The question first asks to show that $\sqrt{3}$ has a root in $\mathbb{Q}(\sqrt{3})=\{a+b\sqrt{3}|a,b \in \mathbb{Q}\}$ but not in $\mathbb{Q}(\sqrt{2})$. This is I solved by assuming, towards a contradiction, that $\exists a,b\in \mathbb{Q}$ such that $a+b\sqrt{2} = \sqrt{3}$, so $b = \frac{\sqrt{3} - a}{\sqrt{2}}$. Clearly this is a contradiction since $\frac{\sqrt{3}}{\sqrt{2}}$ is irrational, and $\frac{a}{\sqrt{2}}$ is either zero or irrational, so their difference must be irrational.  Then the queestion asks me to explain why this implies that $\mathbb{Q}(\sqrt{2})$ is not isomorphic to $\mathbb{Q}(\sqrt{3})$. In previous questions, I showed that $\mathbb{Q}(\sqrt{2})$ is isomorphic to $\mathbb{Q}[x]/\mathopen{&lt;}x^{2}-2\mathclose{&gt;}$, where the latter is the set of congruence classes modulo $x^2 - 2$, and the same for root 3. I'm guessing that we come to some contradiction if we assumed that a field not containing $\sqrt{3}$ is isomorphic to $\mathbb{Q}[x]/\mathopen{&lt;}x^{2}-3\mathclose{&gt;}$, but I don't see how.   Any hints or suggestions would be appreciated. Thanks!
18159
In this  we saw isomorphism of vector spaces over $\mathbb{Q}$. Just came across this question:   Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?   I know these as $\mathbb{Q}$-Vector spaces, are isomorphic from the  linked post. But as fields are they isomorphic? I neither know how to prove it nor how to disprove it.
58495
Prove: $\sqrt{2}+\sqrt[3]{5}$ is irrational    I have tried to look at $1+\sqrt{2}+\sqrt[3]{5}$ but I can see how to continue
92917
I need to prove that the roots $\in \Bbb R$ of $x^3+x+1=0$ aren't rational. Obviously, it's easy to use the rational root theorem to prove that there are not rational solutions to this equation, but i want a different approach. I saw a similar question here, but it didn't provide my "solution".  My try  $x^3+x+1=0$  $x^3+x=-1$  $x(x^2+1)=-1$  Here we have two options (product of two numbers to obtain a negative):  $1.$ $x \gt 0$ $∧$ $x^2+1 \lt 0$ (not possible in $\Bbb R)$  $2.$ $x \lt 0$ $∧$  $x^2+1 \gt 0$   So, we use $2.$ to prove that $x \ne 0$, then:  $x(x^2+1)=-1$  $x^2+1=\frac{-1}{x}$  $x^2=\frac{-1}{x}-1$  $x^2=-(\frac{x+1}{x})$  $x=\pm \sqrt {-(\frac{x+1}{x})}$  Then, $\frac{x+1}{x} \ge 0$ because $x \lt 0$ (the equality occurs when $x=-1$, but this doesn't satisfy the original polynomial equation) but here i'm missing the cases $-1 \lt x \lt 0$ (i don't know how to use this to prove that the root is irrational)  This implies that $2$ of the roots are complex, but there are $3$ roots to a third degree polynomial equation.  And here i'm stuck, because i don't know how to prove that the last solution is irrational.  Any hints?  Is there anyway to prove that the last root is irrational?  Is my proof good so far?
30253
Let $\mathbb{Q}(\sqrt{d}) = \{a + b \sqrt{d}: a,b \in \mathbb{Q} \}.$ Show that $\mathbb{Q}(\sqrt{d})$ is a field.    Everything seems obvious except for existence of inverses in the multiplicative group of $\mathbb{Q}(\sqrt{d}).$  Suppose $\alpha \in \mathbb{Q}(\sqrt{d}) \Longrightarrow \alpha = a + b \sqrt{d}.$ To prove that $\mathbb{Q}(\sqrt{d})$ contains multiplicative inverses, observe that for a non-square $d$ and $a - b\sqrt{d} \ne 0$  one has $$\frac{1}{\alpha} = \frac{1}{a+b\sqrt{d}} \cdot \frac{a - b \sqrt{d}}{a - b \sqrt{d}} = \frac{a}{a^2-b^2d}- \frac{b}{a^2-b^2d} \sqrt{d}.$$  If $a,b \in \mathbb{Q} \Longrightarrow a^2,b^2 \in \mathbb{Q} \Longrightarrow b^2 d \in \mathbb{Q}$ (since $d \in \mathbb{Z}$) $\Longrightarrow a^2-b^2d \in \mathbb{Q} \Longrightarrow \frac{a}{a^2-b^2d}, \frac{b}{a^2-b^2d} \in \mathbb{Q}.$ Hence $\frac{1}{\alpha} \in \mathbb{Q}(\sqrt{d}).$   Is this correct?
62207
$$\sqrt{ 3x }= x + \sqrt {3}$$  this is what i tried   $$\sqrt{ x }= (x + \sqrt {3})^2\\ = x^2 + 3 $$  Give x in the form  $$A \sqrt {B} + C $$ Can you show me how this is done step by step.   The answer I have in the book is:  $$\frac {1}{2} \sqrt{3} + \frac {3}{2} $$
16666
Prove that for for all $p,q\in \mathbb{Z}$, $q&gt;0$ we have: $$ \left| \sqrt{2}-\frac{p}{q} \right| &gt; \frac{1}{3q^2}. $$  To be honest, I do not know where to start - any help would be appreciated.
79051
Let $\alpha$ be a root of $x^3 + x + 1 \in \mathbb{Z}_2[x]$ and $\beta$ a root of $x^3 + x^2 + 1 \in \mathbb{Z}_2[x]$. Then we know that $$\mathbb{Z}_2(\alpha) \simeq  \frac{\mathbb{Z}_2[x]}{(x^3 + x + 1)} \simeq \mathbb{F}_8 \simeq \frac{\mathbb{Z}_2[x]}{(x^3 + x^2 + 1)} \simeq \mathbb{Z}_2(\beta). $$   I need to find an explicit isomorphism $\mathbb{Z}_2(\alpha) \to \mathbb{Z}_2(\beta). $   I was thinking of finding a basis for $\mathbb{Z}_2(\alpha)$ and $\mathbb{Z}_2(\beta)$ over $\mathbb{Z}_2$.   I let $\left\{1, \alpha, \alpha^2\right\}$ and $\left\{1, \beta, \beta^2\right\}$ be these two bases. Now suppose I have a field morphism $$ \phi: \mathbb{Z}_2(\alpha) \to \mathbb{Z}_2(\beta) $$ which maps $1$ to $1$. how can I show that the image of $\alpha$, i.e. $\phi(\alpha)$, completely determines this map?
29865
The contradiction method given in certain books to prove that sqare root of a prime is irrational also shows that sqare root of $4$ is irrational, so how is it acceptable? e.g. Suppose $\sqrt{4}$ is rational, $$\begin{align}        \sqrt{4} &amp;=p/q        \qquad\text{where pand q are coprimes} \\         4 &amp;=p^2/q^2\\       4q^2&amp;=p^2       \tag{1} \\          4&amp;\mid p^2\\          4&amp;\mid p\\      \text {let }p&amp;=4m  \qquad\text{for some natural no. m}  \\        p^2&amp;=16m^2\\       4q^2&amp;=16m^2   \qquad\text{(from (1) )}\\        q^2&amp;=4m^2\\          4&amp; \mid q^2\\          4&amp;\mid q \end{align} $$ but this contradicts our assumption that $p$ and $q$ are coprime since they have a common factor $p$. Hence  $\sqrt{4}$ is not rational. But we know that it is a rational. Why?
123218
I am having a problem understanding this proof.   Suppose that $\sqrt{2}$ was rational.   Then there would exist positive integers $a,b$ such that $\sqrt{2}=\frac{a}{b}$. Consequently, the set $S=\{k\sqrt{2} : k \text{ and } k\sqrt{2}$ are positive integers $\}$ is nonempty set of positive integers. Therefore, by the well-ordering property, $S$ has a smallest element, say, $s = t \sqrt{2}$.   We have  $s \sqrt{2}-s = s\sqrt{2}-t\sqrt{2}=(s-t)\sqrt{2}$.  Because $s\sqrt{2} = 2t$ and $s$ are both integers,  $s\sqrt{2} - s = s\sqrt{2}-t\sqrt{2} = (s-t)\sqrt{2}$  must also be an integer.   Furthermore, it is positive, because $s\sqrt{2}-s = s(\sqrt{2}-1)$ and $\sqrt{2} &gt; 1$.   It is less than $s$, because $\sqrt{2} &lt; 2$ so that $\sqrt{2}-1 &lt; 1$. This contradicts the choice of $s$ as the smallest positive integer in $S$. It follows that $\sqrt{2}$ is irrational.  My main problem here is how they got the set $S$. My second confusion is the algebra part $s \sqrt{2}-s$. How did they come up with this?   If someone can explain this proof to me it would be great! thank you
81628
Here's my proof for the irrationality of $\sqrt{6}$: Proof: Let $\sqrt{6}$ be a rational number. Then $\frac{a}{b}$ = $\sqrt{6}$ , where $a$ and $b$ are coprime. Therefore,  $a^2 = 6 b^2$. Here $a^2$ is divisible by both $2$ and $3$. Which means, $a$ too is divisible by both $2$ and $3$ (fundamental theorem of arithmetic). Thus, $6b^2=(3\cdot2\cdot k)^2$ ; where $k$ is an integer. $6b^2=36k^2$ or $b^2=6k^2$. Here $b^2$ is divisible by both $2$ and $3$. Which means, $b$ too is divisible by both $2$ and $3$. Therefore, $a$ and $b$ have common factors, which is a contradiction. Therefore $\sqrt{6}$ is not rational.  Is the above proof correct? Is there an easier way to prove this?
104050
My problem is "Prove $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$ is a simple extension"  I met a similar problem "Prove $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$ is a simple extension" and I showed that it is equal to Prove $\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}$. I let $u=\sqrt{2}+\sqrt{3}$ and deduce that $u-\sqrt{2}=\sqrt{3})$, then square it and prove $\sqrt{2} \in Q(\sqrt{2}+\sqrt{3})$. However, the same technique cannot be applied in the first problem since we have 3 square root.  I actually found a proof for the first problem, by letting $u=\sqrt{2}+\sqrt{3}+\sqrt{5}$ and calculating $u,u^2,..., u^5$ and prove $\sqrt{2},\sqrt{3}\in Q(u)$. Though it's not so nice. Looking for a simpler answer
117563
There is a very simple proof by means of divisibility that $\sqrt 2$ is irrational. I have to prove that $\sqrt 3$ is irrational too, as a homework. I have done it as follows, ad absurdum:  Suppose $$\sqrt 3= \frac p q$$  with $p/q$ irreducible, then  $$\begin{align} &amp; 3q^2=p^2 \\ &amp; 2q^2=p^2-q^2 \\ &amp;2q^2=(p+q)(p-q) \\ \end{align}$$  Now I exploit the fact that $p$ and $q$ can't be both even, so it is the case that they are either both odd, or have different parity. Suppose then that $p=2n+1$ and $q=2m+1$  Then it is the case that  $$\begin{align} &amp;p-q=2(n-m) \\ &amp;p+q=2(m+n+1) \\ \end{align}$$  Which means that  $$\begin{align} &amp;2q^2=4(m-n)(m+n+1) \\ &amp;q^2=2(m-n)(m+n+1) \\ \end{align}$$  Then $q^2$ is even, and so is then $q$, which is absurd. Similarly, suppose $q=2n$ and $p=2m+1$.  Then $p+q=2(m+n)+1$ and $p-q=2(m-n)+1$. So it is the case that  $$\begin{align} &amp;2q^2=(2(m-n)+1)(2(m+n)+1)\\ &amp;2q^2=4(m^2+m-n^2)+1 \\ \end{align}$$  So $2q^2$ is odd, which is then absurd.    Is this valid?
54528
Show that the set $\mathbb{Q}(\sqrt{p})=\{a+b\sqrt{p}; a,b,p\in\mathbb{Q},\sqrt{p}\notin \mathbb{Q}\}$ is a field  I am having a little trouble proving that for each non-zero $b\in\mathbb{Q}(\sqrt{p})$, there is $d\in\mathbb{Q}$ such that $b\cdot d=1$.  My attempt:  Let $b=a_1+b_1\sqrt{p}$  Let $d=a_2+b_2\sqrt{p}$  We want $$b\cdot d=(a_1+b_1\sqrt{p})(a_2+b_2\sqrt{p})$$ $$=a_1a_2+a_1b_2\sqrt{p}+a_2b_1\sqrt{p}+b_1b_2p$$  $$=(a_1a_2+b_1b_2p)+(a_1b_2+a_2b_1)\sqrt{p}$$ $$=1$$  Am I following the right approach? What should be my next step
85167
I'm currently reading Marcus' "Number Fields," and I'm having difficulty proving the following result:  Corollary 2.2: Let $m$ be a squarefree integer. The set of algebraic integers in the quadratic field $\mathbb{Q}(\sqrt{m})$ is $$ \left\{ a + b \sqrt{m} : a,b \in \mathbb{Z} \right\} $$ if $m \equiv 2,3 \pmod{4}$; and$$ \left\{ \frac{a+b\sqrt{m}}{2} : a,b \in \mathbb{Z} ~,~ a \equiv b \pmod{2} \right\}$$ when $m \equiv 1 \pmod{4}$. $\square$  It is easy enough to see that the claimed sets do, indeed, consist of algebraic integers. The other half of the proof proceeds something like this: suppose that $\alpha = r + s \sqrt{m}$ is an element of $\mathbb{Q}(\sqrt{m})$. Then $\alpha$ is a root of the polynomial $$ x^2 - 2rx + r^2 - s^2m $$ In particular, this means that $\alpha$ is an algebraic integer if and only if both $2r$ and $r^2 - s^2 m$ are integers. At this point, the proof stops, and says that this implies the result. And therein lies the problem: I don't see how to make the connection. I do know, for example, that if one of $r,s$ is an integer, then the other one must be, too, i.e., either both are integers, or both are not. However, I cannot get any further than that. Can anyone give me a gentle nudge in the right direction? Thank you very much!
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.
61797
I aim to prove that $\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ is Galois over $\Bbb Q$. To do this, I aim to show that it is the splitting field of the minimal polynomial of $\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$.  Thee minimal polynomial is $144-288 x^2+144 x^4-24 x^6+x^8$. And all its roots are:$\pm \sqrt{(2\pm \sqrt{2})(3\pm \sqrt{3})}$  I do  notice this post:   But I cannot fully understand how I can do that. My questions are:   Call $\sqrt{(2+ \sqrt{2})(3+ \sqrt{3})}=\alpha$. How can we show that $\pm \sqrt{(2\pm \sqrt{2})(3\pm \sqrt{3})}\in\Bbb Q(\alpha)$? There is a comment which mentioned a trick to check if we have found all the roots.     Part 3 is not tedious at all, just apply the automorphism $\sqrt{2}\rightarrow -\sqrt{2}$ to the polynomial evaluated at $α$ to see that it remains zero. Repeat for $\sqrt{3}$.    I would like to learn about the trick mentioned in the comment but I cannot fully understand it. S ocould someone please show me how to do that?  And now I need to determine the Galois group of $\Bbb Q$. Is there any rather simple way to see what the Galois group is?   Thanks for any help!  EDIT: Now I had stucked on it for some time and I think I do need help...  We know that the automorphism is uniquely determined by where it sent the generater to. So we only need to determine its effect on $\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$.   As $(\sigma(\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})))^2=\sigma((\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})}))^2)=\sigma(\mathbb{Q}((2+\sqrt{2})(3+\sqrt{3})))=6+2\sigma(\sqrt{3})+3\sigma(\sqrt{2})+\sigma(\sqrt{6})$. I need to determine where to send $\sqrt{2},\sqrt{3},\sqrt{6}$.   But I only know that I can send $\sqrt{2}\mapsto\pm\sqrt{2}$, same to $\sqrt{3},\sqrt{6}$. But there are then too many possiblities. So what should I do now to rule out some possiblities?  Now I realize that maybe I need not determine where to send $\sqrt{6}$ to because once we determine $\sqrt{2},\sqrt{3}$ we have already know that where to send $\sqrt{6}$...  Is that correct?
76055
Suppose I want to find the minimal polynomial of the number $\sqrt[3]{2} + \sqrt{3}$.   Now that means I want to find a  unique polynomial that is irreducible over $\Bbb Q$ such that $f(x)=0$. Now I know that because   $\sqrt[3]{2} + \sqrt{3}$ belongs to $\Bbb Q( \sqrt[3]{2} , \sqrt{3})$  it might be degree $2$, $3$ or $6$ and does not belong to $\Bbb Q( \sqrt[3]{2})$ so it cannot be of degree $3$ or  $\Bbb Q( \sqrt{3})$ so it cannot be of degree $2$.  So it is of degree $6$. I think my sayings are a bit intuitive and not formal and lack rigorous. Couldn't it belong to another extension of degree $2$? or $3$? I can't answer that. Why checking only those $2$ is enough? Or is it wrong at all to say that?