Iyengar
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 Aug 5 comment How does a Class group measure the failure of Unique factorization? @KevinCarlson : Great, I understood my mistake now. Oh, so now, you mean that taking out all the principal fractional ideals is what the class group does right ? If so why Principal fractional ideals ? Aug 4 awarded Suffrage Aug 4 comment Area of a polygon I salute your sincerity in keeping Community wiki, which shows that you are not interested in reputation . Aug 4 comment Area of a polygon @t.b. : You are exactly right , But OP could have added a reference , about the book in which he has seen the formula, in order to facilitate others . Aug 4 comment Area of a polygon Its very hard to figure out the answer, without knowing whether you are looking at a regular or irregular polygon . But I think that typing " derive area of polygon " in Google may fetch you lots of links. Prefer Google before moving your question to Math.SE. Thank you. Aug 4 comment How does a Class group measure the failure of Unique factorization? Thank you for your answer. But what does the exact quotient group mean ? I had a tough time in understanding what does one get when one looks at quotient group. Suppose we are having { Fractional Ideals of F } / { Principal fractional ideals of F }. Then what are we looking at exactly ? . Every ideal is a principal ideal , i.e every ideal can be written as $I=\alpha.(1)$ . Doesn't it imply that every ideal is principal ? Forgive me if I am wrong. Aug 4 comment How does a Class group measure the failure of Unique factorization? Thanks a lot for your edit Mr. M Turgeon 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 these solutions of $2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}$ 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 these solutions of $2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}$ 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..