Linked Questions

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1answer
54 views

$\sum_{i=1}^n {(a_i\sqrt{b_i})} \ne 0$

In a surd $a\sqrt{b}$   ($b \in \mathbb{Z^+}$)   the value of $b$ can assumed to be a square-free integer ($b = p_1p_2\dots p_k$, where $p_i$ are distinct primes), since otherwise a ...
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1answer
235 views

How to get this equation solved?

I came across this equation $$\sqrt a-\sqrt b=\sqrt 7-\sqrt 5$$ And you have to find the value of '$a$' and '$b$' when both of them are primes. The solution was $a=7, b=5$. Now, my question is, ...
1
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1answer
141 views

How to convert $\Bbb Q(\sqrt 2,\sqrt 3)$ to $\Bbb Q(\alpha)?$

I have a basic question about algebraic field extensions: How can I convert a multiple extension like $\mathbb{Q}(\sqrt{2},\sqrt{3})$ to a single (elementary) field extension (like $\mathbb{Q}(\sqrt{...
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1answer
33 views

Identifying squares in field extension

I'm trying to solve problem 4.11 of http://homepages.warwick.ac.uk/~masda/MA3D5/Galois.pdf. The problem is, given $a,b\in k$ a field, and assuming that $\sqrt{a}\in k(\sqrt{b})$ prove that, either $\...
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1answer
65 views

How to prove that $\sqrt{p_n} \notin \mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_{n-1}})$

I want to prove $\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_{n-1}})=\mathbb{Q}(\sum\limits_{i =1}^n\sqrt{p_i})$, where $p_i$ are different prime integers. But I must solve this problem firstly: $\sqrt{p_n} ...
0
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1answer
378 views

root of prime numbers are linearly independent over $\mathbb{Q}$ [duplicate]

How can we prove by mathematical induction that $1,\sqrt{2}, \sqrt{3}, \sqrt{5},\ldots, \sqrt{p_n}$ ($p_n$ is the $n^{\rm th}$ prime number) are linearly independent over the rational numbers ? $\...
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1answer
196 views

$\sqrt 1+\sqrt 2 +\sqrt 3 +\cdots +\sqrt {2009}$ change a sign to be rational [closed]

I have this problem: $$\sqrt 1+\sqrt 2 +\sqrt 3 +\cdots +\sqrt {2009}$$ Prove that you need to change ONLY a sign (to convert a $+$ to $-$) of a single square root, for the sum to be rational. EDIT:...
3
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0answers
73 views

Linear independence in $\mathbb{R}$ [duplicate]

Possible Duplicate: The square roots of the primes are linearly independent over the field of rationals I would like to prove that the family $\{\sqrt{p}, p\text{ prime number} \}$ is linearly ...
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0answers
34 views

Is $\{\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\dots\}$ l.i. over $\mathbb{Q}$? [duplicate]

In order to prove that $[\mathbb{R}:\mathbb{Q}] = \infty$, I was trying to construct an infinite linear independent subset of $\mathbb{R}$. Before noticing that $\{1,\pi,\pi^2,\dots\}$ does the job, I ...
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0answers
47 views

If $k_1, \ldots, k_n$ are non-square, pairwise coprime, then $\sqrt {k_n} \not \in \mathbf{Q}(\sqrt {k_1}, \ldots, \sqrt {k_{n-1}})$ [duplicate]

Seems intuitive. Like the fact that $\sqrt 3 \not \in \mathbf{Q}(\sqrt 2)$. But how to approach actually proving it? The proof of this fact doesn't seem to generalize well.
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0answers
142 views

Characterizing $\operatorname{Gal}(\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_n})/\mathbb{Q})$ for $p_i$ primes?

For what $n$ is $$ \operatorname{Gal}(\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_n})/\mathbb{Q}) $$ known, where the $p_i$ are primes? By Kummer theory, I think that $$ \operatorname{Gal}(\mathbb{Q}(\sqrt{...
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0answers
89 views

Linear independence of square root of square free numbers [duplicate]

Possible Duplicate: The square roots of the primes are linearly independent over the field of rationals I am reading a research article in which there is a theorem regarding square roots of ...
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0answers
16 views

Is $\mathbb Q(\sqrt 2) \times \mathbb Q(\sqrt 3)=\mathbb Q(\sqrt 2,\sqrt 3)$ if I prove $\sqrt 2,\sqrt 3$ are L.I. over $\mathbb Q$? [duplicate]

I proved that $\{1,\sqrt 2\}$ and $\{1,\sqrt 3\}$ are respective bases of $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$ over $\mathbb Q$. I want to show in some sense that since $\sqrt 2,\sqrt 3$ are ...
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0answers
127 views

Prove linear independence of roots of square-free numbers over $\mathbb{Q}$ using induction

I want to prove that this system $\{1,\sqrt{m_1},\sqrt{m_2},...,\sqrt{m_n}\}$, where all $m_i$ are different square-free natural numbers, is linear independent over $\mathbb{Q}$. My teacher has asked ...
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0answers
67 views

How to prove $a_{1}\sqrt[b_{1}]{c_{1}}+a_{2}\sqrt[b_{2}]{c_{2}}+…+a_{n}\sqrt[b_{n}]{c_{n}}$ is irrational?

Let's define the number $$A=a_{1}\sqrt[b_{1}]{c_{1}}+a_{2}\sqrt[b_{2}]{c_{2}}+.....+a_{n}\sqrt[b_{n}]{c_{n}}$$ where $a_{1}, a_{2}, ..., a_{n}$ are positive integers and $b_{1}, b_{2}, ..., b_{n}, ...

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