Linked Questions

126
votes
6answers
28k 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?
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 ...
32
votes
3answers
4k 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}[...
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 ...
26
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.
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....
13
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 ...
13
votes
2answers
257 views

Proving that if $x_1,\dots,x_n$ are rational numbers and $\sqrt{x_1}+\dots\sqrt{x_n}$ is rational, then each $\sqrt{x_i}$ is rational as well

I'm having a hard time with the following problem: Let $x_1,x_2...x_n$ be rational numbers. Prove that if the sum $\sqrt{x_1}+\sqrt{x_2}+...+\sqrt{x_n}$ is rational, then all $\sqrt{x_i}$ are ...
10
votes
5answers
216 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: ...
9
votes
2answers
762 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 ...
9
votes
2answers
95 views

Are square root binomials unique?

In Euclid we find the notion of a binomial, its simply a sum $s = \sqrt{a}+\sqrt{b}$ of two square roots $\sqrt{a}$ and $\sqrt{b}$. Lets say such a sum is simple iff $a$ and $b$ are positive non-zero ...
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} | ...
7
votes
2answers
131 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 ...
7
votes
1answer
527 views

Is the sum of the square roots of all natural numbers up to n whole for any value of n other than 1?

For the summation $\sum_{n=0}^x \sqrt{n}$ are there any values of $x$ where the summation equals a whole number other than 1?

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