$\sqrt{2} + \sqrt[3]{3}$ is irrational ?  These are my steps -  $\sqrt{2} + \sqrt[3]{3} = a$  $3 = (a-\sqrt{2})^{3}$  $3 = a^{3} -3a^{2}\sqrt{2} + 6a -2\sqrt{2}$  $3a^{2}\sqrt{2}+2\sqrt{2} = a^{3}+6a-3$  $\sqrt{2}(3a^{2}+2) = a^{3}+6a-3$  Then, $\sqrt{2}$ in the left side is irrational , and mulitply irratinal with rational is irrational. The right side is rational. So,  $irrational \neq rational$.  This is a good proof ?
121556
Goal: Prove that $5^{1/3}+7^{1/2}$ is irrational.   Idea: We can prove this is irrational by supposing it is rational and finding a contradiction. So, $5^{1/3}+7^{1/2} = p/q$ where $p$ and $q$ are integers that have no factors in common other than $1$. The issue here is that I can not seem to find a way manipulate this equation to a form where I can find contradiction. I tried taking each side to the power of 6... That was an unmanageable mess.  Any ideas?
92812
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 hoping to prove this using methods from pre-calculus level algebra.
29297
So while doing all sorts of proving and disproving statements regarding irrational numbers, I ran into this one and it quite stumped me:  Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational.  I tried all the usual suspects like playing with $\sqrt[3]{2} + \sqrt[3]{4} = \frac{a}{b}$ for $a,b\in \mathbb{Z}$ , but got nowhere.  I also figured maybe I should play with it this way:  $2^\frac{1}{3} + 4^\frac{1}{3}=2^\frac{1}{3} + (2^2)^\frac{1}{3}=2^\frac{1}{3} + 2^\frac{2}{3}=2^\frac{1}{3} + 2^\frac{1}{3}\times 2^\frac{1}{3}=2^\frac{1}{3}(1+2^\frac{1}{3})$  But there I got stumped again, because while $1+2^\frac{1}{3}$ is irrational, nothing promises me that $2^\frac{1}{3} \times (1+2^\frac{1}{3})$ is irrational, and I feel like trying to go further down this road is moot.  So what am I missing (other than sleep and food)? What route should I take to  prove this? Thanks in advance!
51542
Why $\sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }}$ is a rational number?
104097
We're all familair with this beautiful proof whether or not an irrational number to an irrational power can be rational. It goes something like this:  Take $(\sqrt{2})^{\sqrt{2}}$  If it's rational, then you proved it, if it's irrational, take $((\sqrt{2})^{\sqrt{2}} ){^\sqrt{2}} = 2$ and you've proved it.  I'm wondering if you can raise $\pi$ or $e$ to a certain non-trivial real power to make it rational? And if not, where is the proof that it can't be done?   p.s. - I almost left out the real part, but then I realized that $e^{i\pi} = -1$.
45633
I'm reading R. Courant &amp; H. Robbins'  for fun. I'm on page $60$ and $61$ of the second addition. There are three exercises on proving numbers irrational spanning these pages, the last is as follows.     Exercise $3$: Prove that $\phi=\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational. Try to make up similar and more general examples.   My Attempt:     Lemma: The number $\sqrt{2}+\sqrt{3}$ is irrational. (This is part of Exercise 2.)   Proof: Suppose $\sqrt{2}+\sqrt{3}=r$ is rational. Then $$\begin{align} 2&amp;=(r-\sqrt{3})^2 \\ &amp;=r^2-2\sqrt{3}+3 \end{align}$$ is rational, so that $$\sqrt{3}=\frac{r^2+1}{2r}$$ is rational, a contradiction. $\square$  Let $\psi=\sqrt{2}+\sqrt{3}$. Then, considering $\phi$, $$\begin{align} 5&amp;=(\phi-\psi)^2 \\ &amp;=\phi^2-\psi\phi+5+2\sqrt{6}. \end{align}$$  I don't know what else to do from here. My plan is/was to use the Lemma above as the focus for a contradiction, showing $\psi$ is rational somehow.  Please help :)  Thoughts:  The "try to make up similar and more general examples" bit is a little vague.    The question is not answered  as far as I can tell.
110389
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?
10813
The following is a Homework Question that I've been working on and I would like some feedback on my answer: Prove or find a counterexample: For all real numbers $x$ and $y$ it holds that $x + y$ is irrational if, and only if, both $x$ and $y$ are irrational.  So I've got (or at least like to think I've got) a counter example.  Let $x = 1$ and $y = \sqrt{2}$.   Thus $x$ is rational and $y$ is irrational. Adding $x$ and $y$ gives us $1 + \sqrt{2}$ which cannot be simplified any further (right?) and is an irrational number.  Thereby proving that $x + y$ can be irrational without both x and y being irrational.   Is this okay? Any feedback is greatly appreciated thank you!
107414
I know  that uses $x={\sqrt{2}}^{\sqrt{2}}$ to prove that there must exist an irrational to an irrational power that evaluates to a rational. But I don't know if $x$ itself is known to be irrational or rational.  Could someone enlighten me? Thanks!
48026
This question is from Apostol's Mathematical Analysis. I have a solution for it at   The question is:  Prove that $\sqrt {n − 1} +\sqrt {n + 1}$ is irrational for every integer n ≥ 1.  And the proof I have is  Now, I can't understand how to get h=$3/2$ and k=$1/2$ ?  Please help.
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!
121472
A popular proof of the irrationality of $\sqrt2$ is to first assume that the number is rational. This means that $\sqrt2=a/b$ where $a$ and $b$ are integers. Another assumption is that $a$ and $b$ are coprime. It turns out that this leads to a contradiction. And the conclusion is that $\sqrt2$ is not rational (because the assumption of rationality led to a contradiction). But what about the second assumption? Logically, the conclusion could also be that $a$ and $b$ are not coprime..  How can this be resolved?  Edit:  Thank you all for your comments. It cleared up my mind.  Three aspects of this kind of proof are: mathematical, logical, and didactical.  Mathematically, it is the case that any rational number can be written as a fraction with coprime numerator and denominator. But this point is often described in a way that is either literally a logical assumption, or in a way that is natural to interpret as a hidden logical assumption. In this case, the proof will have a flawed logic, because any of the two assumptions could be the cause of the contradiction. Didactically, this can cause confusion in the mind of the student, who might not be so accepting of the proofs conclusion. I used to be that student.  So, how to resolve it? Many of you provided explanations on the mathematical aspect, which are all correct. But my main point is not of not understanding the proof, but of the use of flawed logic which leads to not understanding the proof. I am sure that a lot of teachers think or say something along the lines of WLOG, but for many students, this makes the proof weaker, because it is not so convincing. When we present this proof to our students, we need to have this in mind.  Thank you.
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
31336
In a sequence $a_1, a_2,\dots$ of real numbers it is observed that $a_p=\sqrt{2}$, $a_q=\sqrt{3}$, and $a_r=\sqrt{5}$, where $1\leq p&lt;q&lt;r$ are positive integers. Then $a_p$, $a_q$, $a_r$ can be terms of (A) an arithmetic progression (B) a harmonic progression (C) an arithmetic progression if and only if $p$, $q$, and $r$ are perfect squares (D) neither an arithmetic progression nor a harmonic progression  I have tried using the definition of AP and argued that if the first option is true then $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}=\frac{q-p}{r-q}$  is true which implies that the left hand side is an rational number. I don't know whether the left hand side is a rational number or not. I am stuck here. Also, I don't know how correct this approach is. How do I approach and solve this problem?
122604
How can I prove      $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}=2$$   I don't know which method can be used for this?
98195
Use the unique factorization for integers theorem and the definition of logarithms to prove that $\log_3 (7)$ is irrational.     I am taking a beginners fundamental mathematics module, no advanced stuff please. Thanks!   My attempt.  Suppose for a contradiction that it is rational, that is $\log_3(7)=\frac{a}{b}$ for some $a,b\in R$ where $b\neq0$. Therefore, by the definition of logarithms, $7=3^{\frac{a}{b}}$. By the theorem of unique factorization, $7=1*7$ is unique.  Ok I'm plain stuck! Any help please?
23768
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?