This occurred to me while looking at my answer to      Show that   $\ln(\ln(n))$   is not rational   (and is undoubtedly transcendental)   for all integers   $n \ge 2$.   I am sure that this is true and have no idea how to prove it.  If it were, then $n = e^{e^{a/b}}$, and I don't see how to manipulate this to get a contradiction.  Gelfand-Schneider  doesn't seem to apply.  There may be a theorem that $e^{e^r}$ is transcendental for rational $r$, but I don't know it.  If you can do this, then do the $k$-fold iterated log, $\ln_k$, to show that $\ln_k(n)$ is irrational (and, again, undoubtedly transcendental) for all $n$ large enough so that $\ln_k(n)$ is real.
120562
Is there $n\in \mathbb{N}$ such that $\ln(\ln(n)) \in \mathbb{Q}$? If such $n$ exists, we will get $$\ln(\ln(n))  = \frac{p}{q}, \quad p, q \in \mathbb{Z}.$$  Hence we will get $n = e^{e^{p/q}},$ where the question about the nature of  $e^{e^{p/q}}$ haven't been answered yet.   Is there any other direction ?  Thank you in advance.
16317
Let $X$ be transcendental over a field $F$, and let $E$ be a subfield of $F(X)$ properly containing $F$. Prove that $X$ is algebraic over $E$.   Could we maybe use also the following??   Let $f(x) \in E \Rightarrow f(x) \in F(X) \Rightarrow f(x)=\frac{h(X)}{g(X)}$, where $g(X) \neq 0, h,g \in F[x]$.   Or doesn't this help ??
36779
I found the following proof arguing for the irrationality of $\sqrt{2}$.     Suppose for the sake of contradiction that $\sqrt{2}$ is rational, and choose the least integer $q &gt; 0$ such that $(\sqrt{2} - 1)q$ is a nonnegative integer. Let $q'::= (\sqrt{2} - 1)q$. Clearly $0 &lt; q' &lt; q$. But an easy computation shows that $(\sqrt{2} - 1)q'$ is a nonnegative integer, contradicting the minimality of $q$.   Clearly, this is a vague proof in that it expects the reader to work out the math involved on his own. I took that as an exercise and worked out the following:  Assuming $\sqrt{2}$ to be rational, let $$\sqrt{2} = \frac{p}{q}$$ where $p$ and $q$ are least possible integers.  Clearly then, $\sqrt{2} - 1$ would also be rational, $$\sqrt{2} - 1 = \frac{p-q}{q}$$ where both $(p-q)$ and $q$ are also integers.  Multiplying both sides by $q$, we get $$(\sqrt{2} - 1)q = (p-q) = q'$$ where $q'$ is an integer. Using, $2&gt;\sqrt{2} &gt; 1$ , we get $$1&gt;\sqrt{2}-1&gt;0$$ proving that $q'$ is a non-negative integer.  As stated in the proof, computing $(\sqrt{2}-1)q'$ we get, $$(\sqrt{2}-1)q' = \frac{(p-q)^2}{q}$$which is not necessarily an integer (in fact necessarily not an integer). However, the proof claims it to be.  Am I missing something or something is wrong with how I proceeded?  Here is a link to the textbook where I found this proof: , (p 32, problem 1.14).  P.S. This is my first post here and my first time with TeX. I'd also like to know if there is anything wrong with the way my question is presented.
8141
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.
67796
Consider the Riemann $\zeta$ function. We know that $\zeta(2n)$ is a rational multiple of $\pi^{2n}$ (in particular is transcendental). We also know that $\zeta(3)$ is irrational, and we expect $\zeta(n)$ to be irrational (if not even transcendental) for every $n\in\mathbb{N}$, or at least I would be amazed if - say - $\zeta(5)$ turned out to be rational.     Are there known instances of rational numbers appearing where we would not have expected them?   Of course the notion of "expectation" here is very subjective, so this is a soft question just out of curiosity, since I have the feeling that usually (in my limited experience always) complicated expressions yield irrational ($\mathbb{C}-\mathbb{Q}$) numbers.  As a non-example, we have the series $\sum_{n=1}^\infty 2^{-n}=1$. A series is complicated enough (compared to say a finite arithmetic expression), but of course here we have the explicit formula for geometric series so the resulting $1$ is not really a big surprise.
91259
Let $q$ be a prime and $n\ge 2$ an integer. Moreover, define $f_q(n)$ as the number of $q$-th residues modulo $n$. Is it true that if $K$ is a positive constant then there exist infinitely many $n$ such that $$ f_q(n)\le Kn? $$    I would like to add that the result holds if $q=2$, indeed it is known that $$ \limsup_{n \to \infty}\frac{f_q(n)}{n}=\frac{1}{2} \,\,\text{ and }\,\,\liminf_{n \to \infty}\frac{f_q(n)}{n}=0. $$
4819
I've been reading through a course on exponential functions, starting from integer-valued exponents to rational ones as in: $x^r$ from $r\in \Bbb{N}$ to $\Bbb{Z}$, and combining them to rigorously construct for $r\in\ \Bbb{Q}$. Still, this book adresses high-schoolers, and therefore "summons" an extension of the exponential notation for real exponents. Is there any formal basis for this extension? Could it be related to the density of $\Bbb{Q}$ in $\Bbb{R}$ ? Or the limit of a sequence $(x^{(r_n)})$ where $(r_n)_{n=1}$ is a sequence of rationals that converge to $r$ a non-rational exponent?
81312
Let $n &gt; 0$ and let $a_{n-1}, ..., a_0$ be integers. Prove that, if $x^n + a_{n-1}x^{n-1} + ... + a_0 = 0$ has a rational root, then such root is an integer.   I understand that posting questions without showing your work is frowned upon, but I don't know where to start with this one.. thanks in advance.
8257
How can I prove that $\log^k(n) = O(n^\epsilon)$? (for any $\epsilon &gt; 0 $ and for any integer and positive $k$).  I tried to do it by - definition , and to prove that exists $c,n_0 &gt; 0$ so that for any $n &gt; n_0 $ exists: $c \cdot n^\epsilon &gt; \log^k(n) $, but I don't have idea how to continue for this point.    I will be happy to listen ideas :).
57490
This question is motivated from an exercise from Rudin. The exercise says that prove that set of all algebraic numbers is countable.   Proof: We know that a number $z$ is called algebraic if it is the root of a polynomial $a_0z^n+a_1z^{n-1}+\cdots +a_n=0$, where $a_i$'s $\in \Bbb{Z}$. Since there are only finitely many equations of the form $n+|a_0|+|a_1|+\cdots+|a_n|=N$ for a positive integer $N$ this implies there are only finitely many such polynomials. Now since each of these polynomials have finitely many roots therefore there are countably many algebraic numbers.   Now the next exercise in the process asks us to prove that there is a real number which is not algebraic which is very trivial. So this proves that there exists transcendental numbers. Also this shows that there are uncountably many of them. So that means there are fairly many of them.   But where are they? Now if there were uncountably many of them then it should be fairly easy to find them but it is not . Because we can get all sorts of wonky numbers being the root of polynomials.    mentions all the transcendental numbers and the numbers which might be transcendental. These do not seem to be many. So what about the rest of the transcendental numbers. Where are they? And why is it so difficult to find them and prove that they are transcendental?   A side question: Is there some nice property that the class of all these transcendental numbers might satisfy . By a nice property I mean that if I take all transcendental numbers and form a group then what about the isomorphisms on this group. Similarly what about if I take all transcendental numbers and form a topological space?  Note: Now if this is a question that has already been asked then please do notify me in the comments I will delete this question
73678
A friend of mine told me that $X \cong Y \Rightarrow 2^X \cong 2^Y$ ($X$ and $Y$ being sets), which is very easy to prove, but he was wondering about the converse in ZF, i.e., can one take logarithms? Since the (apparently) simpler question of whether it is possible to divide by a natural number is not particularly trivial without assuming the axiom of choice (see Doyle, Conway: ), I would imagine that this problem doesn't have an easy answer either.
122250
This is exercise 1.9 in Lie Groups, Lie Algebras and Representations - Hall.     Suppose $a$ is an irrational real number.  Show that the set $E_a$ of the numbers of the form $e^{2\pi i n a}$, $n \in \mathbb{Z}$, is dense in the unit circle $S_1$.  Hint: Show that if we divided $S^1$ into $N$ equally sized "bins" of length $2\pi/N$, there is at least one bin that contains infinitely many elements of $E_a$.  Then use the fact that $E_a$ is a subgroup of $S^1$.   My proof of this proposition is as follows.  Since $a$ is irrational, you can determine that the set of rotations $E_a$ is infinite.  Since $S^1$ is compact we can find two $r_1, r_2$ that are within $\epsilon$ of each other.  Then $r_1^{-1}r_2$ is a small rotation of size $\epsilon$.  Now, $r_1^{-1}r_2$ generates rotations that are within $\epsilon$ distance of any point of $S^1$.    That said, I don't believe that the hint suggested in the problem uses that technique.  My knowledge of algebra is not all that strong so I was hoping someone could shed some light on what is being suggested there.
95943
Could you please confirm if this proof is correct?  Theorem: If $q \neq 0$ is rational and $y$ is irrational, then $qy$ is irrational.  Proof: Proof by contradiction, we assume that $qy$ is rational. Therefore $qy=\frac{a}{b}$ for integers $a$, $b \neq 0$. Since $q$ is rational, we have $\frac{x}{z}y=\frac{a}{b}$ for integers $x \neq 0$, $z \neq 0$. Therefore, $xy = a$, and $y=\frac{a}{x}$. Since both $a$ and $x$ are integers, $y$ is rational, leading to a contradiction.
89219
Could someone provide, please, a proof of the theorem below?  "Being $x$ and $b$ integers greater than $1$, which can not be represented as powers of the same basis (positive integer) and integer exponent, then the logarithm of $x$, in base $b$, is an irrational number."  Well... Assuming that the logarithm is a rational number $p/q$ ($p$ and $q$ are relatively prime integers), I know I can write $x^q = b^p$, but I can not conclude from this fact that $b$ and $x$ are powers of a same integer (and with integer exponent).
105272
It seems like $$\log n \leq \sqrt n \quad \forall n \in \mathbb{N} .$$ I've tried to prove this by induction where I use $$ \log p + \log q \leq \sqrt p \sqrt q $$ when $n=pq$, but this fails for prime numbers. Does anyone know a proof?
99040
Suppose I was working in radians and took  $$\sin \left(\frac{a\pi}{b}\right)$$  where both $a$ and $b$ are integers.  Is there a proof that the output of this function cannot be transcendental? Or, conversely, that   $$\arcsin \left(t \right)$$  where $t$ is some transcendental number, e.g., $e$, cannot be a rational angle $\left(\frac{a\pi }{b}\right)$?  Sadly, I couldn't get very far with this problem with my maths teacher so any information on the topic would be very useful.
13721
$$\dfrac{1}{2}\ln(x+7)-(2 \ln x+3 \ln y)$$  Our professor let's us solve this but i do not understand how $\ln$ works. He says it has same properties with $\log$ but i still don't get it. What's the difference of the two?