# Linked Questions

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

I was trying to prove that $\sqrt {2} ,\sqrt {4}$ 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 ...
6answers
655 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 ...
3answers
858 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 ...
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 $\mathbb{R}^+$) ...
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 ...
4answers
84 views

### Are these two fields the same?

I wanted to know if the field $\mathbb{Q}(i\sqrt{7}) = \mathbb{Q}(\sqrt{7}, i)$ are the same. I don't think they are because $i \notin \mathbb{Q}(i\sqrt{7})$?
1answer
247 views

### Infinitely many transcendental numbers over Q

My previous question was not well-framed so I will ask again: Can you explicitly produce an infinite set of real numbers which is algebraically independent over $\mathbb Q$?
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Let $p_1,\ldots,p_n\in\mathbb{N}$ be different prime numbers, it can be shown that $[\mathbb{Q}(\sqrt{p_1},\ldots,\sqrt{p_n}):\mathbb{Q}]=2^n$ and in any case it is clearly finite since $[\mathbb{Q}(\... 1answer 365 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 ... 1answer 188 views ###$\sqrt{m_1}+\sqrt{m_2}+ \cdots + \sqrt{m_n}$is Irrational If$m_1 , m_2, \cdots m_n$are natural numbers where at least one of them is not a perfect square, then how do I prove that the sum $$\sqrt{m_1}+\sqrt{m_2}+ \cdots + \sqrt{m_n}$$ is irrational? I'm ... 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,(\...

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