## BR Maths Corner-1

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Eric Demopheles
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### Re: BR Maths Corner-1

kasthuri wrote:
Of late my interest in logic is increasing, especially on computability logic. I guess there are not many good books out there for meta-math and other logics. I don't want basic logic books, which I can find. I need little bit advanced. Do you have any recommendations?

This should do:

Jon Barwise: "Handbook of Mathematical Logic"

It is a general reference.

Another one, if you by "computability logic" you mean you are interested in the work of Kurt Gödel and others:

Jean Heijenoort: "From Frege to Gödel".

Amber G.
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### Re: BR Maths Corner-1

Eric Demopheles wrote:An achievement of a Chinese-born mathematician in remarkable circumstances:

https://www.simonsfoundation.org/quanta/20130519-unheralded-mathematician-bridges-the-prime-gap/

Thanks. He got 2014 Frank Nelson Cole Prize in Number Theory for that.
Also in addition to above, there are NYTimes and other popular articles.. which may be of interest..eg.
http://www.nytimes.com/2013/05/21/science/solving-a-riddle-of-primes.html

Eric Demopheles
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### Re: BR Maths Corner-1

Amber G. wrote:He got 2014 Frank Nelson Cole Prize in Number Theory for that.

Ah, yes. He got the Cole prize. For me the circumstances of the discovery was the most striking, more than the discovery itself.

In this connection the following essay, on 'Two cultures of mathematics', is of interest:

https://www.dpmms.cam.ac.uk/~wtg10/2cultures.pdf

Eric Demopheles
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### Re: BR Maths Corner-1

Speaking of primes, there was a nice proof rather recently in 2005 that there are infinitely many of them. This is understandable for everyone:

http://primes.utm.edu/notes/proofs/infinite/Saidak.html

The original proof is of course cited to Euclid:

http://primes.utm.edu/notes/proofs/infi ... clids.html

I just love the modification Kummer made to it:

http://primes.utm.edu/notes/proofs/infi ... mmers.html

Let me hope that the readers of this thread get some experience in reading actual mathematical proofs in "pure" mathematics, and not for the sake of applications, from this simple example.

Vayutuvan
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### Re: BR Maths Corner-1

Eric Demopheles ji: Thanks. Both Saidak's and Kummer's proofs are exquisite. Made my day.

Amber G.
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### Re: BR Maths Corner-1

Thanks ^^^ Along the same lines another cool proof is just use Fermat's numbers (2^(2^n)+1)..
For example 3,5,17,257 ... (These are 2^1+1, 2^2+1, 2^4+1, 2^8+1 etc..)

And just like above argument, each of these numbers have no common factors and hence must have different prime divisors.
(So each such number will have at least one new prime as a factor)

(Easy to see that F_n = F_1*F_2*F_3*... F_(n-1) + 2)
(Eg 257 = 3*5*17+2 ,.. hence any factor of 257 can not be a factor of 3 or 5 or 7)

BTW, yest the prime pages (Link given above by Eric Demopheles ji) is quite cool, and this link has been given quite a few times before in this dhaga... there are many neat facts/proofs etc are in those pages which are fun to read even for non-experts.

Eric Demopheles
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### Re: BR Maths Corner-1

matrimc-ji, Amber G.ji, Thanks for your kind words. I did not know that this link was given earlier; sorry about the repetition.

Now, something a little more substantive(analytical): There are more primes than there are perfect squares. In fact, in a meaning that can be expressed more precisely, there are many, many, many more primes than there are perfect squares.

The way to make this precise is through the following two lemmas:

Lemma 1: The harmonic series diverges.

That is to say, the series 1 + 1/2 + 1/3 + 1/4 + 1/5 + ....... diverges.

Proof:

Easiest way to see is to visualize this sum as a triangles of heights 1, 1/2, 1/3, .... placed on the x-axis, next to next, of width 1, and this can be bound from below by the integral of (1/x )dx from 1 to infinity.

This proof with illustration and more details, and also another proof using the comparison test, are available at:

http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Divergence.

Lemma 2: The series 1 + 1/4 + 1/9 + 1/16 + 1/25 + ..... , i.e., the sum of reciprocals of the perfect squares, converge.

Proof:
http://en.wikipedia.org/wiki/Basel_problem#The_Riemann_zeta_function

So, now we have enough background to make our theorem interesting:

Theorem: The sums of reciprocals of primes diverges.

Proof:
http://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes#Proofs

___________________________________________________

It is in this sense that there are many, many, many more primes than there are perfect squares. Please note that this whole line of thought is related to the Riemann Zeta function which in fact came from a "proof" of Euler that there are infinitely many primes. Maybe let it be for another day. Btw, all this math typing in plain ASCII is bit restricting. Is there a possibility for using something like jsMath or MathJax in the forum(a long shot; still just noting the possibility).

Amber G.
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### Re: BR Maths Corner-1

Now, something a little more substantive(analytical): There are more primes than there are perfect squares. In fact, in a meaning that can be expressed more precisely, there are many, many, many more primes than there are perfect squares.

Legendre's Conjecture (One of Landau's problem) says that there is at least 1 prime between n^2 and (n+1)^2. (It is still a famous unsolved problem but most think that it is true).. In that case obviously there are more primes than perfect square. (In fact on average there are as many primes between n^2 and (n+1)^2 as there are primes between 1 and n so this gives some order of magnitude ratio)

Another very interesting proven theorem is (Bertand's - if you want to check out wiki) is there is always a prime between n and 2n.

...

Theorem: The sums of reciprocals of primes diverges.

This is extremely slow diverging series. Even if one takes number of terms equal to all the elementary particles in the whole universe., the sum is still less than 6, but eventually it will reach infinity.....

By the way, Mertens constant, related to this shot to fame when a some guy was trying to calculate using a Intel processor they found a bug (Pentium FDIV bug) in the processor chip ..it was a very big news because millions of PC used this chip and no one ever doubted that chip will give a error. (Interested people can see, for example: http://en.wikipedia.org/wiki/Pentium_FDIV_bug

Riemann Zeta function which in fact came from a "proof" of Euler that there are infinitely many primes. Maybe let it be for another day.

You may enjoy in this dhaga ..http://forums.bharat-rakshak.com/viewtopic.php?p=1289492#p1289492
or..this http://forums.bharat-rakshak.com/viewtopic.php?p=1292292#p1292292
Where I gave an interesting "show off proof" that there are infinite primes.
Whis is ..http://forums.bharat-rakshak.com/viewtopic.php?f=2&t=4201&p=1289773&hilit=zeta#p1289773

Btw, all this math typing in plain ASCII is bit restricting. Is there a possibility for using something like jsMath or MathJax in the forum(a long shot; still just noting the possibility).

May be brf moderators can provide a simple LaTex interface...

Vayutuvan
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### Re: BR Maths Corner-1

Yes to LaTex and also display of simple line graphics.

Eric Demopheles
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### Re: BR Maths Corner-1

matrimc wrote:Yes to LaTex and also display of simple line graphics.

If I understand right, this forum is based on phpBB. Maybe this is helpful:

https://www.phpbb.com/community/viewtop ... &t=2133985

Amber G.ji,

Thanks for the links. It seems I missed a lot of pages of very interesting discussion on primes on this thread.

What I was aiming here was to make a slow buildup to the statement of the magnificent prime number theorem, with the proper background for appreciating it.

Amber G. wrote:This is extremely slow diverging series. Even if one takes number of terms equal to all the elementary particles in the whole universe., the sum is still less than 6, but eventually it will reach infinity.....

Yes, it is very slow.

Here, note that the summation of 1/n is itself already very slow to diverge. I know a chap who didn't believe it when he heard, and was bothered enough to write a program and let it run for many days on his computer. Needless to say, the sum is slow growing in the order of O(log n), and as days passed, he was less and less convinced. [:D]

But yes, the sum of reciprocals of primes is even slower. A known lower bound for it involves terms like log log(n).

Still, the divergence is a significant fact when confronted with the convergence of the sum of 1/n^2 .. Or indeed the sum of 1/n^a where a is any real number >1. This can be seen by the comparison test, by comparing with an appropriate geometric series.

ArmenT
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### Re: BR Maths Corner-1

Eric Demopheles wrote:
Nilesh Oak wrote:Do we any evidence for existence, usage or inkling of complex numbers in ancient Indian mathematics/astronomy/trigonometry..etc., anytime before say 1500 CE ?

Inkling. Ah. Good word.
<snipping stuff out for brevity>

Actually, from a philosophical viewpoint, the passage from real numbers to complex numbers is an easier leap(assuming the learner is well-versed in algebra) than the passage from natural numbers to negative numbers, because for the latter, the enormously important(philosophically important) notion of zero is first necessary. That step, although very simple, took a long long time for civilization to arrive at.

Hindsight is always 20-20. In modern times, the concept of complex numbers may be perfectly logical, but in ancient times, it was not always so. In fact, Bhaskaracharya wrote in his book that a square root of a negative number was absurd and a later commentator of his works presented a proof that a negative number cannot have a square root.
http://forums.bharat-rakshak.com/viewtopic.php?p=1492065#p1492065

Of course, in those days, they didn't encounter any sort of calculation where the concept of imaginary numbers were required. Hence, authors of that era simply declared that there cannot be a square root of a negative number and left it at that.

Eric Demopheles
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### Re: BR Maths Corner-1

ArmenT wrote:Hindsight is always 20-20. In modern times, the concept of complex numbers may be perfectly logical, but in ancient times, it was not always so. In fact, Bhaskaracharya wrote in his book that a square root of a negative number was absurd and a later commentator of his works presented a proof that a negative number cannot have a square root.
http://forums.bharat-rakshak.com/viewtopic.php?p=1492065#p1492065

Of course, in those days, they didn't encounter any sort of calculation where the concept of imaginary numbers were required. Hence, authors of that era simply declared that there cannot be a square root of a negative number and left it at that.

That is true. There is no notion of "complex numbers" explicitly at the time of Bhaskaracharya. Indeed complex numbers were given respectability only by the time of the work of Gauss.

But philosophically it could be argued that the ground was already prepared to some extent. After all complex numbers is a construct created to meet the needs of algebra, to always solve polynomial algebraic equations. The preliminaries of solving such equations were already there. That is what I meant. Here I took advantage of the OP's use of the word 'inkling', to push through something that I must concede that literally speaking is not present in any way at all.

Eric Demopheles
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### Re: BR Maths Corner-1

Amber G. wrote:Legendre's Conjecture (One of Landau's problem) says that there is at least 1 prime between n^2 and (n+1)^2.

Legendre is a great arithmetician. As his brilliance was overshadowed in the presence of the great Gauss, people do not know him much.

Let me here make a note to the Quadratic Reciprocity Law regarding congruences of prime numbers that was conjectured and half-proved by him. This is arguably the golden theorem of number theory.

This is truly an *arithmetical* theorem, compared to the prime number theorem which incorporates some mathematical analysis too. Proofs and deeper elaborations have occupied arithmeticians for centuries; e.g., class field theory came up in this route.

SaiK
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### Re: BR Maths Corner-1

matrimc wrote:Yes to LaTex and also display of simple line graphics.

matrimc garu, the other way around would be use symbols that ye-all speaketh - just write the latex as code... of course readabiliy issues, but one might strengthen latex understanding if that may be the intended side-effect.

example:

Code: Select all

    E_0 &= mc^2                              \\    E &= \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}

ref: wiki for above snippet

Amber G.
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### Re: BR Maths Corner-1

But few may like to see the above not as:

Code: Select all

[code]    E_0 &= mc^2                              \\    E &= \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}[/code]

But as:

Amber G.
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### Re: BR Maths Corner-1

A short comment:
Eric Demopheles wrote:Let me here make a note to the Quadratic Reciprocity Law regarding congruences of prime numbers that was conjectured and half-proved by him. This is arguably the golden theorem of number theory.
.

Yes, Quadratic reciprocity theorem is quite important, and remarkable, but by all accounts, in my opinion, it is a little harder theorem (relatively speaking). Not too many people know about it. In fact 99.9% or more UG math majors (no exaggeration, I have asked and NONE of many math professors, let alone students, I asked understood it). Most (as in 99+% of math majors in a typical college) do not even know what quadratic residue is.

A little easier to understand,is Fermat's little theorem ( or Euler's extension) which most bright high school students can understand it they see the proof .. yet again 99% of College (or High school students) do not read about it. In India I have seen it covered in graduate level courses only..Though problems based on this appear in math contests like IMO/RMO all the time)

(We have covered this topic, a few times in this dhaga.. and I have posed some interesting problems based on it)

For example to prove "11111111111" is not a prime (without using a computer or proving "11111... 41 times" even using a computer ) is not too hard using this theorem.

****

For those who would like to know "Quadratic_reciprocity" in simple terms, here it is:

(If you know modulo math, please read on else ignore the rest of the post)

Given a and b are two primes, both of them are not of the form 4k-1
(Otherwise a small complexity creeps in which I am ignoring here, see any standard reference for that case)

If x^2 = a has a solution modulo b
then x^2 = b has a solution modulo a
(And vice-ver-sa)
(The theorem is very remarkable, but not that easy to prove using simple high-school math)

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### Re: BR Maths Corner-1

Eric Demopheles wrote:Still, the divergence is a significant fact when confronted with the convergence of the sum of 1/n^2 .. ...

This also has been mentioned a few times in this dhaga.. in fact the sum of 1/n^2.., that is
S=1+1/4+1/9+1/16+1/25 ...
is some times not covered in many undergraduate math courses... The sum, of course is, (pi^2/6),

This is what fascinated my son. (You just see squares of 1,2,3.. all natural numbers and wow, suddenly pi creeps in).. he got hooked on Math at very young age and this equation was definitely the main cause ( he remembers this as the cause for getting hooked). He learned calculus and many other fun things trying to understand how pi creeps in the sum. (His interest in math is still there.. he did his PhD in Theoretical Physics, which some say is nothing but math with purpose.. )

****

As I have mentioned before, lot of Ramanujan's work was with zeta function.. (which sum of such kind of a series)..

One of silly but really fun thing is zeta (-1) which is (-1/12), and in some weird sense
sum of (1+2+3+4+5+...) is (-1/12)

(These sums are called "ramanujan sums"

(NO it does not make sense to many, but "1 + 2 + 3 + 4 + ⋯ = −1/12" was presented in one of the Ramanujan's book).. If you don't believe me , check out wiki:
http://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_%E2%80%A6

(go a few paragraphs below under Heuristics)
Here is letter from Ramanujan to Hardy:
"Dear Sir, I am very much gratified on perusing your letter of the 8th February 1913. I was expecting a reply from you similar to the one which a Mathematics Professor at London wrote asking me to study carefully Bromwich's Infinite Series and not fall into the pitfalls of divergent series. … I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + · · · = −1/12 under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal. I dilate on this simply to convince you that you will not be able to follow my methods of proof if I indicate the lines on which I proceed in a single letter. …

Nandu
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### Re: BR Maths Corner-1

Came across this recently on those Ramanujan sums.
http://terrytao.wordpress.com/2010/04/1 ... tinuation/

Eric Demopheles
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### Re: BR Maths Corner-1

Amber G. wrote:(His interest in math is still there.. he did his PhD in Theoretical Physics, which some say is nothing but math with purpose.. )

Indeed. Homological mirror symmetry and other similar recent stuff from physics contributes very deeply and seriously to pure mathematics. Such kind of influence of physics in mathematics is always most welcome and groundbreaking.

(NO it does not make sense to many, but "1 + 2 + 3 + 4 + ⋯ = −1/12" was presented in one of the Ramanujan's book).. If you don't believe me , check out wiki:
http://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_%E2%80%A6

If I am not mistaken, such things were already considered by Euler.

About quadratic reciprocity: it is about pure number theory, whereas the series etc. involve mathematical analysis. The reciprocity theorem is undoubtedly an arithmetical theorem. The Fermat's little theorem, while interesting, is not *deep enough* and easy to learn. Quadratic reciprocity's proof involves some nontrivial work and such serious type of number theory must be promoted rather than easier and simple tricks.

Amber G.
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### Re: BR Maths Corner-1

Eric Demopheles wrote:http://en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_%E2%80%A6

If I am not mistaken, such things were already considered by Euler.

Not really.

It is true that Euler was one of the first (or a giant) who studied these zeta functions (Many still call them Euler-Riemann-zeta functions) ( Euler–Riemann zeta function, ζ(s), is a function of a complex variablethat analytically continues the sum of the infinite series (mentioned above). The series converges when the real part of s is greater than 1 (BTW, zeta function plays a very important role, not only in analytic number theory but has applications in physics, statistics.. etc. (For example, random matrices (prof Mehta, Wigner, Dyson etc), just to give one example, uses it heavily)

But the critical point is: as a function of a real argument, was studied by Euler without using complex analysis (It was not developed at that time). Riemann extended the Euler definition to a complex variable, proved its meromorphic continuation and established a relation between zeros ζ(s) ...and the distribution of prime numbers etc..And that is where all the "fun" and "crazy" stuff appears..

Hardy, Littlewood and Ramanujan contributed the major part later...

Let me leave this post with all the history..from the ballad .. (Courtesy University Wisconsin Math dept)

Enjoy!

Where are the zeros of zeta of s?
G.F.B. Riemann has made a good guess,
They're all on the critical line, sai he,
And their density's one over 2pi log t.

This statement of Riemann's has been like trigger
And many good men, with vim and with vigor,
Have attempte to find, with mathematical rigor,
What happens to zeta as mod t gets bigger.

The efforts of Landau and Bohr and Cramer,
And Littlewood, Hardy and Titchmarsh are there,
In spite of their efforts and skill and finesse,
(In) locating the zeros there's been no success.

In 1914 G.H. Hardy did find,
An infinite number that lay on the line,
His theorem however won't rule out the case,
There might be a zero at some other place.

Let P be the function pi minus li,
The order of P is not known for x high,
If square root of x times log x we could show,
Then Riemann's conjecture would surely be so.

Related to this is another enigma,
Concerning the Lindelof function mu(sigma)
Which measures the growth in the critical strip,
On the number of zeros it gives us a grip.

But nobody knows how this function behaves,
Convexity tells us it can have no waves,
Lindelof said that the shape of its graph,
Is constant when sigma is more than one-half.

Oh, where are the zeros of zeta of s?
We must know exactly, we cannot just guess,
In orer to strengthen the prime number theorem,
The integral's contour must not get too near 'em.

Amber G.
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### Re: BR Maths Corner-1

xpost from Physics dhaga:

SaiK wrote:http://www.theguardian.com/science/2013/dec/13/michael-green-cambridge-fundamental-physics-prize
was watching the repeat show on the breakthrough prize presentation awards on sci chi.. man they make it bigger than academy and grammies.. never any sdres could expect such tfta presentations like this.

perhaps the best way to make them real heroes!

Rony
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### Re: BR Maths Corner-1

Dr. M.D Srinivas talks about how evolution of Mathematics in India is different from Mathematics in Europe. He traces the history of Indian mathematics for last two thousand years and shows that how it evolves independently of European Mathematics. He claims that the way mathematics in India evolved teaches very different notions of fundamental of mathematics and its objective in India.

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### Re: BR Maths Corner-1

Nice video...thanks for posting. A lot of ancient hindu mathematicians, astronomers and their works are discussed, along with the sanskritic description of the solutions. Covers a broad range of topics in math (the Indian approach) rather than focus on one area and go deep. Panini and Bhaskara in the same talk.
Prof. M.D. Srinivas http://cpsindia.org/index.php/research/ ... d-srinivas

One thing was intriguing. He seems to suggest that some of the earlier models of astronomy were based on an earth-centric system, and then someone came along and observed paths of mercury and venus and went the heliocentric route. Not sure if understood correctly....

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### Re: BR Maths Corner-1

Okay, Little quiet here, so let me post two elementary problems. These require nothing more than elementary arithmetic. Use any kind of logic (Vedic, complex numbers or even group theory )... Try it out on your Math professors or school age children... You may decide to try them yourself

(Numbers a, and b are all integers)

1. You are give a and b such that

if 13 divides (3a-2b) prove that 13 will also divide (2a+3b)

2. You are give a and b such that

if 13 divides (a^2+b^2) prove that 13 will also divide (2a+3b)(3a+2b)
Last edited by Amber G. on 21 Feb 2014 19:41, edited 2 times in total.

Vayutuvan
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### Re: BR Maths Corner-1

OK I got the first one.
Hint: 3a - 2b has to be divisible by 6.

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### Re: BR Maths Corner-1

matrimc wrote:OK I got the first one.
Hint: 3a - 2b has to be divisible by 6.

Why??? a=7,b=4, then (3a-2b) is 13, it is not divisible by 6 ?

Vayutuvan
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### Re: BR Maths Corner-1

OK. you are right. I found a mistake in my unnecessarily long non-proof. Back to the pen and paper.
Last edited by Vayutuvan on 21 Feb 2014 00:50, edited 1 time in total.

Vayutuvan
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### Re: BR Maths Corner-1

I think I got it. Eliminate b from both equations.

13a = 39 n + 2 x
where
3a - 2b = 13 n
2a + 3b = x
a an integer

Vayutuvan
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### Re: BR Maths Corner-1

Another easier proof follows from two propositions. let | stand for "divides" and ~| for "does not divide"
schema: x|a is x divides a; similarly x~|a is x "does not divide" a. All variables are from the set of natural numbers N.
(Correcting the statement as per AmberG's counter-example).

p and q are primes.

Prop1. if p~|a and p|ab then p|b. Try proving this. Not a couple of lines of proof but easy enough.

From the above, both problems AmberG posed can be proved almost in a couple of steps.

I suppose some argument on prime numbers being the basis for finite fields (I think it is sufficient to have pair wise relative primes in the basis) can be made from which the above follows as a special case. A wild guess I have vague details in my mind. Need to formalize a little to present as a general method (but may be already there in some books in abstract form).
Last edited by Vayutuvan on 21 Feb 2014 21:30, edited 2 times in total.

Amber G.
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### Re: BR Maths Corner-1

Matrimc - Saw the proof of the first part. Very good.

There is a theme, a beautiful concept, in the above seemingly ordinary problem...
(I will point it out, but let us hope others notice it too)

It seems that some times, simple things gets unnecessary complicated.... OTOH many nice concepts can be introduced with simple problems..

I liked the following quote (specially when I try to spark interest in math from young people)

The essence of mathematics is not to make simple things complicated, but to make complicated things simple.

More of this later.
***
BTW The following is NOT (always) correct. As you will see false assumptions (In math anything which is not
ALWAYS true should not be a basis)
matrimc wrote:Another easier proof follows from two propositions. let | stand for "divides" and ~| for "does not divide"
schema: x|a is x divides a; similarly x~|a is x "does not divide" a. All variables are from the set of natural numbers N.

Prop. if c~|a and c|ab then c|b

For example, c=8, a=2,b=4 you see that 8 does divide (ab) but it does not divide 2 or 4.
(It happens to be true - one of the most important theorems in number theory if c is prime
but not in all cases (of natural numbers, as you said)

***
I will leave you with a few thoughts ...
Check out -Brahmgupta identity (basis of the above problems)
(This will also answer some question about if Indian Mathematicians were aware of complex numbers)
Please notice - 13 = (2+3i)(2-3i) ..

Amber G.
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### Re: BR Maths Corner-1

As I said before, above problems have a theme. They are easy to prove, if anyone wants to see an elementary - ordinary kind of solution, one can peek below --
(If you want to work it out for yourself - don't peek until you want too)

If 13 divides (3a-2b) => 13 divides 5(3a-2b)=15a-10b = 13(a-b)+(2a+3b)
=> 13 divides (2a+3b)

Similarly
If 13 divides (a^2+b^2) => 13 divides 6(a^2+b^2) =(2a+3b)(3a+2b)-13ab
=> 13 divides (2a+3b)(3a+2b) rest is easy QED.

Vayutuvan
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### Re: BR Maths Corner-1

AmberG: Thanks for the correction. I corrected. I was also thinking in the lines of (a^2+b^2) can be expresses as (a+ib)(a-ib) but not got anywhere. Will follow your hint and see if there is a proof in that direction.

Amber G.
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### Re: BR Maths Corner-1

matrimc wrote:AmberG: Thanks for the correction. I corrected. I was also thinking in the lines of (a^2+b^2) can be expresses as (a+ib)(a-ib) but not got anywhere. Will follow your hint and see if there is a proof in that direction.

Hello matrimc ji - If you have not done that already.. try looking at multiplication rules..

(3+2i)(a+ib) = (3a-2b) + i(2a+3b)
(and of course you know that 13=(3+2i)(3-2i)

***
Brahmgupta identity (http://en.wikipedia.org/wiki/Brahmagupta's_identity)
Take a special case n=1

(Some times the n=1 case is called Brahmagupta–Fibonacci identity)

(This is why some say some important concepts of complex numbers were studied by Brhamguta/Bhaskara et el)

In any case, 13=3^2+2^2
and 13(a^2+b^2) =(3^2+2^2)(a^2+b^2)= (3a-2b)^2 + (2a+3b)^2

LHS is divisible by 13, so if 3a-2b is divisible by 13, (2a+3b) also has to be.

***

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### Re: BR Maths Corner-1

Hello all,

I'm looking for a collaborator on a paper to be delivered (hopefully) at the WAVES 2014 conference in Dartmouth. The event is organized by Rajiv Malhotra and others, and explores multi-disciplinary issues involved with construction of an Indian historical and social narrative by Western academia.

The objective of the paper will be to review and critique the methodology used and conclusions reached by various publications in population genetics, pertaining to the regional/caste-based genetic stratification of Indian populations, and the origin of such populations, weighing the Aryan Migration theory against other theories such as Out-of-India.

I'm trained as a molecular geneticist, so pop gen is not my primary forte, at least not the aspect of it that goes into heavy stats. I understand just enough of the maths to suspect that something is black in the lentils when analysis is done-- choice of samples, assumptions used in constructing prior distributions via Bayesian analysis, application of Bessel functions to creating such distributions, techniques used to determine principal components etc-- but not enough to rigorously critique these possible flaws. I can bring clarity to any issues involving genetics, but I need someone to bring the math skills.

If anyone is interested in collaborating, and has sufficient background in stats to be familiar with the derivation, underlying assumptions and correct applications of the sort of equations laid out in this paper http://www.genetics.org/content/158/2/897.short please contact me at rudradev DAWT brf ATT gmail DAWT com. Thanks.

Vayutuvan
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### Re: BR Maths Corner-1

Srikumar and others interested in P-NP

There is new book called

Golden Ticket by Lance Fortnow chair if CS at Georgia Tech. which going from the blurb in the jacket is a nonTech intro to this problem including the origins etc.

Fortnow has a blog on algorithms including this problem which is very good. Even though I have not read this book yet I am guessing that it would be good as prof. Fortnow is a good writer. Give it a try. I will review once I read the book.

Agnimitra
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### Re: BR Maths Corner-1

X-post from GDF:

History of Indian Mathematics - a series of Youtube lectures at IIT Bombay:

Mathematics in India - From Vedic Period to Modern Times - by Prof. M.D.Srinivas,Prof.M.S.Sriram & Prof.K.Ramasubramanian,Department of mathematics,IIT Bombay.

SaiK
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### Re: BR Maths Corner-1

mat garu, did not understand your proof propositions. of course only glanced it from a moorkh perspective.

Vayutuvan
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### Re: BR Maths Corner-1

SaiK

Are you talking about how to prove this prop?

Prop1. if p~|a and p|ab then p|b. Try proving this. Not a couple of lines of proof but easy enough.

The proof is actually a couple of lines if you use another theorem the which states that GCD of two numbers can be expressed as a linear combination of the two numbers (always working in Z including negative numbers).

i.e. if g is the gcd of a and b then there exist two numbers x and and y such that

xa+yb = g

Now for the proof of the prop1.

Since p~|a then we have gdc(p,a) = 1

So we have two numbers x and y such that px + ay = 1

Multiply on both sides with b we get

pxb + aby = b

Since p|ab we can write ab = pn for some n.

pxb + pny = b
p(xb+ny) = b

Since LHS is divisible by p, rhs too is dvisible by p. QED.

Prop1 is called Euclid's lemma. More generally if p a prime divides a_1 a_2 ... a_n then p divides at least one a_i, 1 <= i <= n.

Nandu
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### Re: BR Maths Corner-1

Oopsie: Missed that p has to be prime. My apologies.
Last edited by Nandu on 23 Mar 2014 06:58, edited 1 time in total.

Vayutuvan
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### Re: BR Maths Corner-1

Nandu
As AmberG noted, (and I subsequently corrected my erroneous statement of prop1), when p is a prime, then prop1 is true which is known as Euclid's Lemma. Then the given proof is a correct proof of the lemma.

By the way the same proof works for the following too.

If p and a are co-prime and p|ab then p|b.
Proof: By definition gcd(p,a) = 1 and proceed as in Euclid's Lemma proof above.
Last edited by Vayutuvan on 23 Mar 2014 09:11, edited 1 time in total.