Forum: Math
2016-12-21, 21:16
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Replies: 7
Views: 1,397
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
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Forum: Math
2016-12-18, 16:01
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Replies: 7
Views: 1,397
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.
...
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Forum: Math
2010-07-26, 21:41
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Replies: 817
Views: 49,145
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Forum: Math
2010-07-26, 21:40
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Replies: 817
Views: 49,145
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Forum: Math
2010-07-26, 21:28
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Replies: 817
Views: 49,145
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...
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Forum: Math
2010-07-26, 21:19
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Replies: 817
Views: 49,145
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...
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Forum: Math
2010-07-26, 20:40
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Replies: 817
Views: 49,145
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...
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Forum: Math
2010-07-24, 15:52
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Replies: 817
Views: 49,145
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....
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Forum: Math
2010-07-23, 18:44
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Replies: 817
Views: 49,145
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Forum: Math
2010-07-23, 17:54
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Replies: 817
Views: 49,145
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...
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Forum: Math
2010-07-23, 17:01
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Replies: 817
Views: 49,145
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Forum: Math
2010-07-23, 10:30
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Replies: 817
Views: 49,145
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Forum: Math
2010-07-21, 22:06
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Replies: 817
Views: 49,145
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Forum: Math
2010-07-21, 18:22
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Replies: 817
Views: 49,145
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)....
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Forum: Math
2010-07-21, 11:14
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Replies: 817
Views: 49,145
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...
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Forum: Math
2010-07-21, 08:38
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Replies: 817
Views: 49,145
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)...
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Forum: Math
2010-07-17, 15:24
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Replies: 817
Views: 49,145
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Forum: Math
2010-07-17, 10:34
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Replies: 817
Views: 49,145
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Forum: Math
2010-07-16, 13:35
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Replies: 817
Views: 49,145
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...
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Forum: Math
2010-07-15, 13:38
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Replies: 817
Views: 49,145
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,...})...
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Forum: Math
2010-07-13, 20:38
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Replies: 817
Views: 49,145
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Forum: Math
2010-07-12, 14:54
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Replies: 817
Views: 49,145
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Forum: Math
2010-07-12, 13:36
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Replies: 817
Views: 49,145
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)...
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Forum: Math
2010-07-12, 12:11
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Replies: 817
Views: 49,145
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Forum: Math
2010-07-12, 09:15
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Replies: 817
Views: 49,145
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