The square root of a number \(a\), denoted \(\sqrt{a}\), is the number \(b\) such that \[b^{2} = a\text{ and }b\ge 0.\]

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

2017年2月2日 · 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. …

2016年9月16日 · sqrt(2)^sqrt(2) is irrational. ""^oosqrt(2) is rational. While I haven't found a proof that sqrt(2)^(sqrt(2)^sqrt(2)) is irrational, here are answers to the first and third parts. …

2015年1月16日 · 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 …

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 ...

2015年10月13日 · x = 1 You're dealing with radical terms, which means that you need to make sure that the expressions that are unde the square root are always positive. This means that …

2016年6月28日 · 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, …

Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their …

Here we used $\sqrt{50}=5\sqrt{2},\sqrt{25}=5$, and a quick formula for the inverse of a $2\times 2$ matrix: