Let K be any quafratic extension of $\mathbb{R}$. So, $K=\mathbb{R}(\alpha )$, as deg $\frac{k}{\mathbb{R}}=2$, the elements $1,\alpha ,{\alpha}^{2}$ are linearly depend over $\mathbb{R}$. Thus $\mathbb{E}\text{}a,b,c\in \mathbb{R}$, with ${\alpha}^{2}+b\alpha +c=0$, by quadratic formula. $\alpha =\frac{-b+-\sqrt{{b}^{2}-4ac}}{2a}$, with $d={b}^{2}-4aac<0\text{}if\text{}{b}^{2}-4ac0$, alpha in $\mathbb{R}$, contraditing, $\mathrm{deg}(\frac{K}{\mathbb{R}})=2$ Set, $\beta =\frac{2\alpha +b}{\sqrt{-d}}$

Then beta in K and ${\beta}^{2}=-1$ Using this element , we construct a field isomorphism between K and C, the field of complex numbers.

Now, $K=\mathbb{R}(\alpha )=\mathbb{R}(\beta )$, with ${\beta}^{2}=-1$

So, the map $\varphi :K=\mathbb{R}(\beta )\to \mathbb{C},$ with $\varphi (x+y\beta )=x+iy,\text{}x,\text{}y$ in $\mathbb{R}$ is the required field isomorphism between K and $\mathbb{C}$ Next , we showthat there is no field extension K of degree 3 over R. If $K=\mathbb{R}(\alpha )$ and $\mathrm{deg}(\frac{K}{\mathbb{R}})=3$, then alpha satisfies an equation o the form $a{\alpha}^{3}+b{\alpha}^{2}+c\alpha +d=0,$ with $a,b,c,d\in \mathbb{R},$ and $f(x)=a{x}^{3}+b{x}^{2}+cx+d$ is irreducible over $\mathbb{R}$ Completing the proof by arriving at a contradiction. Complex root of $a{x}^{3}+b{x}^{2}+cx+d=0$ with $a,b,c,d\in \mathbb{R},$ occur in pairs. Hence, $a{x}^{3}+b{x}^{2}+cx+d=0$ has a real root, which contradicts f(x) being irreducible over $\mathbb{R}$.

This proves that there is no filed K with $\mathrm{deg}(\frac{K}{\mathbb{R}})=3$