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real analysis - Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 ... Inspired by Ramanujan's problem and solution of $\\sqrt{1 + 2\\sqrt{1 + 3\\sqrt{1 + \\ldots}}}$, I decided to attempt evaluating the infinite radical $$ \\sqrt{1 ...
solution verification - Proving $\sqrt{4+2\sqrt{3}} - \sqrt{3}$ is ... 29 Jul 2024 · $$\sqrt{4 + \sqrt{3}} - \sqrt{3} = 1, \tag1 $$ from which the problem can be completed by an alternative route. As a clarification, (1) above does not prove that $~m = n.~$ Instead, it proves that if $~\displaystyle \sqrt{4 + \sqrt{3}} - \sqrt{3}~$ is …
What does the small number on top of the square root symbol … $\begingroup$ Minor point: I notice quite a few elementary algebra books as well as some writers here taking the view that the n-th root of x is defined as x to the power 1/n.
Why is the square root of a number not plus or minus? For example, $\sqrt{4}$. I've asked a bunch of people and I get mixed answers all the time, as to whether it is $-2$ and $+2$ or just $+2$. How about if there's a negative in front of the square root sign, for example, $-\sqrt{4}$? Would that still be plus or minus or just minus?
How can I show that $\\sqrt{1+\\sqrt{2+\\sqrt{3+\\sqrt\\ldots}}} I would like to investigate the convergence of $$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt\ldots}}}}$$ Or more precisely, let $$\begin{align} a_1 & = \sqrt 1\\ a_2 ...
number theory - Find the value of $x$ such that $\sqrt {4+\sqrt {4 ... 6 Jul 2015 · $$\begin{align}\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}&=x&\Longleftrightarrow \\ \left(\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}\right)^2&=x^2&\Longleftrightarrow\\ 4+\sqrt ...
What is $\\sqrt{i}$? - Mathematics Stack Exchange $\begingroup$ There are only two square roots of ii (as there are two square roots of any non-zero complex number), namely $\pm(1+i)/\sqrt{2}$.
algebra precalculus - Calculating $ x= \sqrt {4+\sqrt {4-\sqrt { 4 ... If $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\sqrt{4-\sqrt{ 4+\sqrt{4-\dots}}}}} $ then find value of 2x-1 I tried the usual strategy of squaring and substituting the rest of series by x again but could not ...
Why sqrt(4) isn't equall to-2? - Mathematics Stack Exchange It's just notation most likely. Yes, $(-2)^2 = 4$, but often the $\sqrt{4}$ symbol is reserved for the positive square root, so $\sqrt{4} = 2$. If you want the negative square root, that would be $-\sqrt{4} = -2$. Both $-2$ and $2$ are square roots of $4$, but the notation $\sqrt{4}$ corresponds to only the positive square root.
False Proof that $\\sqrt{4}$ is Irrational - Mathematics Stack … 19 Apr 2016 · However, when I apply this proof format to $\sqrt{4} $ (which is clearly an integer and thus rational) I get the following: Say $ \sqrt{4} $ is rational. Then $\sqrt{4}$ can be represented as $\frac{a}{b}$ , where a and b have no common factors .