All known proofs that the circle cannot be squared are based on Lindemann's theorem that $\pi$ is not analgebraic number.  But this seems to be a case of using an atomic bomb to kill a fly.    What really has to be proved is that $\pi$ is not the root of an algebraic equation with rational coefficients whose degree is a power of $2$.  Lindemann proves it for all degrees, which is much much more than what is needed for the classical construction problem.    It is not too hard to prove that $\pi$ is not the root of a quadratic equation, and perhaps some clever induction argument could carry the day.  The point is that a direct proof for degree $2^n$ has not only never been published, it does not even seem to have been noticed that this much less powerful result is all that is needed.
122586
Is there a direct proof that $\pi$ is not constructible, that is, that squaring the circle cannot be done by rule and compass?  Of course, $\pi$ is not constructible because it is transcendental and so is not a root of any polynomial with rational coefficients. But is there a simple direct proof that $\pi$ is not a root of polynomial of degree $2^n$ with rational coefficients?  The kind of proof I seek is one by induction on the height of a tower of quadratic extensions, one that ultimately relies on a proof that $\pi$ is not rational. Does any one know of a proof along these lines or any other direct proof?  I just want a direct proof that $\pi$ is not constructible without appealing to transcendence.
119563
I'm a bit embarrassed that I've had difficulty on getting around this one:  $$e^{2\pi i x}$$  Solving it by itself, we can reduce it down to $(e^{2\pi i})^x = 1^x$ such that $e^{2\pi i x} = 1$ for all $x$.   However, directly plugging in non-integer rational numbers, chiefly $x = 1/2$, we get results that does not stay true to the above equality.   I'm a bit perplexed of trying to explain the different results for this one. I admit that I haven't had too much experience with complex numbers compared to other fields.
37931
How can I prove that $\sqrt{2}+\sqrt[3]{3}$ is algebraic?  I have been trying various things, and have failed as of now. I know that the degree of the polynomial satisfying this equation has to be $\leq 6$. But that is all I know. Clearly, the simple idea of building the polynomial from the root itself, like in the case of $\sqrt{2}+\sqrt{3}$ does not work here.   Thanks in advance!
13085
It can be proved that there are two irrational numbers $a$ and $b$ such that $a^b$ is rational (see ) and that for each irrational number $c$ there exists another irrational number $d$ such that $c^d$ is rational (see ).  My question is: Is there an irrational number $a$ such that $a^a$ is rational (and how could you prove that)?
17849
$\pi$ can be represented as $C/D$, and $C/D$ is a fraction, and the definition of an irrational number is that it cannot be represented as a fraction.   Then why is $\pi$ an irrational number?
90736
I have seen a proof of $\pi$ being transcendental by conclude that transcendental number powered by algebraic number must be transcendental and algebraic number powered by algebraic number must be algebraic.  So, from the equation $e^{i\pi} = -1$ and $i$ is algebraic so $\pi$ must be transcendental.  But then why is $e^{\pi}$ transcendental?
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.
33284
In my Discrete Math class, we need to prove there aren't nonzero rational numbers $a$ and $b$ such that $5a^2 = 3b^2$.  I thought about using The Fundamental Theorem of Arithmetic but can't think of a way to apply it. How can I approach this proof in a good manner?
59256
Let $n$ be a positive integer. I know the traditional proof that $e$ is irrational. How do we show that $e^n$ is irrational in some sort of similar line? I am of course assuming it is but I would be astounded if not. It occurs to me that since $e$ is transcendental, of course $e^n$ is irrational, but I don't want to use that fact.  Googling gives me something for $e^2$, but I could not easily find anything for $e^3$.
64402
Im asked to prove that a polynomial of an odd degree has a root. Im going to use IVT to prove this, but I'm wondering if I can use the assumption that an odd degree polynomial has either two cases:  $$ \lim_{x\to\infty}h(x)=+\infty \\ \lim_{x\to-\infty}h(x)=-\infty \\ $$ $$and$$ $$ \lim_{x\to\infty}h(x)=-\infty \\ \lim_{x\to-\infty}h(x)=+\infty \\ $$ Then by this there exists numbers $a,b$ such that $$ h(a)&lt;0&lt;h(b) $$ $$or$$ $$ h(b)&lt;0&lt;h(a) $$ Which I think would prove that odd degreed $h(x)$ has at least a root. Or is there a more convincing way to show that there are numbers in $h(X)$ greater than zero and less than zero?  edit: (sorry about the bad limit notation)
7010
I have read a few proofs that $\sqrt{2}$ is irrational.  I have never, however, been able to really grasp what they were talking about.  Is there a simplified proof that $\sqrt{2}$ is irrational?
47757
Im trying to solve a problem X.12.6 from Sarason's Complex Function Theory:  So I tried using Rouché's theorm, on the function $g(z)=z^n$, on the domain $\{z:z&lt;\sqrt{1+|c_{n-1}|^2+...+|c_0|^2}\}$ and using the fact that in that open circle $g$ has $n$ roots ($0$ is zero of order $n$) and then get by the thorem that the polynomial $f$ has n roots in that circle. The problem is that I cant show that $|f(z)-g(z)|&lt;|g(z)|$ on the boundary of the circle. Can you please give me a hint how to show that?
71373
If $n\geq2$, then $n!$ is not a perfect square. The proof of this follows easily from Chebyshev's theorem, which states that for any positive integer $n$ there exists a prime strictly between $n$ and $2n-2$. A proof can be found .  Two weeks and four days ago, one of my classmates told me that it's possible to prove that $n!$ is never a perfect square for $n\geq2$ without using Chebyshev's theorem. I've been trying since that day to prove it in that manner, but the closest I've gotten is, through the use of the prime number theorem, showing that there exists a natural number $N$ such that if $n\geq N$, $n!$ is not a perfect square. This isn't very close at all. I've tried numerous strategies over the past weeks and am now trying to use the Sylow theorems on $S_n$ to somehow show that $|S_n|$ can't be square (I haven't made any progress).  Was my classmate messing with me, or is there really a way to prove this result without Chebyshev's Theorem? If it is possible, can someone point me in the right direction for a proof?  Thanks!
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?
8142
I'm looking for an understandable proof of this theorem, and also a complex one involving beautiful math techniques such as analytic number theory, or something else. I hope you can help me on that. Thank you very much
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.
49923
While I understand the proof of the power rule for logarithms for positive integers, I can not prove them for fractional and negative powers. Also curious to know what would a proof for irrational powers look like?