meta user
network profile
| bio | website | wallenborn.net |
|---|---|---|
| location | Bonn, Germany | |
| age | ||
| visits | member for | 1 year, 2 months |
| seen | 2 hours ago | |
| stats | profile views | 36 |
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Jun 27 |
comment |
algorithm for solving diagonal quadratic equations over real or complex numbers I added a link to the paper. It is a rather algebraic context. What you say is, concerning my first follow-up question, equivalent to extending the alphabet to $\mathbb F$ and adding special states for adding and multiplying two cells to the set of states of the turing machine, right? |
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Jun 27 |
revised |
algorithm for solving diagonal quadratic equations over real or complex numbers added 86 characters in body |
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Jun 27 |
asked | algorithm for solving diagonal quadratic equations over real or complex numbers |
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Jun 25 |
awarded | Scholar |
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Jun 25 |
accepted | Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$ |
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Jun 25 |
comment |
Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$ So either $\dim(P)<2$ or ($(b.b)=0$ and $(a.b)=0$). The later implies that $b\in\mathop{\rm rad}$. Now if $b=0$, $\dim(P)<2$ and if $b\neq 0$, $P$ is degenerate. Thank you very much! |
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Jun 25 |
comment |
Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$ looking at this element (let's call it $c=(b.b)a-(a.b)b$) i get $c.a=0$ and $c.b=0$ yielding that $c$ is in the radical of $P$. This implies that $c$ is zero or $P$ is degenerate. In the later case we are done. If $c$ is zero, we have that $(b.b)a-(a.b)b=0$ which is a linear combination of $0$ in $P$. So either $\dim(P)<2$ or $(b.b)=0$ and $(a.b)=0$. |
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Jun 25 |
awarded | Editor |
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Jun 25 |
comment |
Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$ of course "is zero"... sorry (edited). |
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Jun 25 |
revised |
Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$ added 2 characters in body |
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Jun 25 |
asked | Two elements $a$ and $b$ of a quadratic space generate a nondegenerate two dimensional subspace if and only if $(a\cdot a)(b\cdot b)\neq(a\cdot b)^2$ |
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Jun 10 |
awarded | Supporter |
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May 28 |
awarded | Teacher |
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May 28 |
answered | Is there a rational univariat polynomial of degree 3 with 3 irrational roots? |
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May 27 |
awarded | Student |
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May 27 |
comment |
Is there a rational univariat polynomial of degree 3 with 3 irrational roots? @RagibZaman yes, that's absolutely right and it works out perfectly. Thank you all! I'll post the answer to this question in around 6 hours, when the software here allows me to. |
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May 27 |
comment |
Is there a rational univariat polynomial of degree 3 with 3 irrational roots? @vgty6h7uij: That's a good hint, I'll look at the two examples $x^3+x^2-2x-1\ $ and $x^3+x^2-3x-1\ $ and check whether none of the three real roots is rational. |
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May 27 |
comment |
Is there a rational univariat polynomial of degree 3 with 3 irrational roots? @QiaochuYuan the polynomial $f(x)=x^3+x+1$ has two complex roots. So it is not an example for a polynomial with 3 irrational (but real) roots. link |
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May 27 |
comment |
Is there a rational univariat polynomial of degree 3 with 3 irrational roots? @Micah yes I know that theorem, but it only states necessary conditions on rational roots, right? But I'm asking for irrational roots. |
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May 27 |
asked | Is there a rational univariat polynomial of degree 3 with 3 irrational roots? |
