Question

Prove that $\sqrt{\mathbf{\pi }}$is algebraic over$\mathbf{\mathbb{Q}}\left(\pi \right)$.

Short answer

It is proved that $\sqrt{\pi }$is algebraic over $\mathrm{\mathbb{Q}}\left(\pi \right)$.

Step-by-step solution

Describe the concept of simple extension

An element u of an extension field F is said to be algebraic over F if that element u is the root of some nonzero polynomial in F[x].

Prove that π is algebraic over ℚ(π).

The given term$\sqrt{\pi }$ and K is an extension field of the field F .

Let,

$f\left(x\right)=x^2-\pi \in Q\left(\pi \right)\left[x\right]$

This is the required polynomial.

Find the roots of the polynomial as follows.

$$\begin{gathered} f\left(x\right)=0 \\ {x}^2-\pi =0 \\ {x}^2=\pi \\ x=\sqrt{\pi } \end{gathered}$$

Thus, the root of$f\left(x\right)$ is $\sqrt{\pi }$. Therefore,$\sqrt{\pi }$ is algebraic over$\mathrm{\mathbb{Q}}\left(\pi \right)$ .

Hence, it is proved that $\sqrt{\pi }$is algebraic overrole="math" localid="1659162881722" $\mathrm{\mathbb{Q}}\left(\pi \right)$ .