1,595 reputation
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location India
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visits member for 3 years, 8 months
seen Dec 5 at 9:08

I am Iyengar, and I hail from South India, from a rural city. I have completed my engineering in computer science, but I was always interested in advanced physics, mathematics, philosophy and cognition and evolutionary theories. Because I barely have any teacher along with me to teach me, and no proper library, I did quite a bit of self learning from the internet, and roughly learned many diverse concepts, starting from Quantum mechanics, to Algebraic geometry, Galois theory, philosophy and cognition. I didn't have a path so I dumped down all the knowledge that was available before me. I had generated some ideas to solve Millennium prize problems and some ideas in Quantum mechanics, which have to be made concrete, and since I didn't have a proper channeling, all my ideas and work went unnoticed.

Nevertheless I continue to go further, seeking knowledge. Knowledge is power.


Aug
4
comment Karatsuba Multiplication
@mixedmath : Perfect answer, I was going to type what you have added in the comment, and you saved my effort of typing again , Thanks . But to add something, David, this types of strategies fall under something called " [Divide and conquer strategies ](en.wikipedia.org/wiki/Divide_and_conquer_algorithm) , where you divide the initial problem into pieces and later on assemble them into a the original problem. Rest of the thing is neatly explained in mixedmath's version.
Aug
4
revised How does a Class group measure the failure of Unique factorization?
added something
Aug
4
comment How does a Class group measure the failure of Unique factorization?
Thanks a lot for your answer. +1
Aug
4
comment How does a Class group measure the failure of Unique factorization?
Good answer, Thank you sir +1
Aug
4
asked How does a Class group measure the failure of Unique factorization?
Aug
3
comment Original works of great mathematician Évariste Galois
Great answer indeed. Thank you. @WillJagy : Thank you for your links sir.
Jul
28
accepted Can the order of learning be changed?
Jul
28
comment Are the solutions of $x^{x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}}}=2$ correct?
@AndréNicolas : Is it advisable if I can ask another separate question stating all these things, so that you can answer that elaborately . But anyway your comment is quite good and I thank you for that.
Jul
28
comment Are the solutions of $x^{x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}}}=2$ correct?
@AndréNicolas : But sir, a small doubt. How can one believe that $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^\cdots }}= 2 ? $. Why should one believe that, and what is the intuition . Can you throw some light ?
Jul
27
comment Obtain a contradiction
@GerryMyerson : Yes, thank you for teaching me the responses trick, which I don't know previously. Its true that professor wanted me to solve but I was stuck some-where. I can't proceed further. I never asked you to solve it completely , but instead I wanted someone to post some ideas/hints in that direction.
Jul
27
comment Can dividing two rational numbers yield an integer?
@MistyD : Simply to state $a=n*b$ then it means $ b | a $ for some $n \in \mathbb{z}$
Jul
27
comment Can dividing two rational numbers yield an integer?
@MistyD : .. Contd . 1) Start with $ K= \dfrac{a}{b}$. 2)$ M= \rm{G.C.D}(a,b) $3 ) $K=\dfrac{\dfrac{a}{M}}{\dfrac{b}{M}}$. Recursively repeat until it yields an integer 4) If it don't yield an integer take the new numerator and new denominator, if their G.C.D is 1 , then its division will not be an integer. Or else it would be an integer.
Jul
27
comment Can dividing two rational numbers yield an integer?
@MistyD : That don't work always. I wanted to edit further, but suddenly my internet was interrupted. Anyway you will not get an integer always when G.C.D is not 1. Suppose $\dfrac{42}{56}=\dfrac{3}{4} \notin \mathbb{Z}$ . But $\rm{G.C.D(42,56)}=14$. So here are few steps you need to work out. 1) See that numerator is always greater or equal to denominator. 2) Reduce the fraction to the least form by applying recursive cancellations. A pseudo code can be as follows. Contd..
Jul
27
comment Can dividing two rational numbers yield an integer?
@MistyD : Its not the point of integers or rationals. Its the point of G.C.D . If the numerator and denominator have a G.C.D of 1, they wont yield integers and leave you with some decimal part
Jul
27
comment Obtain a contradiction
@OldJohn : Its simply a Diophantine equation. I was asked to solve it by a professor. I don't think there will be some motivation behind choosing some equations. For example, why does we need to know $x^n+y^n \neq z^n $ when $n\gt2$ ? .
Jul
27
comment Obtain a contradiction
@GerryMyerson : I have now added the motivation sir. I beg you to use '@' while posting comments, I didn't see the comment from many weeks, and that's why I didn't edit it so far. Now I have done by your kind suggestion. Thank you for that.
Jul
27
revised Obtain a contradiction
added something
Jul
27
comment Obtain a contradiction
@GerryMyerson : There is no motivation sir. My plan was to prove that the equation has no solutions.
Jul
27
revised Obtain a contradiction
added something
Jul
21
comment find valuations
These Articles A, B, C and D may be of some use to you.