$\begingroup$ There are only two square roots of ii (as there are two square roots of any non-zero complex number), namely $\pm(1+i)/\sqrt{2}$.

Here is an intuitive proof. Let $\sqrt{a},\sqrt{b}$ be the sides of a right-angles triangle. Then, the hypothenuse is of length $\sqrt{a+b}$ by Pythagorean theorem. From the triangle inequality, we get that the hypothenuse is smaller than the sum of the side lengths, which proves \begin{align} \sqrt{a+b} \leq \sqrt{a}+\sqrt{b}. \end{align}

So, my friend show me prove that $1=-1$ by using this way: $$1=\\sqrt{1}=\\sqrt{(-1)\\times(-1)}=\\sqrt{-1}\\times\\sqrt{-1}=i\\times i=i^2=-1$$ At first sight, I ...

Jun 24, 2014 · As we have the unwritten index $2$ for the sqare root, we multiply it by the index of the root inside the first root. $$\sqrt{\sqrt{x}} = \sqrt[2\times2]{x}= \sqrt[4]{x}$$ Another example is: $$\sqrt[3]{\sqrt[4]{x}} = \sqrt[3\times4]{x}= \sqrt[12]{x}$$

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Jul 9, 2023 · It's not wrong to write $\sqrt{\sqrt{x}}$ in the same way that it's not wrong to write $\sqrt{\sqrt{\frac{2x^1}{2}}}$, but it's generally preferable to write expressions as simply as possible. The phrase "square root" means the same thing as "one-half power".

Jun 28, 2016 · Consider $\sqrt{2}^{\sqrt{2}}$. If it is rational, we are done. If it is irrational, consider $$ \left(\sqrt{2}^{\sqrt{2}}\right)^\sqrt{2} = \sqrt{2}^2=2.$$ However, as Cheng notes, this doesn't tell us whether $\sqrt{2}^{\sqrt{2}}$ itself is rational or not. So which is it?

Feb 2, 2017 · Then $$(\psi-\sqrt{5})^2=(\sqrt{2}+\sqrt{3})^2 \\ \psi^2-2\sqrt{5}\psi+5=5+2\sqrt{6} \\ \psi^2=2\sqrt{6} +2\sqrt{5}\psi\\ $$ Square it one more time, and you reach the contradiction. For the general case, here is a nice trivial solution from Kvant: prove the following lemma:

Jan 16, 2015 · The reason that often only the Taylor series for $\sqrt{1 + x}$ is given in the books is that – for the square-root function – the general case can easily be reduced to the special case: $$ \sqrt {\mathstrut x} = \sqrt {\mathstrut x_0 + x - x_0} = \sqrt {\mathstrut x_0}\sqrt {1 + \frac {\mathstrut x-x_0}{x_0}} $$ and now you can use the ...

$\begingroup$ 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.