I can't seem to figure out how to input a root that isn't square in Ubuntu's default calculator. I know I can input 4^(1 / 3) but I'd prefer to input them in a more.. "natural" way.  (I can't think of a good way to put it.)  Is there a  way to work with non-square roots in the advanced mode of the calculator without using exponents?
120849
I don't know which buttons I have to click or what key combination to type to calculate a cube root.
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!
30432
I feel like this is an extremely simple question but somehow I cannot solve it.      Is it possible to unnest $$\sqrt{1+\sqrt2}\quad?$$    This was motivated by the .   If we let $x^2=1+\sqrt2\implies x^4-2x^2-1=0$ and after solving with substitution $u=x^2$, the argument is circular.
59728
Why does the calculator do a cc (counterclockwise) rotation for positive trig ratios instead of clockwise,  and a clockwise rotation for negative sine &amp; tan instead of cc and a counterclockwise rotation for negative cos ratios instead of a clockwise   ie. in degree mode  $\cos^{-1}(-5/12)=114.62$  $\sin^{-1}(-5/12)=-24.62$  $\tan^{-1}(-5/12)=-22.61$  Is it maybe picking the value that involves the least amount of computing power? or is it a matter of convention? or am I overlooking something?
43762
I know there is something wrong with this but I don't know where. It's some kind of a math fallacy and it is driving me crazy. Here it is: $$-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1?$$
6221
It is perhaps a bit embarrassing that while doing higher-level math, I have forgot some more fundamental concepts. I would like to ask whether the square root of a number includes both the positive and the negative square roots.  I know that for an equation $x^2 = 9$, the solution is $x = \pm 3$. But if simply given $\sqrt{9}$, does one assume that to mean only the positive root? And when simply talking about the square root of a number in general, would one be referring to both roots or just the positive one, when neither is specified?
83192
My question looks quite obvious, but I'm looking for a strict proof for this:     Why can't the sum of two square roots of non-perfect squares be an integer?   For example: $\sqrt8+\sqrt{15}$ isn't an integer. Well, I know this looks obvious, but I can't prove it...
87190
Does math have an incorrect notation / syntax? I don't mean writing misaligned notation (google), but when you take something like a number to powers to powers to powers, $${{2^2}^2}^3$$ (I was told this is incorrect notation by a teacher). Is it really incorrect, or does it just need to be simplified with parentheses? Do people write maths like this?  a radical expression with the root being a radical expression? $$\sqrt[\sqrt{2^3}]{2}$$
27856
A previous problem had us solving $\sum_{n=0}^{\infty} \frac{1}{4^n}$ which I calculated to be $\frac{4}{3}$ using a bit of mathematical manipulation.   Wonderful. Thank you for all the prompt responses.  Could anyone suggest an alternate technique that does not involve differentiation?
1747
Is there any rule for powers so that i can compare which one is greater without actually calculating? For example    54^53 and 53^54  23^26 and 26^23 3^4 and 4^3 (very simple but how without actually calculating)
43390
Okay so I am in college and in the notes it shows this example:  // Example: Compute the square root of 3 modulo 143  3 modulo 143 = 3 mod (11*13)  Then he jumps to this:   √3 (mod 11) = ± 5  √3 (mod 13) = ± 4  Using the Chinese Remainder Theorem, we can calculate the four square roots as 82, 126, 17 and 61.  //  The lecturer never makes anything clear even though it is our first encounter with module arithmetic. I don't know were the 4 or the 5 come from or the √3 . Can anyone explain to be how he got  √3 (mod 11) = ± 5  √3 (mod 13) = ± 4   And also how does he use this information in conjunction with the Chinese Remainder Theorem arrive at the square roots. I would really appreciate if anyone could help me out here
90180
Are there some books or articles regarding "polynomials" with non-integer (real) exponents, i.e., $$f(x)=a_1x^{e_1}+a_2x^{e_2}+\dots+a_nx^{e_n},$$ where $e_1,e_2,\dots$ are any real numbers (and $x$ being also real)?  I am mostly interested in theorems regarding the location of the roots, number of the roots, bounds, etc., of such "polynomials". Thanks.
14652
If I want to find the continued fraction of $\sqrt{n}$ how do I know which number to use for $a_0$? Is there a way to do it without using a calculator or anything like that? What's the general algorithm for computing it? I tried to read the wiki article but was overwhelmed and lost. I tried Googling but couldn't find a website that actually explained this question.  If anyone has a good site that answers these questions either, please let me know. Thanks!
95318
I'm trying to show that if $\alpha$ is the square root of a non-perfect square $d$, then for any rational $p/q$ there's a positive real $x$ such that $|\alpha - p/q| &gt; x/q^2$. I'm having trouble starting on the problem, could someone give a hint? Thanks
95910
I am from a physics background and my mathematics is not very good, so pardon my insolence with the question.  Editing based on the comments : We know that $\pi$ in decimal (i.e. base 10) is transcendental. Is it possible to have a radix base in which $\pi$ can be rational in that base ?  PS : My question is in general for any transcendental number.
85824
Ok, so I've been playing around with radical graphs and such lately, and I discovered that if the   nth x = √(1st x √ 2nd x ... √nth x);   Then  $$\text{the "infinith" } x = x$$  Example:   $$\sqrt{4\sqrt{4\sqrt{4\sqrt{4\ldots}}}}=4$$ Try it yourself, type calc in Google search, hit  then a number, such as $4$, and repeat, ending with $4$, (or press the buttons instead).  I'm a math-head, not big enough though, I think this sequence is divergent or convergent or whatever, too lazy to search up the difference.  However, can this be explained to me? Like how the Pythagorean Theorem can be explained visually.
52743
When solving different equations, I have realised, that some roots containing only arithmetic operations and square roots (4th, 8th roots too, because they can be represented using only square roots) can be converted to nested square roots form. Examples (these are roots of equations of 2nd, 4th, 4th and 8th degree): $$\sqrt{2}+\sqrt{3}=\sqrt{5+\sqrt{24}}$$ $$\sqrt{2}+\sqrt{3}+\sqrt{6}=\sqrt{15+\sqrt{160+\sqrt{6912+\sqrt{18874368}}}}$$ $$1+\sqrt{2}+\sqrt{3}+\sqrt{6}=\sqrt{21+\sqrt{413+\sqrt{4656+\sqrt{16588800}}}}$$ $$\sqrt{2}+\sqrt{3}+\sqrt{5}=\sqrt{14+\sqrt{140+\sqrt{4096+\sqrt{8847360}}}}$$ However, I have failed to convert to such form following root (8th degree equation): $$3+\sqrt{2}+\sqrt{3}+\sqrt{5}$$ Performing any operations with it, number of square roots inside increases, what makes me think that converting that root is impossible.  So, question: Can it be done with that root and with what roots in general?  Some forms I was able to get: $$\sqrt{19+6 \sqrt{2}+6 \sqrt{3}+6 \sqrt{5}+2 \sqrt{6}+2 \sqrt{10}+2 \sqrt{15}}$$ $$\sqrt{19+2\left(\sqrt{33+6 \sqrt{30}}+\sqrt{37+6 \sqrt{30}}+\sqrt{51+6 \sqrt{30}}\right)}$$  If one don't know how I got those expressions, here you are an example.  $$\sqrt{2}+\sqrt{3}+\sqrt{5}=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=$$ $$=\sqrt{10+2 \left(\sqrt{15}+\sqrt{6}+\sqrt{10}\right)}=\sqrt{10+2 \left(\sqrt{15}+\sqrt{\left(\sqrt{6}+\sqrt{10}\right)^2}\right)}=$$ $$=\sqrt{10+2 \left(\sqrt{15}+\sqrt{16+4 \sqrt{15}}\right)}=\sqrt{10+2 \left(\sqrt{15}-a+a+\sqrt{16+4 \sqrt{15}}\right)}=$$ $$=\sqrt{10+2a+2 \left(\sqrt{\left(\sqrt{15}-a\right)^2}+\sqrt{16+4 \sqrt{15}}\right)}=$$ $$=\sqrt{10+2a+2 \left(\sqrt{15+a^2-2a \sqrt{15}}+\sqrt{16+4 \sqrt{15}}\right)}=$$ $$[2a=4 \Rightarrow a=2]$$ $$=\sqrt{14+2 \left(\sqrt{19-4\sqrt{15}}+\sqrt{16+4 \sqrt{15}}\right)}=$$ $$=\sqrt{14+2 \sqrt{\left(\sqrt{19-4\sqrt{15}}+\sqrt{16+4 \sqrt{15}}\right)^2}}=$$ $$=\sqrt{14+2 \sqrt{35+4 \sqrt{16+3 \sqrt{15}}}}=\sqrt{14+\sqrt{140+\sqrt{4096+\sqrt{8847360}}}}$$