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### An element not in a field extension [duplicate]

Possible Duplicate: Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$? Consider the field extension $\mathbb{Q}(\sqrt2)$. I want to show that $\sqrt5 \notin \mathbb{Q}(\sqrt2)$. If this were ...
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### $\sqrt 1+\sqrt 2 +\sqrt 3 +\cdots +\sqrt {2009}$ change a sign to be rational [closed]

I have this problem: $$\sqrt 1+\sqrt 2 +\sqrt 3 +\cdots +\sqrt {2009}$$ Prove that you need to change ONLY a sign (to convert a $+$ to $-$) of a single square root, for the sum to be rational. EDIT:...
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### 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 ...
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### 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 ...
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### 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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### 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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### How to get this equation solved?

I came across this equation $$\sqrt a-\sqrt b=\sqrt 7-\sqrt 5$$ And you have to find the value of '$a$' and '$b$' when both of them are primes. The solution was $a=7, b=5$. Now, my question is, ...
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### 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 $i \notin \mathbb{Q}(i\sqrt{7})$?
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### 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 ...