# Linked Questions

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### 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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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 $\mathbb{Q}(\sqrt{... 5answers 114 views ### Let$a_0+a_1x+…+a_nx^n$be a non zero polynomial with integer coefficients.if$p(√2+√3+√6)=0$, the smallest possible value of n is? Question Let$a_0+a_1x+....+a_nx^n$be a non zero polynomial with integer coefficients.if$p(√2+√3+√6)=0$, the smallest possible value of n is? Honestly I have no idea how to begin to solve this ... 5answers 220 views ### Using the fact that$\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}$is irrational. Following from the question, I tried: ... 2answers 781 views ### Show that$ 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$,are rational. I know that can be proved, would like to know that there is ... 0answers 51 views ### If$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)$. But how to approach actually proving it? The proof of this fact doesn't seem to generalize well. 3answers 61 views ### Prove that root of number is rational. Consider$ x_1, x_2, ..., x_n \in \mathbb{R}$. We have to prove that each$\sqrt x $is rational if the sum of$\sqrt x_1 + \ldots + \sqrt x_n $is rational. I think that I could prove it using ... 4answers 4k views ### Proving that$\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 ... 6answers 29k 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? 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 ... 1answer 97 views ### Sum of 5 square roots equals another square root. What is the minimum possible value of the summed square root? In the equation$\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}+\sqrt{e}=\sqrt{f}$, each variable is a distinct positive integer. What is the least possible value for$f?$Out of purely trial and error, I have ... 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 ... 2answers 115 views ### Prove that$[ \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}]=8.$I have to solve the following exercise: Compute$[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}]$and$\operatorname{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}).$Here my attempt: ... 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. 3answers 88 views ### Irrationality of$(a_1+\sqrt{b_1})(a_2+\sqrt{b_2})$Sorry, for a rather silly question. Suppose$a_1$,$b_1$,$a_2$,$b_2$are integers, all different from zero, while$b_1$and$b_2\$ are co-prime positive integers, neither being a complete square. ...

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