Linked Questions

27
votes
5answers
5k views

Can a finite sum of square roots be an integer? [duplicate]

Can a sum of a finite number of square roots of integers be an integer? If yes can a sum of two square roots of integers be an integer? The square roots need to be irrational.
6
votes
1answer
5k views

Showing field extension $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})/\mathbb{Q}$ degree 8 [duplicate]

Possible Duplicate: The square roots of the primes are linearly independent over the field of rationals I am trying to classify the Galois group of the field extension $\mathbb{Q}(\sqrt{2}, \sqrt{...
3
votes
3answers
283 views

Why is $n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \sqrt{5} + n_4 \sqrt{7} $ never zero? [duplicate]

Here $n_i$ are integral numbers, and not all of them are zero. It is natural to conjecture that similar statement holds for even more prime numbers. Namely, $$ n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \...
4
votes
1answer
394 views

Is there a way to show the sum of any different square root of prime numbers is irrational? [duplicate]

Is there a way to show the sum of any different square root of prime numbers is irrational? For example, $$\sqrt2+\sqrt3+\sqrt5 +\sqrt7+\sqrt{11}+\sqrt{13}+\sqrt{17}+\sqrt{19}$$ should be a irrational ...
0
votes
1answer
421 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 ? $\...
2
votes
1answer
315 views

$\sqrt{p_1}$ is not in $Q[\sqrt{p_2},…,\sqrt{p_n}]$ [duplicate]

How to show $\sqrt{p_1}$ is not in $Q[\sqrt{p_2},...,\sqrt{p_n}]$ if $p_1,...,p_n$ are distinct primes? Intuitively, this is pretty clear, but it makes me very uncomfortable to just believe. Any idea ...
2
votes
1answer
146 views

Exercise about field extensions [duplicate]

Consider $a_1,\ldots,a_n\in \mathbb Z$. i) Suppose $a_1,\ldots, a_n$ are pairwise relatively prime. I have to see by induction on n that $[\mathbb Q(\sqrt a_1,\ldots,\sqrt a_n):\mathbb Q]=2^n$ Once ...
1
vote
0answers
91 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 ...
3
votes
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 ...
1
vote
0answers
53 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.
-1
votes
0answers
49 views

How to prove $\sqrt{p}$ and $\sqrt{q}$ are linearly independent over the rationals, where $p,q$ are primes? [duplicate]

How do I prove that if we have two distinct prime numbers $p$ and $q$, then $\sqrt{p}$ and $\sqrt{q}$ will be linearly independent over the field of rational numbers?
1
vote
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 ...
128
votes
6answers
30k views

Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?

It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?
21
votes
6answers
4k views

Is there a way to write an infinite set that contains only irrational numbers without integer multiples?

Is there a way to write an infinite set that contains only irrational numbers without integer multiples? The infinite set must not contain integer multiples of any other members of that set. For ...
19
votes
3answers
5k views

How to prove $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$?

How do I prove that $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? $\mathbb{Q}$ is the rational field. I want to know the detail about the proof. Thanks in advance....
27
votes
3answers
2k views

Is it possible for integer square roots to add up to another?

I initially was wondering if it were possible for there to be three $x,y,z \in \mathbb{Q}$ and $\sqrt{x},\sqrt{y},\sqrt{z} \notin \mathbb{Q}$ such that $\sqrt{x} + \sqrt{y} = \sqrt{z}$. I had ...
40
votes
4answers
4k views

Proving that $\left(\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right)=2^n$ for distinct primes $p_i$.

I have read the following theorem: If $p_1,p_2,\dots,p_n$ are distinct prime numbers, then$$\left(\mathbb Q\left[\sqrt p_1,\dots,\sqrt p_n\right]:\mathbb Q\right)=2^n.$$ I have tried to prove a ...
14
votes
5answers
9k views

Infinite Degree Algebraic Field Extensions

In I. Martin Isaacs Algebra: A Graduate Course, Isaacs uses the field of algebraic numbers $$\mathbb{A}=\{\alpha \in \mathbb{C} \; | \; \alpha \; \text{algebraic over} \; \mathbb{Q}\}$$ as an example ...
33
votes
3answers
5k views

How to prove that $\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots,\sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}]$, for $p_i$ prime?

This is Exercise 18.14 from Algebra, Isaacs. $p_{1}\ ,\ p_{2}\ ,\ ... p_{n}$ are different prime numbers. How to show that $$\mathbb{Q}[\sqrt{p_{1}}, \sqrt{p_{2}}, \ldots, \sqrt{p_{n}} ] = \mathbb{Q}[...
5
votes
7answers
691 views

Proving that $\sqrt[3] {2} ,\sqrt[3] {4},1$ are linearly independent over rationals

I was trying to prove that $\sqrt[3] {2} ,\sqrt[3] {4}$ and $1$ are linearly independent using elementary knowledge of rational numbers. I also saw this which was in a way close to the question I was ...
5
votes
6answers
711 views

Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$?

This isn't a homework question but one I found online. Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$? We just started doing field theory in my class and I want extra practice, but I ...
4
votes
3answers
911 views

The sum of square roots of non-perfect squares is never integer [duplicate]

My question looks quite obvious, but I'm looking for a strict proof for this: Why can't the sum of two square roots of non-perfect squares be an integer? For example: $\sqrt8+\sqrt{15}$ isn't an ...
7
votes
2answers
1k views

Is $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$?

Is $\mathbf{Q}(\sqrt{2},\sqrt{3}) = \mathbf{Q}(\sqrt{6})$? If $\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\mid a,b,c,d\in\mathbf{Q}\}$ and $\mathbf{Q}(\sqrt{6})= \{a+b\sqrt{6} | ...
10
votes
5answers
241 views

Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.

Question: Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational. Following from the question, I tried: ...
3
votes
3answers
4k views

Finding a basis for $\Bbb{Q}(\sqrt{2}+\sqrt{3})$ over $\Bbb{Q}$.

I have to find a basis for $\Bbb{Q}(\sqrt{2}+\sqrt{3})$ over $\Bbb{Q}$. I determined that $\sqrt{2}+\sqrt{3}$ satisfies the equation $(x^2-5)^2-24$ in $\Bbb{Q}$. Hence, the basis should be $1,(\...
2
votes
3answers
456 views

Showing that $\sqrt5$ is not in $\mathbb{Q}(\sqrt7)$

How can I prove that $\sqrt5$ is not in $\mathbb{Q}(\sqrt7)$ ? I can only think of trying to write $\sqrt5 = a+b\sqrt7$ (where $a,b$ are in $\mathbb{Q}$), but I can't think of a good reason that ...
4
votes
2answers
1k views

Linear independence of roots over Q

Let $p_1,\ldots,p_k$ be $k$ distinct primes (in $\mathbb{N}$) and $n>1$. Is it true that $[\mathbb{Q}(\sqrt[n]{p_1},\ldots,\sqrt[n]{p_k}):\mathbb{Q}]=n^k$? (all the roots are in $\mathbb{R}^+$) ...
10
votes
2answers
842 views

Show that $ a,b,c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q \implies \sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q $

Assume that $a,b,c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q$ are rational,prove $\sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q$,are rational. I know that can be proved, would like to know that there is ...
6
votes
1answer
515 views

Elementary proof for $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ where $p_i$ are different prime numbers. [duplicate]

Take $p_1, p_2, \ldots, p_n, p_{n+1}$ be $n+1$ prime numbers in $\mathbb{P} \subseteq \mathbb{N}$. $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ seems to be quite ...
7
votes
2answers
133 views

Is $(5+\sqrt[3]2)^n$ ever an integer for $n \in \Bbb Z \setminus \{0\}$?

In general, I would like to prove that if $m>2$ is an integer, then $(5+\sqrt[m]2)^n$ is never an integer (unless for $n=0$). First, I'm interested in the simple case $m=3$ (I already solved it ...

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