Linked Questions

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
710 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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