Show $\sqrt[4]{2}$ on horizontal axis

Question

It is very basic question ,to show $\sqrt[2]{2}$ on $x$ axis .we do like below

but,how can we show $\sqrt[4]{2}$ on $x$ axis ?(like we do for $\sqrt2)$
I am thankful if you guide me .
$\bf{Remark}:$ may be a question for many people .

Answer 1

It's possible to repeatedly square root any constructible number. So far, you know how to construct the point located at $A(\sqrt 2, 0)$. Now construct the point $B(-1, 0)$ and connect the two points (note that they pass through the origin $O(0, 0)$). Construct a semi-circle above the $x$-axis whose diameter is $AB$ and let $C(0, k)$ be its $y$-intercept. We claim that $OC = k = \sqrt[4]{2}$.

Indeed, observe that triangles $AOC$ and $COB$ are similar, so we have: $$ \frac{k}{1} = \frac{\sqrt 2}{k} \iff k^2 = \sqrt 2 \iff k = \sqrt[4]{2} $$

Answer 2

The following steps will work . . . \begin{align*} 1.\;\,&\text{Construct a segment of length$\,\sqrt{2}$.}\\[4pt] 2.\;\,&\text{Construct a segment $AB$ of length$\,\sqrt{2} + {\small{\frac{1}{4}}}$.}\\[4pt] 3.\;\,&\text{Construct a circle with $AB$ as a diameter.}\\[4pt] 4.\;\,&\text{Construct a point $P$ on $AB$ such that $PB$ has length$\,\sqrt{2} - {\small{\frac{1}{4}}}$.}\\[4pt] 5.\;\,&\text{Construct a circle with center $B$, and radius $BP$.}\\[4pt] 6.\;\,&\text{Let $C$ be one of the two points where the circles intersect.}\\[10pt] \end{align*} \begin{align*} \text{Then}\;\;|AC|^2 &= |AB|^2 - |BC|^2 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\, \\[4pt] &=\left(\sqrt{2} + {\small{\frac{1}{4}}}\right)^2 -\left(\sqrt{2} - {\small{\frac{1}{4}}}\right)^2\\[4pt] &= \left(2+{\small{\frac{1}{2}}}\sqrt{2}+{\small{\frac{1}{16}}}\right) - \left(2-{\small{\frac{1}{2}}}\sqrt{2}+{\small{\frac{1}{16}}}\right)\\[4pt] &=\sqrt{2}\\[8pt] \text{hence}\;\;|AC|\;&= \sqrt[4]{2},\;\text{as required.}\\[4pt] \end{align*}