
numbers
714 and 715, besides having the Pomerance property that 714 x 715 is
the product of the first seven primes
714 + 715 = 1429, which was the year Columbus stumbled onto America.
:^)
1429 is also a "backwards-forwards-sideways prime": 1429, 9241, 1249,
9421, 4129, 4219 are all prime numbers.
Finally, a PAIR of consecutive integers that has the property that the
sum of the prime factors of one equals the sum of the prime factors
of the other:
714 = 2 x 3 x 7 x 17 2 + 3 + 7 + 17 = 29
715 = 5 x 11 x 13 5 + 11 + 13 = 29
is called a "Ruth Aaron pair". Paul Erdos proved that there are
INFINITELY many Ruth-Aaron pairs.
Do you know ANY Ruth-Aaron pair greater than (714, 715)? :-)
-- Bob. |
|
| Back to top |
|
|
gerard46
science forum beginner
Joined: 24 Mar 2005
Posts: 46
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: All Integers are Interesting (with Proof)
|
|
|
| Reef Fish wrote:
|> Mike Terry wrote:
|>> Frank J. Lhota wrote:
|>>> Reef Fish wrote:
|>>> Actually, that statement can only be PROVED if the number is an
|>>> INTEGER.
|>> Not necessarily. Using the Axiom of Choice, every set can be well ordered,
|>> that is, every set S has a linear ordering such that every non-empty subset
|>> of S has a minimal member. Your proof can be modified to show that any set
|>> that can be well ordered has no uninteresting members.
|> This only works if the well ordering is a "particular" well ordering that is
|> interesting in some way... :)
| Even THEN, the set of all intergers, well ordered into its natural
| order,
| which is rather interesting, has no minimal member.
|
| And if anyone else is going to be pedantic about this, I'll have to
| bring my
| friend Godel here to lecture you on HIS principle, which pulled the rug
| out of 20 years of hard labor by Russell and Whitehead on Principia
| Mathematica!
|
| While we are at it, here are still other interesting facts about the
| numbers
| 714 and 715, besides having the Pomerance property that 714 x 715 is
| the product of the first seven primes
|
| 714 + 715 = 1429, which was the year Columbus stumbled onto America.
| :^)
Before he was even born? _________________________________________Gerard S.
| 1429 is also a "backwards-forwards-sideways prime": 1429, 9241, 1249,
| 9421, 4129, 4219 are all prime numbers.
|
| Finally, a PAIR of consecutive integers that has the property that the
| sum of the prime factors of one equals the sum of the prime factors
| of the other:
|
| 714 = 2 x 3 x 7 x 17 2 + 3 + 7 + 17 = 29
| 715 = 5 x 11 x 13 5 + 11 + 13 = 29
|
| is called a "Ruth Aaron pair". Paul Erdos proved that there are
| INFINITELY many Ruth-Aaron pairs.
|
| Do you know ANY Ruth-Aaron pair greater than (714, 715)? |
|
| Back to top |
|
|
Reef Fish
science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 200
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: All Integers are Interesting (with Proof)
|
|
|
gerard46 wrote:
| Quote: | |
| While we are at it, here are still other interesting facts about
the
| numbers
| 714 and 715, besides having the Pomerance property that 714 x 715
is
| the product of the first seven primes
|
| 714 + 715 = 1429, which was the year Columbus stumbled onto
America.
| :^)
Before he was even born?
_________________________________________Gerard S. |
Yes! That's why he discovered it in his reincarnation in 1492! :-)
Actually, it is now commonly accepted (outside of the USA) that America
was discovered before Columbus was born!
http://www.apol.net/dightonrock/CodFish/discovery_of_north_america.htm
LOL! You caught my hidden Dyslexic Devil!
Actually there are many theories about who First discovered America.
One theory was that a Chinese explorer discovered America during
1421-1423.
| Quote: | http://english.people.com.cn/200203/06/eng20020306_91553.shtml
|
The Portuguese, unaware of the Chinese claim, claimed they discovered
America in 1924. Here's another link about the same claim:
| Quote: | http://www.thornr.demon.co.uk/kchrist/portam.html#COLOM
|
All of this made 1429 all the more interesting as a number, doesn't it?
-- Bob.
| Quote: |
| 1429 is also a "backwards-forwards-sideways prime": 1429, 9241,
1249,
| 9421, 4129, 4219 are all prime numbers.
|
| Finally, a PAIR of consecutive integers that has the property that
the
| sum of the prime factors of one equals the sum of the prime factors
| of the other:
|
| 714 = 2 x 3 x 7 x 17 2 + 3 + 7 + 17 = 29
| 715 = 5 x 11 x 13 5 + 11 + 13 = 29
|
| is called a "Ruth Aaron pair". Paul Erdos proved that there are
| INFINITELY many Ruth-Aaron pairs.
|
| Do you know ANY Ruth-Aaron pair greater than (714, 715)? |
|
|
| Back to top |
|
|
Reef Fish
science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 200
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: All Integers are Interesting (with Proof)
|
|
|
Mike Terry wrote:
| Quote: | "Frank J. Lhota" <NOSPAM.Lhota.adarose@verizon.net> wrote in message
news:GyRbe.7643$WX.1703@trndny01...
"Reef Fish" <Large_Nassau_Grouper@Yahoo.com> wrote in message
news:1114374431.963748.290770@o13g2000cwo.googlegroups.com...
Actually, that statement can only be PROVED if the number is an
INTEGER.
Not necessarily. Using the Axiom of Choice, every set can be well
ordered,
that is, every set S has a linear ordering such that every
non-empty
subset
of S has a minimal member. Your proof can be modified to show that
any set
that can be well ordered has no uninteresting members.
This only works if the well ordering is a "particular" well ordering
that is
interesting in some way...
|
Even THEN, the set of all intergers, well ordered into its natural
order,
which is rather interesting, has no minimal member.
And if anyone else is going to be pedantic about this, I'll have to
bring my
friend Godel here to lecture you on HIS principle, which pulled the rug
out of 20 years of hard labor by Russell and Whitehead on Principia
Mathematica! :-)
While we are at it, here are still other interesting facts about the
numbers
714 and 715, besides having the Pomerance property that 714 x 715 is
the product of the first seven primes
714 + 715 = 1429, which was the year Columbus stumbled onto America.
:^)
1429 is also a "backwards-forwards-sideways prime": 1429, 9241, 1249,
9421, 4129, 4219 are all prime numbers.
Finally, a PAIR of consecutive integers that has the property that the
sum of the prime factors of one equals the sum of the prime factors
of the other:
714 = 2 x 3 x 7 x 17 2 + 3 + 7 + 17 = 29
715 = 5 x 11 x 13 5 + 11 + 13 = 29
is called a "Ruth Aaron pair". Paul Erdos proved that there are
INFINITELY many Ruth-Aaron pairs.
Do you know ANY Ruth-Aaron pair greater than (714, 715)? :-)
-- Bob. |
|
| Back to top |
|
|
Reef Fish
science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 200
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: All Integers are Interesting (with Proof)
|
|
|
gerard46 wrote:
| Quote: | |
| While we are at it, here are still other interesting facts about
the
| numbers
| 714 and 715, besides having the Pomerance property that 714 x 715
is
| the product of the first seven primes
|
| 714 + 715 = 1429, which was the year Columbus stumbled onto
America.
| :^)
Before he was even born?
_________________________________________Gerard S. |
Yes! That's why he discovered it in his reincarnation in 1492! :-)
Actually, it is now commonly accepted (outside of the USA) that America
was discovered before Columbus was born!
http://www.apol.net/dightonrock/CodFish/discovery_of_north_america.htm
LOL! You caught my hidden Dyslexic Devil!
Actually there are many theories about who First discovered America.
One theory was that a Chinese explorer discovered America during
1421-1423.
| Quote: | http://english.people.com.cn/200203/06/eng20020306_91553.shtml
|
The Portuguese, unaware of the Chinese claim, claimed they discovered
America in 1924. Here's another link about the same claim:
| Quote: | http://www.thornr.demon.co.uk/kchrist/portam.html#COLOM
|
All of this made 1429 all the more interesting as a number, doesn't it?
-- Bob.
| Quote: |
| 1429 is also a "backwards-forwards-sideways prime": 1429, 9241,
1249,
| 9421, 4129, 4219 are all prime numbers.
|
| Finally, a PAIR of consecutive integers that has the property that
the
| sum of the prime factors of one equals the sum of the prime factors
| of the other:
|
| 714 = 2 x 3 x 7 x 17 2 + 3 + 7 + 17 = 29
| 715 = 5 x 11 x 13 5 + 11 + 13 = 29
|
| is called a "Ruth Aaron pair". Paul Erdos proved that there are
| INFINITELY many Ruth-Aaron pairs.
|
| Do you know ANY Ruth-Aaron pair greater than (714, 715)? |
|
|
| Back to top |
|
|
Bob Pease
science forum beginner
Joined: 29 Apr 2005
Posts: 47
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: All Integers are Interesting (with Proof)
|
|
|
"Reef Fish" <Large_Nassau_Grouper@Yahoo.com> wrote in message
news:1114374431.963748.290770@o13g2000cwo.googlegroups.com...
| Quote: |
Pavel314 wrote:
"Reef Fish" <Large_Nassau_Grouper@Yahoo.com> wrote in message
news:1114361157.935875.229730@z14g2000cwz.googlegroups.com...
So, what's so interesting about numbers like 714 or 715?
Well, on April 8, 1974, Hank Aaron's 715th home run eclipsed
Babe Ruth's 1935 record of 714. Pomerance, a Georgia Asst Prof,
also noticed that 714x715= 2x3x5x7x11x13x17, the product of the
first 7 primes!
714 is one of my favorite numbers. Besides being the number of home
runs hit
by Babe Ruth, it was also the badge number of Joe Friday on the
original
"Dragnet". And 7/14 in American date notation is July 14, Bastille
Day!
Thanks for the 714 X 715 information; I never noticed that before.
I am glad I stumbled into interest and your interesting number.
Hardy (a great mathematician by his own record) rated himself 20 on a
scale of 1 to 100, and the self-taught Indian postal clerk Ramamujan,
100.
When Ramanujan, was lying in a nursing-home with a mysterious illness,
Hardy called to see him. Finding making conversation difficult, Hardy
observed lamely that he had come in Taxi no. 1729, which he considered
to be an uninteresting number. Ramanujan immediately perked up, and
said that on the contrary, 1729 was a very interesting number: It was
the SMALLEST positve integer that could be expressed as the sum of two
cubes in two different ways:
1729 = 1^3 + 12^3 = 9^3 + 10^3
There's no such thing as an uninteresting number.
Actually, that statement can only be PROVED if the number is an
INTEGER.
Proof:
We proceed by the method of reductio ad absurdum
Suppose the set U of un-interesting integers is non-empty. Then, since
it is a set of integers, there is a least member of U, u, say. u has
the property of being the smallest un-interesting integer, which is
interesting --- a logical contradiction. Thus we have disproved the
hypothesis that U is non-empty, hence empty. The set of un-interesting
integers is empty means there are no un-interesting integers.
QED.
|
Oldie, but goodie.
I think this proof could be expanded to any set of numbers which is
denumerable.
Actually, more "interesting" IMO is the following
**Please restrict the following to Restricting to positive integers..**
Suppose that Dustin Hoffman found nothing interesting about anything except
1,2,
Someone points out that the number 3 is the first uninteresting number,
then.
So Dustin admits this and groups all numbers this way.
so if you name any number whatever, it is now interesting to HIM because of
its residue mod 2.
someone else says , "I don't find any numbers interesting except 1."
The argument falls apart here.
The question immediately arises that the concept "Interesting" need to be
examined.
In the context of the original problem it means a UNIQUE property.
The premise is that somehow correspondence to some subset of integers
having a common property would be intrinsically "interesting"
But if this is admitted, then "Even ness" or odness" should do very nicely!
( reducing Dustin's world to binary)
Even *I* think this post is getting odd!!
Bob Pease |
|
| Back to top |
|
|
Jim Spriggs
science forum Guru
Joined: 24 Mar 2005
Posts: 761
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: All Integers are Interesting (with Proof)
|
|
|
Reef Fish wrote:
| Quote: |
...
Actually, that statement can only be PROVED if the number is an
INTEGER.
Proof:
We proceed by the method of reductio ad absurdum
Suppose the set U of un-interesting integers is non-empty. Then, since
|
You need to say _positive_ integers, else U can be
{-1, -2, -3, -4, ...}
| Quote: | it is a set of integers, there is a least member of U, u, say. u has
the property of being the smallest un-interesting integer, which is
interesting --- a logical contradiction. Thus we have disproved the
hypothesis that U is non-empty, hence empty. The set of un-interesting
integers is empty means there are no un-interesting integers.
QED.
-- Bob. |
|
|
| Back to top |
|
|
MDCovney
science forum beginner
Joined: 07 Jun 2005
Posts: 30
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: How do I this exercise???
|
|
|
As noted by John, the recursion is:
P(n) = p*(1-P(n-1)) + (1-p)*P(n-1), i.e., not as written by Neto.
Rewrite as P(n) = p + (1-2p)*P(n-1) and develop new relationship in p and
powers of (1-2p).
P(0) = 1 (0 is an even number and no tosses certainly produces 0 successes)
P(1) = 1-p
P(2) = p + (1-2p)*(1-p)
P(3) = p + (1-2p)*p + (1-2p)^2*(1-p)
P(n) = p * summation of powers of (1-2p)^i from i=0 to n-2 +
(1-p)*(1-2p)^(n-1)
Trick is to collapse the expanding summations. Use same approach as summing
infinite series. Create a second series that when added to or subtracted
from the first produces a sum that includes only a few terms.
Create a second series by multiplying first series by (1-2p) and then
subtract second from first.
(1-2p)*P(n) = p * summation of powers of (1-2p)^(i+1) from i=i to n-2 +
(1-p)*(1-2p)^n
With a bit of careful algebraic manipulation, result will follow.
LHS = P(n) - (1-2p)*P(n) = 2p*P(n)
RHS = p*(1-(1-2p)^(n-1)) + 2p(1-p)(1-2p)^(n-1)
"Neto" <otnetobr@yahoo.com.br> wrote in message
news:1f880084.0501231706.511b6bba@posting.google.com...
| Quote: | Experts,
I am trying do this exercise and I don't... Can anyone help me?
Exercise:
Independent trials that result in a success with probability "p" and a
failure with probability "1-p" are called Bernoulli Trials. Let P(n)
denote the probability that n Bernoulli trials result in a even number
of success (0 being considered as an even number). Show that:
P(n)=p*(1-P(n-1))+(1+p)*P(n-1)
and use this to prove (by induction) that
P(n)= (1+(1-2p)^n)/2
I'm so grateful for help.
Neto |
|
|
| Back to top |
|
|
Phillip L. Emerson
science forum beginner
Joined: 28 Apr 2005
Posts: 5
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: help - maximum likelihood estimator
|
|
|
"john" <john@hudson8889.fsnet.co.uk> wrote in message
news:cogd1q$apj$1@newsg4.svr.pol.co.uk...
| Quote: |
"Henry" <se16@btinternet.com> wrote in message
news:rd4nq0lpnjrt10tqikck5foulh128r6dqs@4ax.com...
On Sun, 28 Nov 2004 12:39:34 -0000, "john"
john@hudson8889.fsnet.co.uk> wrote:
On 21 Oct 2004 04:43:38 -0700, steve_039@yahoo.com (Steve Johnson)
wrote:
suppose we have n samples x_i from a uniform random variable x
with
unknown mean (but known length of the interval of definition).
what
is
the maximum likelihood estimator for the unknown mean?
From an old statistics book by Mood and Greybill I recall that the ML
estimate of the mean is the average of the two min and max samples.
Intermediate values can be discarded as offereing no further
information.
That would be true if the range was unknown - but in this case it is
known, so if it is w then all values between max-w/2 and min+w/2 have
the same likelihood.
The interval w can be fitted anywhere that includes Xmax and Xmin.samples.
The likelihood must be constant for all such locations. So there is no
maximum likelihood point for the mean, only a uniform likelihood for this
range of locations.
A curious circumstance. I suppose a piling up of observations in a small |
region between the two extremes would cast doubt on the assumptions,
as would any pair of observations farther apart than the "known range.
It is logical, but strange, that the likelihhood outside the observed
extremes
is zero. |
|
| Back to top |
|
|
Rusty
science forum addict
Joined: 30 Apr 2005
Posts: 70
|
| Posted: Thu Apr 28, 2005 3:19 pm Post subject:
Re: help - maximum likelihood estimator
|
|
|
| Quote: |
The interval w can be fitted anywhere that includes Xmax and
Xmin.samples.
The likelihood must be constant for all such locations. So there is no
maximum likelihood point for the mean, only a uniform likelihood for this
range of locations.
A curious circumstance. I suppose a piling up of observations in a small
region between the two extremes would cast doubt on the assumptions,
as would any pair of observations farther apart than the "known range.
It is logical, but strange, that the likelihhood outside the observed
extremes
is zero.
|
Not sure I understand what follow the last sentence exactly |
|
| Back to top |
|
|
Google
|
|
| Back to top |
|
|
|
|
The time now is Sun Nov 23, 2008 9:54 am | All times are GMT
|
|
Car Insurance | Shares | Credit Card | Loans | Mortgage Calculator
|
|
Copyright © 2004-2005 DeniX Solutions SRL
|
|
Other DeniX Solutions sites:
Electronics forum |
Medicine forum |
Unix/Linux blog |
Unix/Linux documentation |
Unix/Linux forums
|
Powered by phpBB © 2001, 2005 phpBB Group
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
Forum index
»
Science and Technology
»
Math
»
Probability
