English | Español

Try our Free Online Math Solver!

Online Math Solver

 

 

 

 

 

 

 

 
 
 
 
 
 
 
 
 

 

 

 
 
 
 
 
 
 
 
 

Please use this form if you would like
to have this math solver on your website,
free of charge.





SQARE ROOTS OF INTEGER NUMBER
How to solve Non linear Equation solving ,   free worksheet solving addition and subtraction equations ,    simplifying exponential expressions variable in the exponent,     algebra easy finding greatest common denominator ,   basic algebra power fraction ,   graph equation basic algebra help, rational expressions on-line calculator , slope formula, Quadratic formula, slope intercept formula sheet , non linear equation solver unknowns equations, algebra help calculate greatest common denominator , solving 3rd order quadratic equations,   first order linear partial differential equation solutions ebook , free simultaneous quadratic equation solver  
Thank you for visiting our site! You landed on this page because you entered a search term similar to this: sqare roots of integer number. We have an extensive database of resources on sqare roots of integer number. Below is one of them. If you need further help, please take a look at our software "Algebrator", a software program that can solve any algebra problem you enter!
From: "Achava Nakhash, the Loving Snake" Newsgroups: sci.math Subject: Re: How Do You Prove A Number Is Irrational? Date: Mon, 28 Sep 1998 17:06:39 -0700 Kurt Foster wrote: > [query as to methods of proving irrationality deleted] > > There aren't all that many methods, and, as you suspect, none is > comprehensive. The best known are (1) Unique factorization ("fundamental > theorem of arithmetic"). (2) Irreducibility of polynomials (various > results, Eisenstein's criterion perhaps being best known. Zeroes of > irreducible polynomials of degree > 1 are not rational numbers; if the > zeroes are also real, they are real irrational numbers. (3) Infinite > series which "converge too fast" can be proven to have sums which are > irrational, or even transcendental. > Examples: > > Method (1): log_2(3) = a/b --> 2^a = 3^b, violates unique factorization. > > Method (2): sqrt(2) satisfies x^2 - 2 = 0, x^2 - 2 irreducible by > Eisenstein's criterion. > > Method (3): zeta(3) = SUM, n >=1, 1/n^3 (irrational - Apery, 1979) > > Number with unknown character: Euler's constant "gamma" = .577+ To expand a bit on what was already said, The Eisentein irreducibility criterion says that a polynomial of the form x^n + a_(n-1)x^(n-1) + ... + a_1x + a_0 such that, for some prime p, a_0, a_1, ..., a_(n-1) are divisible by p, but a_0 is not divisible by p^2, is necessarily irreducible over the rational numbers. Hence any of its roots are irrational, especially if they are real. Examples of such polynomials are x^2 - 2, x^3 - 5, x^2 - 7x - 7, and many more. Notice that this technique proves that all sqare roots of square-free integers are necessarily irrational. Of course, the unique factorization theorem (which is used in the proof of the Eisenstein irreducibility theorem) can be used to prove this directly. To elaborate on Kurt's third technique, let us suppose that x = a/b is a rational number. Let us try to approximate it by some other irrational number p/q, where p and q are any integers, not necessarily prime, and obviously such that q is not equal to 0. Also assume that a/b and p/q are not actually equal to each other and that both fractions are presented in lowest terms. Then |a/b - p/q| = |(aq-bp)/bq| >= 1/bq by the Fundamental Theorem of Transcendental Number Theory which says that a non-zero integer must have an absolute value greater than or equal to 1. In particular, if x is rational, there is a constant c, depending only on x, such that for sufficiently large q, |x - p/q| > c/q. Thus if we have an infinite sequence of rational numbers, p_n / q_n and a real number x such that |x - p_n / q_n| 0 and for all sufficiently large n, then x must be irrational. Armed with this load of baggage, it is not too difficult to demonstrate that e = 1/0! + 1/1! + 1/2! + 1/3! + ... is irrational by letting the n'th partial sums be our p_n / q_n and estimating the difference, 1/(n+1)! + 1/(n+2)! + ... to see that our sequence of rationals meets the above sequence criterion. I have been a little sloppy about when things are in reduced terms and so on, but it is quite easy to make this argument rigorous, and it can also be found in several books, some of which have been mentioned in previous messages in this thread. The approach by finding good rational approximations to x is also behind many of the transcendance proofs in the literature. Liouville demonstrated way back when that a real number x that satisfies an irreducible polynomial with integer coefficients of degree n has a number C associated with it such that any rational p/q satisfies |x - p/q| > C / q^n and then used this to demonstrate that numbers of the form 1/2^(0!) + 1/2^(1!) + 1/2^(2!) + ... could not satisfy any irreducible polynomial of degree n with integers coefficients for an n, and hence such numbers must be transcendental. Many years were to pass before mathematicians found arguments of extraordinary cleverness to prove that e and pi were transcendental. In 1909, Thue managed to lower than exponent attached to q in Liouville's theorem, and proved as a consequence of this sharpening that a class of Diophantine equations could have only a finite number of solutions. Research continue to this day to improve upon this result and to extend the significance of this result for the solution of Diophantine equations. I am not move than superficially familiar with any recent developments, but I believe they are rather spectacular. Regards, Achava