BR Maths Corner1
Re: BR Maths Corner1
^^^
Pretty simple sir.
First, some basics:
1. We know that prime #s are only divisible by 1 and themselves (this is the definition of a prime #)
2. Any non primes are divisible by numbers other than 1 and themselves (inferred from step 1)
3. Therefore, any non primes can be decomposed into a product of prime #s. For instance 120 is (1 * 2 * 3 * 5 * 2 * 2). This can be inferred from 1 and recursively from 2.
Now for the proof:
I. Let us assume initially that there are indeed a finite # of primes.
II. Let N be the largest prime number.
III. Consider another number X, which is the product of all prime #s between 2 and N (i.e.) X = 2 * 3 * 5 * 7 * 11 * 13 *...* N
IV. Let Y = X + 1
V. Clearly Y is not divisible by the prime numbers 2, 3, 5, 7, 11, 13 ... N
VI. Therefore Y is prime. It is also larger than N. Note that we can repeat the same arguments from step II downwards using Y instead of N and keep going infinitely.
VII. Therefore our assumption in step I is incorrect and there are infinite # of primes.
Pretty simple sir.
First, some basics:
1. We know that prime #s are only divisible by 1 and themselves (this is the definition of a prime #)
2. Any non primes are divisible by numbers other than 1 and themselves (inferred from step 1)
3. Therefore, any non primes can be decomposed into a product of prime #s. For instance 120 is (1 * 2 * 3 * 5 * 2 * 2). This can be inferred from 1 and recursively from 2.
Now for the proof:
I. Let us assume initially that there are indeed a finite # of primes.
II. Let N be the largest prime number.
III. Consider another number X, which is the product of all prime #s between 2 and N (i.e.) X = 2 * 3 * 5 * 7 * 11 * 13 *...* N
IV. Let Y = X + 1
V. Clearly Y is not divisible by the prime numbers 2, 3, 5, 7, 11, 13 ... N
VI. Therefore Y is prime. It is also larger than N. Note that we can repeat the same arguments from step II downwards using Y instead of N and keep going infinitely.
VII. Therefore our assumption in step I is incorrect and there are infinite # of primes.
Re: BR Maths Corner1
Paging AmberG ji,
In the first page of the thread,you have given the standard proof,that x*y*z*a +1 is either prime or a product of two other new primes.ArmenT omits the second part? Why do you need the part about the product+1 being a product of new primes?
In the first page of the thread,you have given the standard proof,that x*y*z*a +1 is either prime or a product of two other new primes.ArmenT omits the second part? Why do you need the part about the product+1 being a product of new primes?
Re: BR Maths Corner1
^^^ As said above the proof is using the principle of contradiction:
Y = (2.3.5.......N)+ 1
is bigger than N:
Two possibilities: Either
1)case 1 : it is prime,
2) case 2: have a factor(s)
In case 1, contradiction, (Y is larger than N and is prime)
case 2: Its factor can not be 2 (remainder is 1) .. can not be 3 (remainder 1) ... can not be N, so it must be larger than N.
For example 2*3*5*7*11*13*17*19+1 (= 9 699 691 is NOT a prime but is divisible by 347, a prime larger than 19)
BTW, a number of the type 2*3*5*7...+() 1 , if prime is called primorial prime, largest known such numebr is 2*3*5*7....392113+1 which happens to be prime.
(of course other examples of primorial prime primes are 7 (=3*3+1), 31 (=2*3*5+1) and also 5 (=2*31) is a primorial prime prime too)
Y = (2.3.5.......N)+ 1
is bigger than N:
Two possibilities: Either
1)case 1 : it is prime,
2) case 2: have a factor(s)
In case 1, contradiction, (Y is larger than N and is prime)
case 2: Its factor can not be 2 (remainder is 1) .. can not be 3 (remainder 1) ... can not be N, so it must be larger than N.
For example 2*3*5*7*11*13*17*19+1 (= 9 699 691 is NOT a prime but is divisible by 347, a prime larger than 19)
BTW, a number of the type 2*3*5*7...+() 1 , if prime is called primorial prime, largest known such numebr is 2*3*5*7....392113+1 which happens to be prime.
(of course other examples of primorial prime primes are 7 (=3*3+1), 31 (=2*3*5+1) and also 5 (=2*31) is a primorial prime prime too)
Re: BR Maths Corner1
May be of interest, from very readable prime pages from UTM' (same as Armen_T's post)
Euclid's Proof of the Infinitude of Primes (c. 300 BC)
It does mention:
BTW in above one could have used (2*3*5...1 ) in stead of "+1" and still the proof will remain the same.
****************************************************
(Sorry could not resist: One of the funniest/wise guy/show_off proof of this I have seen (Is like using a cannon to swat a fly) follows:
1. "Trivial" to prove that (6/pi^2) = (1(1/2)^2)(1(1/3)^2)(1(1/5)^2) (1(1/7)^2)...
(All prime numbers appear in RHS  and LHS=RHS= 1/(zeta(2)) (see Ramanujan/hardy's papers)
2. If the number of primes are finite, RHS would be rational
3. But we know (and "Trivial to prove ") that pi is irrational.
QED. )
Euclid's Proof of the Infinitude of Primes (c. 300 BC)
It does mention:
and actually puts original Euclid's proof in modern language:It is a common mistake to think that this proof says the product p1p2...pr+1 is prime. The proof actually only uses the fact that there is a prime dividing this product
Theorem.
There are more primes than found in any finite list of primes.
Proof.
Call the primes in our finite list p1, p2, ..., pr. Let P be any common multiple of these primes plus one (for example, P = p1p2...pr+1). Now P is either prime or it is not. If it is prime, then P is a prime that was not in our list. If P is not prime, then it is divisible by some prime, call it p. Notice p can not be any of p1, p2, ..., pr, otherwise p would divide 1, which is impossible. So this prime p is some prime that was not in our original list. Either way, the original list was incomplete.
BTW in above one could have used (2*3*5...1 ) in stead of "+1" and still the proof will remain the same.
****************************************************
(Sorry could not resist: One of the funniest/wise guy/show_off proof of this I have seen (Is like using a cannon to swat a fly) follows:
1. "Trivial" to prove that (6/pi^2) = (1(1/2)^2)(1(1/3)^2)(1(1/5)^2) (1(1/7)^2)...
(All prime numbers appear in RHS  and LHS=RHS= 1/(zeta(2)) (see Ramanujan/hardy's papers)
2. If the number of primes are finite, RHS would be rational
3. But we know (and "Trivial to prove ") that pi is irrational.
QED. )
Re: BR Maths Corner1
Recommend it highly. BTW Kerala school (eg Madhav's) work (arc tan series etc) have been discussed in this thread a few times. Some other works (Eg Brhamgupta's, Bhaskara's) have appeared here in problems or discussion form.
Re: BR Maths Corner1
ArmenT wrote:^^^
Pretty simple sir.
First, some basics:
1. We know that prime #s are only divisible by 1 and themselves (this is the definition of a prime #)
2. Any non primes are divisible by numbers other than 1 and themselves (inferred from step 1)
3. Therefore, any non primes can be decomposed into a product of prime #s. For instance 120 is (1 * 2 * 3 * 5 * 2 * 2). This can be inferred from 1 and recursively from 2.
Now for the proof:
I. Let us assume initially that there are indeed a finite # of primes.
II. Let N be the largest prime number.
III. Consider another number X, which is the product of all prime #s between 2 and N (i.e.) X = 2 * 3 * 5 * 7 * 11 * 13 *...* N
IV. Let Y = X + 1
V. Clearly Y is not divisible by the prime numbers 2, 3, 5, 7, 11, 13 ... N
VI. Therefore Y is prime. It is also larger than N. Note that we can repeat the same arguments from step II downwards using Y instead of N and keep going infinitely.
VII. Therefore our assumption in step I is incorrect and there are infinite # of primes.
Got it, thanks.

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Re: BR Maths Corner1
http://www.nature.com/nature/journal/v464/n7286/full/464165a.html
Genius who shuns the limelight
George Szpiro
BOOK REVIEWED
Perfect Rigor: A Genius and the Mathematical Breakthrough of the Century
by Masha Gessen
Genius who shuns the limelight
George Szpiro
BOOK REVIEWED
Perfect Rigor: A Genius and the Mathematical Breakthrough of the Century
by Masha Gessen
Mathematics rarely makes the news. One exception was the proof of the centuryold Poincaré conjecture in 2003. When Russian mathematician Grigory 'Grisha' Perelman posted his solution on the Internet, people took notice. Experts found that it was correct and Perelman was awarded the highest honour in maths, the Fields Medal, by the International Mathematical Union in 2006. Yet the laureate refused to attend the award ceremony in Madrid, leaving King Juan Carlos, who was to present the medal, waiting in vain. Media interest again soared: here was a longhaired weirdo who had flouted the King, thus confirming every prejudice about mathematicians. Tales of priority disputes and plagiarism quickly followed.
...
Science writer Masha Gessen has researched the reclusive mathematician thoroughly, interviewing his teachers, maths coaches and colleagues in the United States, Russia and Israel. She also consulted psychologists about his behaviour, which, she suggests, has much in common with Asperger's syndrome.
Born in St Petersburg (at that time Leningrad), Perelman became a maths prodigy. At 16 he won a gold medal at the 1982 International Mathematical Olympiad with a perfect score. After studies at Leningrad State University, the Steklov Institute of Mathematics in St Petersburg and a postdoc at the University of California, Berkeley, he vanished for several years. None of his colleagues knew it, but he was working in solitude on a proof of the Poincaré conjecture — a famous problem in topology about threedimensional boundaries of fourdimensional spheres. After reappearing briefly to explain his proof in a whirlwind US tour, he vanished again. His whereabouts today are unknown.
Gessen describes the best and the worst that Soviet maths had to offer; the best being the support given to gifted youths such as Perelman through maths clubs and competitions, the worst being the rampant antiSemitism that pervaded maths departments until perestroika in the late 1980s. Had it not been for the strenuous efforts of some professors to support Perelman, even a prodigy of his stature might have been prevented from pursuing graduate studies because of his Jewish background. Gessen, a selfdescribed 'maths junkie' brought up in Moscow, is ideally placed to tell of discriminatory practices in the Soviet academic world — her parents are Jewish engineers who suffered through such ordeals.
The book falls short in one important respect: like others who have tried, Gessen was unable to contact Perelman in person. Putting on a brave face, she writes that she was not burdened by any allegiance to Perelman's narrative. Unfortunately, lacking tangible facts, she nevertheless speculates on why he shunned both the medal and the public.
...
Gessen, by proffering gratuitous speculations, both misleads the reader and does Perelman grave injustice.
Until 2006, the Poincaré conjecture was one of the most famous open problems in maths; now it is one more theorem. For Perelman, proving the conjecture was sufficient reward in itself — no prize or recognition was needed. Perfect Rigor reminds us that it is journalists and the public who want more.
Re: BR Maths Corner1
AmberG ji,
Thanks a lot.The proof by contradiction has some meat in it after all.For a second,I was seduced by Armen's argument.Perhaps,it was an example like yours that motivated the proof.
Thanks a lot.The proof by contradiction has some meat in it after all.For a second,I was seduced by Armen's argument.Perhaps,it was an example like yours that motivated the proof.
Re: BR Maths Corner1
Krishnapremi, Biswas, ArmenT 
As Krisnapremi noticed, Armen's post has a slight hole in it namely in bolded part:
"
as Y need not be prime (But it will have, at least one factor larger than N.
One interesting and important problem, still open, (no one has published an answer yet  whoever does it first will make a name for herself) is: are there finite (or infinite) prime numbers of the type Y? (As I said in previous post they are called primorial primes  if one wants to google it)
(We know Y is prime for N=2,3,5,7,11 but NOT for 13,17,19,23,29 and (again prime for 31) etc)
For fun, if you like to try your hand at such proofs, here is a similar problem:
There are three classes of prime numbers:
class 1  even prime number(s) (Eg 2)
class 2  odd prime numbers of the form 4k1 (eg 3, 7, 11 etc...)
class 3  odd prime numbers of the form 4k+1 (eg 5,13, 17 etc...)
We know class 1 is finite (only even number is 2
How about class 2 , or class 3? can you prove your answer?
(Hint: Both are infinite, it is little easier (my opinion) to prove for class 2, than class 3. For both, one can use the same kind of argument as ArmenT/Euclid's but it will need a small refinement)
(Though it is, perhaps, fairy easy to look it up in wiki etc.. I will put a proof here if no one else does in next few days)
As Krisnapremi noticed, Armen's post has a slight hole in it namely in bolded part:
"
IV. Let Y = X + 1
V. Clearly Y is not divisible by the prime numbers 2, 3, 5, 7, 11, 13 ... N
VI. Therefore Y is prime. It is also larger than N. Note that we can repeat the same arguments from step II downwards using Y instead of N and keep going infinitely.
as Y need not be prime (But it will have, at least one factor larger than N.
One interesting and important problem, still open, (no one has published an answer yet  whoever does it first will make a name for herself) is: are there finite (or infinite) prime numbers of the type Y? (As I said in previous post they are called primorial primes  if one wants to google it)
(We know Y is prime for N=2,3,5,7,11 but NOT for 13,17,19,23,29 and (again prime for 31) etc)
For fun, if you like to try your hand at such proofs, here is a similar problem:
There are three classes of prime numbers:
class 1  even prime number(s) (Eg 2)
class 2  odd prime numbers of the form 4k1 (eg 3, 7, 11 etc...)
class 3  odd prime numbers of the form 4k+1 (eg 5,13, 17 etc...)
We know class 1 is finite (only even number is 2
How about class 2 , or class 3? can you prove your answer?
(Hint: Both are infinite, it is little easier (my opinion) to prove for class 2, than class 3. For both, one can use the same kind of argument as ArmenT/Euclid's but it will need a small refinement)
(Though it is, perhaps, fairy easy to look it up in wiki etc.. I will put a proof here if no one else does in next few days)
Re: BR Maths Corner1
abhishek_sharma wrote:http://www.nature.com/nature/journal/v464/n7286/full/464165a.html
Genius who shuns the limelight
wrt to prof. Grigory Perelman, met/attended a few of his lectures (in 90's ?  in SUNY before he went to Berkley  He again gave a workshop/lectures in 2003 at SUNY) He declined offers to Princeton etc go back to Russia only to abandon their institutes also.
one funny thing, there was a controversy about Fields medal, because you have to be younger than 40 to win it, and at the time of announcement there were questions about his age, no one knew his "actual" birth date for sure.
BTW, his younger sister is also a good Mathematician.
Re: BR Maths Corner1
Thanks for the moderate enlightenment . I was quoting the proof from memory as how I learned it, so it is always good to be enlightened.

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Re: BR Maths Corner1
Amber G. wrote:wrt to prof. Grigory Perelman, met/attended a few of his lectures (in 90's ?  in SUNY before he went to Berkley  He again gave a workshop/lectures in 2003 at SUNY) He declined offers to Princeton etc go back to Russia only to abandon their institutes also.
one funny thing, there was a controversy about Fields medal, because you have to be younger than 40 to win it, and at the time of announcement there were questions about his age, no one knew his "actual" birth date for sure.
BTW, his younger sister is also a good Mathematician.
Senior members might be aware of this article on Grigory Perelman:
http://www.newyorker.com/archive/2006/08/28/060828fa_fact2
Re: BR Maths Corner1
Thanks for New Yorker Article.
Last birthday my son gave me Donal O'shea's book "The Poincare's conjecture" The book is for general audience and is very readable. Recommend it.
Last birthday my son gave me Donal O'shea's book "The Poincare's conjecture" The book is for general audience and is very readable. Recommend it.
Re: BR Maths Corner1
If I remember correctly, the New Yorker article caused some severe H&D loss for a Chinese professor. There was some big controversy when it was published.
AmberG, I think I've solved the case for 4k  1. Took me a bit of working out, but I think I've got it:
1. Say there are a finite number of primes of type 4k  1
2. Let's say N is the largest number of type 4k  1
3. Let's say X = 4 * (3 * 7 * ... * N)  1
4. Now X can't be divisible by 3, 7 ... N since they'll all leave reminder of 1. So if there are any divisors of type (4k  1), then they must be > N, which means that assumption in step 2 is wrong and N is not the largest prime of type (4k  1) and therefore assumption in step 1 is incorrect.
5. There may be a chance that X is divisible by a number of type (4k + 1). We need to disprove this bit (and this is what took me a little while to work out)
6. Clearly if there are any prime factors of X, they need to all be of type (4k + 1). If there are any factors of type (4k  1) they're covered by our conclusion on step 4. Let's assume there are two prime numbers (4a + 1) and (4b + 1) which divide X initially.
7. (4a + 1) * (4b + 1) => 16ab + 4a + 4b + 1 => 4 (4ab + a + b) + 1
8. Note that the above is of the format 4Y + 1 as well, where Y = (4ab + a + b). Therefore, we conclude that multiplying numbers of format (4a + 1)(4b + 1) will produce a number of format (4k + 1) as well. It follows from this that product of (4a + 1)(4b + 1)(4c + 1)... will also produce a number of type (4k + 1)
9. This proves that X cannot be divisible by a number of format (4k + 1) because a number cannot be of type (4k  1) AND be divisible by (4k + 1) at the same time. This means that assumption of step 5 is incorrect as well.
10. Therefore assumption in step 1 is incorrect and there are infinite primes of type (4k  1)
Please feel free to apply the moderately enlightening clue stick if I've missed something.
AmberG, I think I've solved the case for 4k  1. Took me a bit of working out, but I think I've got it:
1. Say there are a finite number of primes of type 4k  1
2. Let's say N is the largest number of type 4k  1
3. Let's say X = 4 * (3 * 7 * ... * N)  1
4. Now X can't be divisible by 3, 7 ... N since they'll all leave reminder of 1. So if there are any divisors of type (4k  1), then they must be > N, which means that assumption in step 2 is wrong and N is not the largest prime of type (4k  1) and therefore assumption in step 1 is incorrect.
5. There may be a chance that X is divisible by a number of type (4k + 1). We need to disprove this bit (and this is what took me a little while to work out)
6. Clearly if there are any prime factors of X, they need to all be of type (4k + 1). If there are any factors of type (4k  1) they're covered by our conclusion on step 4. Let's assume there are two prime numbers (4a + 1) and (4b + 1) which divide X initially.
7. (4a + 1) * (4b + 1) => 16ab + 4a + 4b + 1 => 4 (4ab + a + b) + 1
8. Note that the above is of the format 4Y + 1 as well, where Y = (4ab + a + b). Therefore, we conclude that multiplying numbers of format (4a + 1)(4b + 1) will produce a number of format (4k + 1) as well. It follows from this that product of (4a + 1)(4b + 1)(4c + 1)... will also produce a number of type (4k + 1)
9. This proves that X cannot be divisible by a number of format (4k + 1) because a number cannot be of type (4k  1) AND be divisible by (4k + 1) at the same time. This means that assumption of step 5 is incorrect as well.
10. Therefore assumption in step 1 is incorrect and there are infinite primes of type (4k  1)
Please feel free to apply the moderately enlightening clue stick if I've missed something.
Re: BR Maths Corner1
ArmenT wrote:Please feel free to apply the moderately enlightening clue stick if I've missed something.
..The proof is valid, and nice, but since you asked I may rewrite this part
because a number cannot be of type (4k  1) AND be divisible by (4k + 1) at the same time.
Take the number 15 (of the type 4k1) and is divisible by 5.
(Replace the above with "If a number is (4k1) type, it has at least one prime factor of (4k1) type)" Of course,
the proof you gave, still remains valid as nothing changes.
(Also, keeping the same steps, you could have also selected what you called Y in your earlier post ("Y" (=X+1) and X was product of 2*3*5....N) ... Y is of the form 4k1, and as you said, must have a factor (at least one) of the type 4k1 which is not in the range of (2...N) )
BTW, there is a beautiful theorem (most likely first mentioned by Madhava) : For any number (positive integer) if p is number of factors of the type (4k+1), and q is number of factors of type (4k1) then 4*(pq) = number of ways that number can be written as sum of two squares (of integers).
For example, take 45, p=4 (1,5,9,45 are factors) and q=2 (3,15 are other type) and there are 8 ways 45 can be written as sum of 2 squares (6^2+3^2) or (3^2+6^2) and each number can be plus or minus (eg (6)^2 + (3^2)) etc,
(The above theorem is little known, it is not very hard, not too easy either  to prove, if one has taken a course in number theory)
Re: BR Maths Corner1
To day is Pi day... See the google logo!
(3.14 It's also happens to be Einstein's birthday )

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Re: BR Maths Corner1
Here is outline/hint for case of class3 in the above problem:
We as usual take X=2*3*5*.......N (where N is the largest prime known etc)
For case 2  As ArmenT had it, one could have used Y=X+1 (see posts above)
(Y is of the form 4k1 and must have at least one factor of the form 4k1) etc...
For case 3, use Hint: Z=X^2+1, all (odd) factors of Z have to be of the form (4k+1) (why?)
(Rest is easy)
class 2  odd prime numbers of the form 4k1 (eg 3, 7, 11 etc...)
class 3  odd prime numbers of the form 4k+1 (eg 5,13, 17 etc...)
We know class 1 is finite (only even number is 2
How about class 2 , or class 3? can you prove your answer?
We as usual take X=2*3*5*.......N (where N is the largest prime known etc)
For case 2  As ArmenT had it, one could have used Y=X+1 (see posts above)
(Y is of the form 4k1 and must have at least one factor of the form 4k1) etc...
For case 3, use Hint: Z=X^2+1, all (odd) factors of Z have to be of the form (4k+1) (why?)
(Rest is easy)

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Re: BR Maths Corner1
wrt to noneuclidean geometry I put a detail post in Physics thread, when someone asked a question about "how universe has a finite volume  yet has no boundary" Seems like the post disappeared in the forum glitch.. hopefully it was read before it got lost.
BTW: From my earlier post:
Any easy proof to validate that all factors of Z are indeed of the form (4k+1) ?
BTW: From my earlier post:
For case 3, use Hint: Z=X^2+1, all (odd) factors of Z have to be of the form (4k+1) (why?)
(Rest is easy)
Any easy proof to validate that all factors of Z are indeed of the form (4k+1) ?
Re: BR Maths Corner1
joshvajohn wrote:India a world mathematics power, says professor Raghunathan
http://www.hindu.com/2010/04/01/stories ... 251200.htm
Rumors is that Kiran Kedlaya is in for Fields Medal. (Guys you heard it here first!). He is under 40, and is an invited speaker. People may recall I had talked about him in our math thread. He has won many IMO medals ... for those who don't know him, he is at present a prof at MIT.
(Of course, It is a guarded secret, and no one is suppose to know till the announcement is made)
Other I think may be Ngo or may be Lurie or Manjul Bhargava.. Lets us wait and see.
Re: BR Maths Corner1
Amber G. wrote:Here is outline/hint for case of class3 in the above problem:class 2  odd prime numbers of the form 4k1 (eg 3, 7, 11 etc...)
class 3  odd prime numbers of the form 4k+1 (eg 5,13, 17 etc...)
We know class 1 is finite (only even number is 2
How about class 2 , or class 3? can you prove your answer?
We as usual take X=2*3*5*.......N (where N is the largest prime known etc)
For case 2  As ArmenT had it, one could have used Y=X+1 (see posts above)
(Y is of the form 4k1 and must have at least one factor of the form 4k1) etc...
For case 3, use Hint: Z=X^2+1, all (odd) factors of Z have to be of the form (4k+1) (why?)
(Rest is easy)
I decided to get off my lazy behind and take a crack at this. Well the real reason was that I came back grumpy after a hard day at work and I figured it would take my mind off of work. Here's my attempt:
1. Let N be the largest known prime of type (4k + 1)
2. Let X = (5 * 13 * 17 ... N)
3. Let Y = 2X and Z = Y^2 + 1. Therefore Z = 4X^2 + 1
4. Clearly Z is not divisible by 5, 13, 17 .... N since they'll all return a reminder of 1
5. So either Z is prime or divisible by a number of (4k  1) or divisible by a number of type (4K + 1) which is greater than N
6. Now in the previous proof, I'd shown that multiplying numbers of the form (4a + 1) * (4b + 1) returns a number of type (4k + 1). Since Z is in the same format, it must either be prime or divisible by numbers of type (4k + 1) which is not in the set (5, 13, 17... N). Either way, this means that there is a prime of type (4k +1 ) which is > N, which means the assumption in step 1 is incorrect and there are infinite # of primes of type 4k + 1
As before, please be gentle with applying the moderately enlightening clue stick.
Re: BR Maths Corner1
^^^ Hello. Something is not clear to me.
say a number is like 21 (or something larger); of the form (4k+1), it does not have *any* factors of the form (4k+1).
(factors are 3 and 7, none of them are of the form 4k+1)
(So, what if *all* the factors of your z are of the form (4k1), and have *no* factors in the range of (5,13,17..N))?
(Difficulty is, while in the case of factors of (4k1), at least one factor has to be of the form of (4k1), while factors of (4k+1) you may have a case where there are no factors are of the form 4k+1).. all you can say is there are even number of factors of the form (4k1))
say a number is like 21 (or something larger); of the form (4k+1), it does not have *any* factors of the form (4k+1).
(factors are 3 and 7, none of them are of the form 4k+1)
(So, what if *all* the factors of your z are of the form (4k1), and have *no* factors in the range of (5,13,17..N))?
(Difficulty is, while in the case of factors of (4k1), at least one factor has to be of the form of (4k1), while factors of (4k+1) you may have a case where there are no factors are of the form 4k+1).. all you can say is there are even number of factors of the form (4k1))
Re: BR Maths Corner1
Here is an outline of proof for the case 3.
It follows the earlier problem posted, about cycle length of (1/p) and numbers like '111111..'
Please see earlier posts few posts above and below this link:
link
we use the fact: if p is prime (>5) , then if the cycle length of decimal representation of 1/p is m then
10^m1 is divisible by p. (or vice versa )
and cycle length m divides (p1) (Eq 2)
Now: we take Z=X^2+1, (where X=2.3.5.7......N .. etc)
Let p is a factor of Z.
Write (1/p) not in decimal but in the base of X
(We know, p divides X^2+1 , so p also divides (X^+1)(X^21)=X^41
so cycle length of (1/p) is 4.
but this should divide (p1) (eq 2) or (p1) is divisible by 4
IOW, p has to be of the form 4k+1
QED.
(One can check this out by trying various values  all (odd) factors of (x^2+1) are always of the form (4k+1) ...)
It follows the earlier problem posted, about cycle length of (1/p) and numbers like '111111..'
Please see earlier posts few posts above and below this link:
link
we use the fact: if p is prime (>5) , then if the cycle length of decimal representation of 1/p is m then
10^m1 is divisible by p. (or vice versa )
and cycle length m divides (p1) (Eq 2)
Now: we take Z=X^2+1, (where X=2.3.5.7......N .. etc)
Let p is a factor of Z.
Write (1/p) not in decimal but in the base of X
(We know, p divides X^2+1 , so p also divides (X^+1)(X^21)=X^41
so cycle length of (1/p) is 4.
but this should divide (p1) (eq 2) or (p1) is divisible by 4
IOW, p has to be of the form 4k+1
QED.
(One can check this out by trying various values  all (odd) factors of (x^2+1) are always of the form (4k+1) ...)

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Re: BR Maths Corner1
Take It to the Limit
By STEVEN STROGATZ
http://opinionator.blogs.nytimes.com/2010/04/04/takeittothelimit/
By STEVEN STROGATZ
http://opinionator.blogs.nytimes.com/2010/04/04/takeittothelimit/
Re: BR Maths Corner1
can anyone post a link to material on 'Vedic Mathematics', my physics teacher takes only seconds to compute huge calculations mentally, he says its Vedic Maths, although my exam is in a week so it wont help me as of now but later on it could be of good use
Re: BR Maths Corner1
NehraA wrote:can anyone post a link to material on 'Vedic Mathematics', my physics teacher takes only seconds to compute huge calculations mentally, he says its Vedic Maths, although my exam is in a week so it wont help me as of now but later on it could be of good use
Have a look at this
www.tifr.res.in/~vahia/danivmsm.pdf
Re: BR Maths Corner1
abhishek beat me to this.
steven strogatz series.
http://opinionator.blogs.nytimes.com/au ... strogatz/
steven strogatz series.
http://opinionator.blogs.nytimes.com/au ... strogatz/
Re: BR Maths Corner1
Just curious, if any BRFite (or their kids etc) participated in US Math Olympiad or Indian Math Olympiad.
Just for fun, the following is a problem from India's INmo  try it if you like it (HIgh school, not too hard)
Given x,y,z are real, find all possible values of x,y and z if both the following eq. are satisfied.
(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)=xyz
and (x^4+x^2y^2+y^4) (y^4+y^2z^2+z^4)(z^4+z^2x^2+x^4) = (xyz)^3
Just for fun, the following is a problem from India's INmo  try it if you like it (HIgh school, not too hard)
Given x,y,z are real, find all possible values of x,y and z if both the following eq. are satisfied.
(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)=xyz
and (x^4+x^2y^2+y^4) (y^4+y^2z^2+z^4)(z^4+z^2x^2+x^4) = (xyz)^3
Last edited by Amber G. on 16 May 2010 21:33, edited 1 time in total.
Re: BR Maths Corner1
^^^^
I'm kinda wiped out right now, but here's a stab at the first one
Given:
(x^2 + xy + y^2)(y^2+yz+z^2)(z^2+zx+x^2) = xyz
We can rewrite the left hand side a bit:
x(x + y + y^2/x) y(y + z + z^2/y) z(z + x + x^2/z) = xyz
and rearrange the terms like this:
xyz (x + y + y^2/x)(y + z + z^2/y)(z + x + x^2/z) = xyz
The trivial solution is x, y or z, or any two, or all three are 0
Another possibility is that: (x + y + y^2/x)(y + z + z^2/y)(z + x + x^2/z) = 1
and that's where I'm not going to attempt any further at this moment
I'm kinda wiped out right now, but here's a stab at the first one
Given:
(x^2 + xy + y^2)(y^2+yz+z^2)(z^2+zx+x^2) = xyz
We can rewrite the left hand side a bit:
x(x + y + y^2/x) y(y + z + z^2/y) z(z + x + x^2/z) = xyz
and rearrange the terms like this:
xyz (x + y + y^2/x)(y + z + z^2/y)(z + x + x^2/z) = xyz
The trivial solution is x, y or z, or any two, or all three are 0
Another possibility is that: (x + y + y^2/x)(y + z + z^2/y)(z + x + x^2/z) = 1
and that's where I'm not going to attempt any further at this moment
Re: BR Maths Corner1
To make it more clear: (Above post edited)  (find x,y,z which satisfy both the equations)

 BRF Oldie
 Posts: 9664
 Joined: 19 Nov 2009 03:27
Re: BR Maths Corner1
Reflections Around the Ramanujan Centenary
By Atle Selberg (Fields Medal Winner)
http://www.ias.ac.in/resonance/Dec1996/pdf/Dec1996Reflections.pdf
By Atle Selberg (Fields Medal Winner)
http://www.ias.ac.in/resonance/Dec1996/pdf/Dec1996Reflections.pdf

 BRF Oldie
 Posts: 9664
 Joined: 19 Nov 2009 03:27
Re: BR Maths Corner1
THE LORD OF THE NUMBERS, ATLE SELBERG. ON HIS LIFE AND MATHEMATICS
http://www.ams.org/journals/bull/20084504/S0273097908012238/S0273097908012238.pdf
http://www.ams.org/journals/bull/20084504/S0273097908012238/S0273097908012238.pdf

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Re: BR Maths Corner1
Now, suppose you were the intelligence officer analyzing the after effects of the battle of Assal Uttar and you have seen the 10 (say) M48 Paki tanks knocked out in the battle , and the 10 tanks have serial numbers 12,40,25, 18,67,7,43,55,83 , what was the likely total number of tanks supplied to the Pakis by the Americans (assuming that Pakis made No M48s and this was the first time M48s in Paki inventory were used and got knocked out).

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Re: BR Maths Corner1
Amber G. wrote:Just curious, if any BRFite (or their kids etc) participated in US Math Olympiad or Indian Math Olympiad.
Just for fun, the following is a problem from India's INmo  try it if you like it (HIgh school, not too hard)
Given x,y,z are real, find all possible values of x,y and z if both the following eq. are satisfied.
(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)=xyz
and (x^4+x^2y^2+y^4) (y^4+y^2z^2+z^4)(z^4+z^2x^2+x^4) = (xyz)^3
How about x=y=z = 1/3 ? I think that is the only non trivial answer.
I got that since (x^2+xy+y^2) * ((x^2  xy+y^2) = (x^4+x^2y^2+y^4) , if you divide the 2nd eqn by the 1st, you basically get ((y/x  1/2)^2 + 3/4) * (((z/y  1/2)^2 + 3/4)*((x/z  1/2)^2 + 3/4) = 1
Now, (a2 +3/4) ((b2 +3/4) (c2+3/4) = 1 ==> least value for a,b,c = 1/4 each .. So if you solve (y/x  1/2)^2 = 1/4, you get x=y (for non trivial value), similarly y=z and x=z . Plugging in ,you get value of 1/3 .
Cant figure out how to prove that these are the only real values though (I think they are, given the constraints).
Re: BR Maths Corner1
vina wrote:Now, suppose you were the intelligence officer analyzing the after effects of the battle of Assal Uttar and you have seen the 10 (say) M48 Paki tanks knocked out in the battle , and the 10 tanks have serial numbers 12,40,25, 18,67,7,43,55,83 , what was the likely total number of tanks supplied to the Pakis by the Americans (assuming that Pakis made No M48s and this was the first time M48s in Paki inventory were used and got knocked out).
I would write a field report saying the 10th serial has been erased and hence would need extra time and budget to estimate and the go off on a vacation to panama.
** you gave only 9 serials
well i dunno the answer. But I was once asked a similar question in an interview
In a party , everyone was given a serial number at the gate. Later you randomly pick one guy and his serial number is 47 . so whats the estimate of total number of ppl in the party. Ofcourse i could not answer and did not get the job
****spoiler below *****
the solution to this question was apparently actually used to estimate german tank numbers
http://www.guardian.co.uk/world/2006/ju ... tvandradio
nowadays i find myself solving more and more of my problems (at work etc ) using google and less using my brains.Like if i get compilation error i usually paste the error in google and read remedies before even lookint at my code. Hell at this stage i would even hire a candidate if he replies "i will google it when i need it".
Re: BR Maths Corner1
I have one questions which again I was asked in the same interview as above , I think i answered the first part correctly but the second part I have no clue. I am answering the first part myself and gurus please correct me.
In a fair there is a game , you roll a dice and whatever the dice shows you get that many Rs . say you throw and get a 5 , you win 5 Rs. So how much it is worth for you to pay to play such a game.
here's my answer .. I figured over a long series of throws, you will win what would be the expected value i.e ( 1+2+3+4+5+6)/6 = 3.5 , so i answered that anything less than 3.5 rs would be okay to pay
now the second part of the question was
if say you are allowed a 2nd throw in case you wanted to change (though you could not revert back to the first value if you get a lesser score 2nd time) how much would you be willing to pay to play such a game .
This part bowled my tiny brain and I gave up ...
In a fair there is a game , you roll a dice and whatever the dice shows you get that many Rs . say you throw and get a 5 , you win 5 Rs. So how much it is worth for you to pay to play such a game.
here's my answer .. I figured over a long series of throws, you will win what would be the expected value i.e ( 1+2+3+4+5+6)/6 = 3.5 , so i answered that anything less than 3.5 rs would be okay to pay
now the second part of the question was
if say you are allowed a 2nd throw in case you wanted to change (though you could not revert back to the first value if you get a lesser score 2nd time) how much would you be willing to pay to play such a game .
This part bowled my tiny brain and I gave up ...

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Re: BR Maths Corner1
^^^
Part one seems fine i.e. use of expected value of 3.5
For the part 2, imho there is no one answer/. I would rather approach this as 6 different scenarios arising out of the 6 different results of the first throw. Say, you throw "1" in first attempt, then for the 2nd throw, you have a prob p=1 of getting atleast what you got in 1st throw & a prob. p=5/6 of getting more (i.e. throwing 2 or more in 2nd throw)
Throw1  Prob. of getting more in throw 2
1  5/6
2  4/6
3  3/6
4  2/6
5  1/6
6  0
For each of these scenarios, there will be different answers.
Lets consider that you got "6" in 1st throw, so there is no way of bettering it in 2nd throw, hence, the 2nd throw has no value. For a score of "5" in 1st throw, you have a probability of 1/6 of bettering your score in 2nd throw (i.e. by throwing a 6), hence the 2nd throw is worth 6*1/6 = Re 1.
Part one seems fine i.e. use of expected value of 3.5
For the part 2, imho there is no one answer/. I would rather approach this as 6 different scenarios arising out of the 6 different results of the first throw. Say, you throw "1" in first attempt, then for the 2nd throw, you have a prob p=1 of getting atleast what you got in 1st throw & a prob. p=5/6 of getting more (i.e. throwing 2 or more in 2nd throw)
Throw1  Prob. of getting more in throw 2
1  5/6
2  4/6
3  3/6
4  2/6
5  1/6
6  0
For each of these scenarios, there will be different answers.
Lets consider that you got "6" in 1st throw, so there is no way of bettering it in 2nd throw, hence, the 2nd throw has no value. For a score of "5" in 1st throw, you have a probability of 1/6 of bettering your score in 2nd throw (i.e. by throwing a 6), hence the 2nd throw is worth 6*1/6 = Re 1.

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Re: BR Maths Corner1
chilarai wrote:In a fair there is a game , you roll a dice and whatever the dice shows you get that many Rs . say you throw and get a 5 , you win 5 Rs. So how much it is worth for you to pay to play such a game.
here's my answer .. I figured over a long series of throws, you will win what would be the expected value i.e ( 1+2+3+4+5+6)/6 = 3.5 , so i answered that anything less than 3.5 rs would be okay to pay
I have to disagree. This is standard option pricing math. In fact, there is a trick question, if suppose a stock exhibits 5050 chance of going up 1% every one minute or down 1% every minute are you in the long run going to be making money, losing money or remain flat ?. (I will give the answer at the end)
The trick part of your question is basically your returns are binomially distributed based on which die turns up from (1 to 6) and for large numbers approximates to normally distributed. So if that is the case, your returns are going to be less than what you calculated by Expected Value of Mean (here you calculated as 3.5). How much will that be ?. From basic math we know Geometric Mean is LESS than Arithmetic mean. It will be actually less by half of variance .. (2.9166/2) ie 1.458. The actual worth for which you should be playing is 3.5  1.458 = 2.04 Rs! .
The moral of the story is volatility (which is a measure of risk) ALWAYS reduces your long term return. That is why you have risk management.
The answer to my question is you will lose money consistently , by how much by 0.005% to be exact.
See, not bad eh, there is some use of YumBeeYeas and Phynance after all see ?

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Re: BR Maths Corner1
if say you are allowed a 2nd throw in case you wanted to change (though you could not revert back to the first value if you get a lesser score 2nd time) how much would you be willing to pay to play such a game .
Now, what you are doing here is modifying the variability of payoff by introducing an option. So, lets see, you have a 50% prob that it is > 3 and 50% less than 3. So if the value of the die is 1 , 2 or 3, you would throw again and if it is above 3, go with what the die gives you. So you basically need to calculate the conditional payoffs now for 1,2 and 3 and throwing again , while calculating the variance . The math is not too difficult and it is quite easy to do in 10 mins max with a piece of paper, so I dont give a quantitative answer, but will give a qualitative one.
The answer is the option has VALUE . So the price you will be willing to pay will be higher than what you did in the first case. It will be higher than in the first case, but will still be lower than the 3.5 expected mean value you calculated (there is still volatility after all).
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