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 Showing results 1 to 25 of 379 Search took 0.04 seconds. Search: Posts Made By: blob100
 Forum: Math 2016-12-21, 21:16 Replies: 7 Views: 992 Posted By blob100 I'm indeed Israeli and jewish (atheist,... I'm indeed Israeli and jewish (atheist, agnostic). Anyway, both Hanukkah and xmas are great! I'm celebrating Hanukkah for the same reasons xilman celebrates xmas. Happy holidays
 Forum: Math 2016-12-18, 16:01 Replies: 7 Views: 992 Posted By blob100 My sincere thanks to this forum Hello everyone, Seven years ago, when I was 14 years old, I took part in this forum on a daily basis. I left when I entered high school and started studying at the university in parallel. ...
 Forum: Math 2010-07-26, 21:41 Replies: 817 Views: 25,468 Posted By blob100 So can you just tell me the solution? So can you just tell me the solution?
 Forum: Math 2010-07-26, 21:40 Replies: 817 Views: 25,468 Posted By blob100 I'll try to express myself: Is the theorem says... I'll try to express myself: Is the theorem says the next: every polynomial with complex coefficients must have one or more complex roots? My ridiculous statement: By the first exercize we found...
 Forum: Math 2010-07-26, 21:28 Replies: 817 Views: 25,468 Posted By blob100 I know this theorem, I just do silly mistakes: ... I know this theorem, I just do silly mistakes: The set of all prime devisors of a is A, of b is B and of c is C. 1) The intersection of A and B is the empty set. The intersection of (BUC) and A is...
 Forum: Math 2010-07-26, 21:19 Replies: 817 Views: 25,468 Posted By blob100 First of all, I'm not smoking. I do understand... First of all, I'm not smoking. I do understand the fundamental theorem of algebra, but this theorem is corresponding complex coefficient polynomials and here you gave a polynomial with natural...
 Forum: Math 2010-07-26, 20:40 Replies: 817 Views: 25,468 Posted By blob100 The seventh problem: I'm needed to find the... The seventh problem: I'm needed to find the roots of the next polynomial: f(x)=x^4+x^3+x^2+x+1=0. We see by the first problem that the coefficients of f(x) are equivallent to those of g(x) and...
 Forum: Math 2010-07-24, 15:52 Replies: 817 Views: 25,468 Posted By blob100 Am I need to prove the next?: If we have... Am I need to prove the next?: If we have (a,b)=1 and b|can then b|c. Where c is an integer. If yes, I'll try again: We have B the set of all prime devisors of b, And A is samely defined for a....
 Forum: Math 2010-07-23, 18:44 Replies: 817 Views: 25,468 Posted By blob100 I know Shanks has discussed it, and it seems to... I know Shanks has discussed it, and it seems to me really intuitive.. 9 is a devisor of 6^2 just becuase 3 is a devisor of 6, we must show the cannonical form as for now: 6=3*2=(the cannonical...
 Forum: Math 2010-07-23, 17:54 Replies: 817 Views: 25,468 Posted By blob100 I'll try to make it much more formally: 1) We... I'll try to make it much more formally: 1) We see that every prime divisor of an is a factor of a, so there is no prime devisor of an which isn't of a. 2) If we have (a,b)=1 then we may say that...
 Forum: Math 2010-07-23, 17:01 Replies: 817 Views: 25,468 Posted By blob100 I'm not stupid, you can't pick b=4 and a=6 when... I'm not stupid, you can't pick b=4 and a=6 when you define (a,b)=1. For b=4 and a=6 we have (a,b)=2. I'll make the proof: For (a,b)=1 and b|anan we have b|an. We see that (a,b)=1, which...
 Forum: Math 2010-07-23, 10:30 Replies: 817 Views: 25,468 Posted By blob100 I can't see any problem here, b|an just becuase... I can't see any problem here, b|an just becuase b isn't a factor of a... Same with a|a0, a isn't a factor of b.
 Forum: Math 2010-07-21, 22:06 Replies: 817 Views: 25,468 Posted By blob100 The proof is given by the rational roots theorem.... The proof is given by the rational roots theorem. Proof: Let f(x) defined as before. By the rational roots theorem, any given rational root a/b of f(x) must agree the next: a|a0 and b|an. *a0, an...
 Forum: Math 2010-07-21, 18:22 Replies: 817 Views: 25,468 Posted By blob100 We would like to prove the next: For a given... We would like to prove the next: For a given monic polynomial f(x)=x^n+...+a0 with integer coefficients, any rational root of it is an iteger. A={a0,a1,...,an} the set of all coefficients of f(x)....
 Forum: Math 2010-07-21, 11:14 Replies: 817 Views: 25,468 Posted By blob100 I'll try again: I'll prove this theorem by... I'll try again: I'll prove this theorem by contradiction: Let f(x)=x^n+...+a0 a monic polynomial. A={a0,a1,...,an} the set of all coefficients of f(x). We would like to prove that a root Z=b/c...
 Forum: Math 2010-07-21, 08:38 Replies: 817 Views: 25,468 Posted By blob100 I'll prove this theorem by contradiction: Let... I'll prove this theorem by contradiction: Let f(x)=x^n+...+a0 a monic polynomial. A={a0,a1,...,an} the set of all coefficients of f(x). We would like to prove there is no such root Z=b/c of f(x)...
 Forum: Math 2010-07-17, 15:24 Replies: 817 Views: 25,468 Posted By blob100 What is the direction? What is the direction?
 Forum: Math 2010-07-17, 10:34 Replies: 817 Views: 25,468 Posted By blob100 Let a polynomial of degree n be f(x)=anx^n+...+a0... Let a polynomial of degree n be f(x)=anx^n+...+a0 with roots R={x1,x2,...,xn}, each rational root is an integer. A={a0,a1,...,an} the set of the whole coefficients (are all integers) of f(x)....
 Forum: Math 2010-07-16, 13:35 Replies: 817 Views: 25,468 Posted By blob100 We have a polynomial f(x) such that Z is a... We have a polynomial f(x) such that Z is a complex root of it. f(x)=anx^n+...+a0. A={a0,a1,...,an} the set of the coefficients of f(x). f(Z)=anZ^n+...+a0=0. We see: The conjugate of (aiZ^i) is...
 Forum: Math 2010-07-15, 13:38 Replies: 817 Views: 25,468 Posted By blob100 This theorem follows by the statement: every... This theorem follows by the statement: every polynomial with a complex root a+bi is divisible (has the root) a-bi too. (a,b are too constants). So every polynomial must have 2m (m={0,1,2,3,...})...
 Forum: Math 2010-07-13, 20:38 Replies: 817 Views: 25,468 Posted By blob100 I want to understand one thing: Isn't the... I want to understand one thing: Isn't the solution of the problem trivially true? I gave it three times...
 Forum: Math 2010-07-12, 14:54 Replies: 817 Views: 25,468 Posted By blob100 Of course, this is why I left this question... Of course, this is why I left this question...
 Forum: Math 2010-07-12, 13:36 Replies: 817 Views: 25,468 Posted By blob100 We try proving the next: f(x)=(x-r)g(x)+f(r). ... We try proving the next: f(x)=(x-r)g(x)+f(r). Let's see f(x)-f(r) as a polynomial with k roots, We try proving it (the polynomial) has one root r which is equivallet to say it (the polynomial)...
 Forum: Math 2010-07-12, 12:11 Replies: 817 Views: 25,468 Posted By blob100 So I'm needed to prove that there is such g(x)? So I'm needed to prove that there is such g(x)?
 Forum: Math 2010-07-12, 09:15 Replies: 817 Views: 25,468 Posted By blob100 Is this the easier proof? I think I gave a much... Is this the easier proof? I think I gave a much simpler way to prove this theorem.
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