No doubt an easy question: I'm trying to follow Wikipedia's (second) proof of the irrationality of $\sqrt{3}$ and it relies on the notion that since $3n^2 = m^2$ is divisible by 3 then so is $m$. Why is this so and can you help me prove it?
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This is a problem from "" (specifically, exercise set 2 #9), which is one of my math texts. Please note that this isn't homework, but I would still appreciate hints rather than a complete answer.  The problem reads as follows:     If 3p2 = q2, where $p,q \in \mathbb{Z}$, show that 3 is a common divisor of p and q.   I am able to show that 3 divides q, simply by rearranging for p2 and showing that  $$p^2 \in \mathbb{Z} \Rightarrow q^2/3 \in \mathbb{Z} \Rightarrow 3|q$$  However, I'm not sure how to show that 3 divides p.    Edit:  Moron left a comment below in which I was prompted to apply the solution to this question as a proof of $\sqrt{3}$'s irrationality. Here's what I came up with...  [incorrect solution...]  ...is this correct?  Edit:  The correct solution is provided in the comments below by Bill Dubuque.
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Please correct any mistakes in this proof and, if you're feeling inclined, please provide a better one where "better" is defined by whatever criteria you prefer.    Assume $2^{1/2}$ is irrational.  $2^{1/3} * 2^{x} = 2^{1/2} \Rightarrow x = 1/6$.  $2^{1/3} * {2^{1/2}}^{1/3} = 2^{1/2}$. if $2^{1/2}$ is irrational, then ${2^{1/2}}^{1/3}$ is irrational.  $2^{1/3} = 2^{1/2} / {2^{1/2}}^{1/3}$.  $2^{1/3}$ equals an irrational number divided by an irrational number.  $2^{1/3}$ is an irrational number.
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The fact that there exists irrational number $a,b$ such that $a^b$ is rational is proved by the law of excluded middle, but I read somewhere that irrationality of $\sqrt{2}^{\sqrt{2}}$ is proved constructively. Do you know the proof?
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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
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If $p$ is a prime number, then $\sqrt{p}$ is irrational.  I know that this question has been asked but I just want to make sure that my method is clear. My method is as follows:  Let us assume that the square root of the prime number $p$ is rational. Hence we can write $\sqrt{p} = \frac{a}{b}$. (In their lowest form.) Then $p = \frac{a^2}{b^2}$, and so $p b^2 = a^2$. Hence $p$ divides $a^2$, so $p$ divides $a$. Substitute $a$ by $pk$. Find out that $p$ divides $b$. Hence this is a contradiction as they should be relatively prime, i.e., gcd$(a,b)=1$.
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Can the expression $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m \in \mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $\sqrt{m} = a-\sqrt{n}, \ \ a \in \mathbb{Q}$, which I don't think is possible, but how to show it?
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We all know that $\sqrt[3]{3}\not\in \mathbb{Q}({\sqrt[3]{2}})$ intuitively. We even know that $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$ for two distinct primes $p, q$. However, I don't know how to prove these things rigorously. In case of $\sqrt{p}\not\in \mathbb{Q}(\sqrt{q})$, we can just set assume $\sqrt{p}=a+b\sqrt{q}$ for some $a, b\in\mathbb{Q}$ and deduce a contradiction. However, in cubic case, expansion of $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^{3}$ is very complicated and even I expand this I couldn't get any contradiction by this.  To be specific, it gives us 3 diophantine equations  $$a^{3}+2b^{3}+4c^{3}+12abc=3$$ $$a^{2}b+2b^{2}c+2c^{2}a=0$$ $$ab^{2}+2bc^{2}+ca^{2}=0$$  and I don't know how to check solvability of this kind of diophantine equations (over $\mathbb{Q}$). Even if there exists a way to check solvability, I cannot convince that it can be used to prove $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$.   So I tried to use Galois theory, but actually I don't know where to go. I assumed $\mathbb{Q}(\sqrt[3]{3})=\mathbb{Q}(\sqrt[3]{2})$ and take a Galois closure, i.e. $K=\mathbb{Q}(\sqrt[3]{3}, w)=\mathbb{Q}(\sqrt[3]{2}, w)$ where $w^{3}=1$ then tried to use field norm maps. We can check that $\beta=\sqrt[3]{3}\cdot\sigma(\sqrt[3]{3})\cdot\sigma^{2}(\sqrt[3]{3})\in \mathbb{Q}$ when $\sigma\in Gal(K/\mathbb{Q}), \sigma(\sqrt[3]{2})=\sqrt[3]{2}, \sigma(w)=w^{2}$. But I cannot get any information from this.   Edit : It would be great if there is a simple algorithm to check whether $\alpha\in K$ or not for a given algebraic number $\alpha$ and a number field $K$.
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I have to prove that any number of the form $3n+2$ has a prime factor of the form $3m+2$. Ive started the proof  I tried saying by the division algorithm the prime factor is either the form 3m,3m+1,3m+2.  Case 1: $3m$      suppose $3m$ does divide $3n+2$. That means $3mr=3n+2$ where $r$ is some integer. But we get $mr-n= 2/3$ and since $mr-n$ is an integer that's a contradiction. Thus $3m$ does not work.  Case 2: $3m+1$ Same argument here  Case 3: $3m+2$ By the fundamental theorem of arithmetic every integer is divisible by a prime. But not sure what to do from here.
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Rationalize   $\dfrac{1}{\sqrt[3]{p^2}+\sqrt[3]{pq}+\sqrt[3]{q^2}}.$  How would I go about doing this without wading through lots of algebra?  Is there a trick similar to how you can multiply by $\dfrac{\sqrt a-\sqrt b}{\sqrt a-\sqrt b}$ with square roots?  Thanks in advance!
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Possible Duplicate:          Will every $n^{th}$ root of $2$ be an irrational number? If yes, how can I prove that?
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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 $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$.
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Prove that $\sqrt[31]{12} +\sqrt[12]{31}$ is irrational.   I would assume that $\sqrt[31]{12} +\sqrt[12]{31}$ is rational and try to find a contradiction.  However, I don't know where to start. Can someone give me a tip on how to approach this problem?
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I understand that the set of irrational numbers with multiplication does not form a group (clearly, $\sqrt{2}\sqrt{2}=2$, so the set is not closed). But is there a proof or a counter-example that the irrationals with addition form (or do not form) a group?  Thank you!  Edit: In particular, I am wondering if the set is closed with respect to addition.
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As the title implies.  I'm proving it by contradiction and I've already ruled out $p \equiv 0$ and $2$ (mod $4$).  I'm having trouble proving that it cannot be $\equiv 3$ (mod $4$). Namely, if $p \equiv 3$ (mod 4), then $m^2 + n^2 \equiv 1$ or $3$ (mod $4$). I know that $m^2 + n^2$ cannot be congruent to $3$ (mod $4$); but what about $1$?  Thank you.
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How can I prove that the sum  $\sqrt 2+\sqrt[3] 3$  is an irrational number ??
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I'm trying to prove that $\sqrt[n]{\frac{s}{t}}$ is irrational unless both s and t are perfect nth powers. I have found plenty of proofs for nth root of an integer but cannot find anything for rationals. Also trying to work up from the proofs I have found is rather difficult.  My lecturer wants me to prove this using uniqueness of prime factorisation but I also have no idea where to start with that.   Any help is greatly appreciated.
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I have just proved the divisibility rule for 3 and 9.  Let $n\in\mathbb{N}$. Let $m$ be a number that appears when you shuffle the digits in $n$. Show that $n-m$ is a multiple of 9.  Can anyone offer some help?