I need to solve the following problem:  Can $\sqrt[3]{2}$ be expressed as a linear combination of roots of unity (with coefficients in $\Bbb Q$)?  I am completely lost because I have no idea of how to start this problem. Can someone help me, please?
91845
Is $\sqrt[3]{2}$ contained in $\mathbb{Q}(\zeta_n)$ for some $n$, where $\zeta_n=e^{2\pi i/n}$?   I think the answer is no, but I can't give a full proof. Assume the contrary, we then have $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{Q}(\zeta_n)$, can I say $\mathbb{Q}(\zeta_n)/ \mathbb{Q}$ is a cyclic extension but $\mathbb{Q}(\sqrt[3]{2}) / \mathbb{Q}$ is not, so this is a contradiction?
73456
If $a,b$ are the roots of the equation $2 x^2 -3 x +1 = 0$, find an equation whose roots are $a/(2b +3)$, $b/(2a +3)$   I was practicing quadratic equation questions online but I am stuck on this question.  I found the values of roots $a= 1$, $b= 1/2$
98573
I'm trying to show that $Q[\sqrt3]$ is a field. I am particularly struggling with showing that every element is a unit.   So for some element, we want to show that for some element $a+b\sqrt3$, there exists $c+d\sqrt3$ in $Q[\sqrt3]$ such that $(a+b\sqrt3)(c+d\sqrt3) = 1$.  Is that the right direction? I am not really sure of how to come up with that inverse though. Any help would be appreciated.
58292
If $\alpha$ is a root of the equation $4x^2+2x-1=0$, then prove that $4\alpha^3-3\alpha$ is the other root. How do I proceed? The sum of the roots, the product of the roots lead me nowhere. Should I find the roots of the equation and substitute in the given two expressions of $\alpha$ and check whether manipulating the first root gives me the second root (which seems much complicated), or is there any other way?
80166
I was reading an example, where it is calculating the square roots of $\sqrt 3 +3i$.  $w=\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)\\=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$  Let $z^2=w \Rightarrow r^2(\cos(2\theta)+i\sin(2\theta))=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$.  But how did they get from $\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$?  And can one just 'let $z^2=w$' as above?  Edit: $w=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})=z^2\\ \Rightarrow z=\sqrt{2\sqrt 3}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})\\ \Rightarrow \sqrt{2\sqrt 3}\frac{\sqrt 3}{2} +i \sqrt{2\sqrt 3} \frac{1}{2}$
41544
Denest $\sqrt {\sqrt[3]{5}-\sqrt[3]{4}}$.   I have tried completing square by several method but all failed. Can anyone help me please? Thank you.  p.s. I'm a poor question-tagger.
71294
The question asks to compute:   $$\sum_{k=0}^{n-1}\dfrac{\alpha_k}{2-\alpha_k}$$   where $\alpha_0, \alpha_1, \ldots, \alpha_{n-1}$ are the $n$-th roots of unity.   I started off by simplifiyng and got it as:  $$=-n+2\left(\sum_{k=0}^{n-1} \dfrac{1}{2-\alpha_k}\right)$$  Now I was stuck. I can rationalise the denominator, but we know $\alpha_k$ has both real and complex components, so it can't be simplified by rationalising. What else can be done?
92469
If $w = \frac {\sqrt2+\sqrt3}{\sqrt5-\sqrt3} $  Express the following in the simplest form (with a rational denominator)  i) $w$  ii) $\frac1w$  I'm confused about (ii) question :/ pls help me.
94937
From a previous question, we have that $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[4]{3})$, and I am assuming that we would need to use this to justify the answer to the question.   Would it be right to use some idea of a splitting field here? Or is it something to do with $\mathbb{Q}(\sqrt[3]{3}, \sqrt[4]{3})$ not containing a primitive 3rd root of unity?   This is a past exam paper question worth 8 marks, so it doesn't seem like it's a few lined answer, but I am a little stuck on where to go.
80032
Why must $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ -- the set of all polynomials in $\sqrt{2}$ and $\sqrt{3}$ with rational coefficients  -- contain multiplicative inverses?   I have gathered that every element of $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$ takes form $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{2}\sqrt{3}$ for some $a,b,c,d \in \mathbb{Q}$, and I have shown that $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are both fields, but it's not clear to me why this allows for general multiplicative inverses to exist in $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$.
7382
Problem: prove $\sqrt{2\sqrt{3\sqrt{\cdots\sqrt{n}}}}&lt;3$ by induction. I tried some, but stopped in $\sqrt[2^n]{n+1}$. Also tried with $2\sqrt{3\cdots}&lt;3^2$ and so on.
36223
Is there any $2 \times 2$ matrix $A \neq I$ such that $A^3=I$?  In my opinion: no. Thank you very much.
74230
In Abbot's Understanding Analysis I am asked to show that $\sqrt{2}+\sqrt{3}$ is an algebraic number. I have shown that those two are algebraic separately (that was simple), but I can't figure out what to do to show that their sum is algebraic, too. For example, I tried $(\sqrt{2}+\sqrt{3})^{2}=5+\sqrt{24}$ and then I tried to think of a polynomial of form $ax^2-bx^{0}=0$ e.g. $c(\sqrt{2}+\sqrt{3})^{2}-c(5+\sqrt{24})=0$ that would work, but couldn't find the $c$ value that would make $b=c(5+\sqrt{24})$ integer, but couldn't find one. Maybe I can somehow conjecture that the sum of two algebraic numbers must be algebraic, too, but I was wondering if there's a way to find a polynomial to show this. Thanks!
79185
Assumption: $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}\ge\frac{2}{3}k\sqrt{k}$   Prove true for $n=k+1$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}(k+1)\sqrt{k+1}$$  I'm upto :$$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}k\sqrt{k}+\sqrt{k+1}$$after which I'm stuck.
70693
How to compute $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}$$ &amp; $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}$$?  I understand that  $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}=(1^{1/2})(2^{1/4})(3^{1/8})\cdots$$  and  $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}=(1+(2+(3+(\cdots))^{1/8})^{1/4})^{1/2} $$  How to Proceed further?
47441
Is there a general or elegant way to approach this problem? One can show that $\sqrt 5+\sqrt[3] 2$ is a root of the hexic $x^6-6x^4-10x^3+12x^2-60x+17$, which should then be its minimal polynomial to get the result. I see no straightforward way to show this is irreducible, though. It has a root in $\mathbf{F}_4$, and to show it's irreducible over $\mathbf{F}_3$ already seems to require a lot of grunt work.  One thing I can do is rule out the possibility that $[\mathbf{Q}[\sqrt 5+\sqrt[3] 2]:\mathbf{Q}]=2$, because $[\mathbf{Q}(\sqrt 5,\sqrt[3] 2):\mathbf{Q}]=6$ and $\mathbf{Q}(\sqrt 5,\sqrt[3] 2)$ is the compositum of $\mathbf{Q}[\sqrt 5+\sqrt[3] 2]$ and $\mathbf{Q}[\sqrt 5]$, where the latter has degree two. But I don't seem to be able to play an analogous trick to rule out $3$, because I know of no general lower bound on the degree of a compositum.   Could there be some way to bring Galois theory into the picture? These extension aren't Galois, and so it's by no means clear to me how to reason from their normal closures, but perhaps that's the direction to go in. Suggestions are very much appreciated.
47198
I thought of this a while ago, but can't make up a proof or a counterexample. Does anyone know more about this?  $$\sqrt{a+b}  \le \sqrt a + \sqrt b , \forall a,b \in \mathbb R_+$$  Moreover, what happens with more variables? Say:  $$\sqrt{x_1+x_2+...+x_n} \le \sqrt x_1 +\sqrt x_2 + ... + \sqrt x_n $$  with $ x_i \in\mathbb R_+ \forall i \in \{1,2,...,n\} $  Or when $i = \mathbb N$  PS: I tried looking on the internet for this but I don't know how I would call this.