Show that the sequence $$\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},...$$ converges and find its limit. I put the sequence in this form , $(x_n)$ where $$\large …

$\frac{2}{\sqrt2}$ is obtained by multiplying the $\sqrt2$ by 1 or $\frac{\sqrt2}{\sqrt2}$. $$\left(\frac{\sqrt2}{1}\right)\left(\frac{\sqrt2}{\sqrt2}\right) = \frac{2}{\sqrt2}$$ This happens …

2018年9月14日 · Since $\sqrt{2}$ is irrational, is there a way to compute the first 20 digits of it? What I have done so far . I started the first digit decimal of the $\sqrt{2}$ by calculating …

2024年11月7日 · Using this theorem we can examine the number $2^\sqrt 2$, known as the Gelfond-Schneider constant. Both $2$ and $\sqrt2$ are algebraic and $\sqrt2$ is irrational, …

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$\alpha^2-2=\sqrt 2\in K$ by closure of multiplication and addition. $$\frac{\sqrt 2}{\sqrt{2+\sqrt{2}}}=\frac{\sqrt 2 \cdot\sqrt{2-\sqrt 2}}{\sqrt{2+\sqrt{2}}\sqrt{2 ...

2022年8月1日 · I'd like to know how can one simplify the following expression $$\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$$ into $$\sqrt{2(2+\sqrt{2})}.$$ Wolfram alpha suggests it …

2016年6月28日 · In §2.2 of her essay on mathematical morality, Eugenia Cheng includes the following example: Why is it possible for an irrational to the power of an irrational to be …

Prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}} \ $ converges to $2$. My attempt I proved that the sequence is increasing and bounded by $2$, can anyone help me show

this is a general way to do it: $\sqrt{2-\sqrt{2}}=\sqrt x+\sqrt y $ then $2-\sqrt 2=2 \sqrt{xy}+x+y$ and then make the rational parts equal to the rational parts as such: