Question

Squaring the Circle Given a circle of radiusr, show that it is impossible to construct by straightedge and compass the side of a square whose area is the same as that of the given circle. You may assume the nontrivial fact that$\mathbf{\sqcap }$is not the root of any polynomial in Q[x] .

Short answer

It is impossible to construct by straightedge and compass the side of a square whose area is the same as that of the given circle. Since, $\prod$ is not the root of any polynomial in Q[x].

Step-by-step solution

Defining the Quadratic Extension.

A number is constructible iff it belongs to any quadratic extension of$\mathrm{Q}\left[\mathrm{x}\right]$ . A number is in quadratic extension of $\mathrm{Q}\left[\mathrm{x}\right]$if it satisfies a polynomial over $\mathrm{Q}\left[\mathrm{x}\right]$which means it is a root of polynomial of $\mathrm{Q}\left[\mathrm{x}\right]$.

Showing II is not a root of any polynomial.

Since, ris constructible number and now draw a circle with radius r . The area of this circle is ${\mathrm{IIr}}^2$. If we construct a square having side whose area is equal to the area of circle, then $s^2={\mathrm{IIr}}^2$, this implies that $\frac{s^2}{r^2}=\mathrm{II}$.

In this case, L.H.S of the equation is constructible whereas R.H.S is not.

Since II is not the root of any polynomial in $\mathrm{Q}\left[\mathrm{x}\right]$as it does not satisfies any polynomial.