Question

$\mathbf{Show}\mathbf{}\mathbf{that}\mathbf{}\sqrt{\mathbf{2}}\mathbf{}\mathbf{is}\mathbf{}\mathbf{not}\mathbf{}\mathbf{in}\mathbf{}\mathbf{Q}\left(i\right)\mathbf{}\mathbf{and}\mathbf{}\mathbf{hence}\mathbf{}\mathbf{C}\mathbf{\ne }\mathbf{Q}\left(i\right)$

Short answer

$\mathrm{C}\ne \mathrm{Q}\left(\mathrm{i}\right)$

Step-by-step solution

Definition of splititing field.

A splititing field of a polynomial with coefficients in a smallest field extension of that field over which the polynomial splits or splits into linear factors.

Showing that  

Let if possible $\sqrt{2}\in \mathrm{Q}\left(\mathrm{i}\right)$. Which can be written as

$\sqrt{2}=\mathrm{a}+\mathrm{ib}$

Square both the side and solve.

$$\begin{gathered} {\left(\sqrt{2}\right)}^2={\left(\mathrm{a}+\mathrm{ib}\right)}^2 \\ 2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ia}b \end{gathered}$$

If $\mathrm{ab}\ne 0\mathrm{then}\mathrm{i}\in \mathrm{Q}$

This is a contradiction as$\mathrm{i}\in \mathrm{C}.\mathrm{So},\mathrm{ab}=0.$

This implies a=0 or b=0

If a=0 then

$$\begin{gathered} 2=-{\mathrm{b}}^2 \\ \sqrt{2}\mathrm{i}=\mathrm{b}\in \mathrm{Q} \end{gathered}$$

Which is not possible. So, again if b=0 then

$$\begin{gathered} 2={\mathrm{a}}^2 \\ \sqrt{2}=\mathrm{a}\in \mathrm{Q} \end{gathered}$$

Which is also not possible since $\sqrt{2}$is an irrational number. Thus $\sqrt{2}\notin \mathrm{Q}\left(\mathrm{i}\right)$and since the$\sqrt{2},\mathrm{i}\in \mathrm{C}\mathrm{but}\sqrt{2},\mathrm{i}\notin \mathrm{Q}\left(\mathrm{i}\right)$

Therefore $\mathrm{C}\ne \mathrm{Q}\left(\mathrm{i}\right)$