Question

$$\begin{gathered} \mathbf{If}\mathbf{}{\mathbf{n}}_{\mathbf{1}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{n}}_{\mathbf{t}}\mathbf{}\mathbf{are}\mathbf{}\mathbf{distinct}\mathbf{}\mathbf{positive}\mathbf{}\mathbf{integers}\mathbf{,}\mathbf{}\mathbf{show}\mathbf{}\mathbf{that}\mathbf{} \\ \mathbf{[}\mathbf{Q}\left(\sqrt{n_1}.........\sqrt{n_t}\right)\mathbf{:}\mathbf{Q}\mathbf{]}\mathbf{\le }{\mathbf{2}}^{\mathbf{t}}. \end{gathered}$$

Short answer

$\left[\mathrm{Q}\left(\sqrt{{\mathrm{n}}_1}......\sqrt{{\mathrm{n}}_{\mathrm{t}}}\right):\mathrm{Q}\right]\le 2^{\mathrm{t}}.$

Step-by-step solution

Definition of extension field.

Let K is an extension field of F. Then K is said to be algebraic extension of F is every element of K is algebraic overF.

Showing that  [Qn1......nt:Q]≤2t.

Let${\mathrm{n}}_1..............{\mathrm{n}}_{\mathrm{t}}\mathrm{are}\mathrm{distinct}\mathrm{primes}\mathrm{and}\mathrm{x},\mathrm{y}\in \left[\mathrm{Q}\left(\sqrt{{\mathrm{n}}_1}........\sqrt{{\mathrm{n}}_{\mathrm{t}-1}}\right)\right]$.So that

$\sqrt{{\mathrm{n}}_{\mathrm{t}+1}}=\mathrm{x}+\mathrm{y}.\sqrt{{\mathrm{n}}_{\mathrm{t}}}$

Inductive hypothesis provides $\left[\mathrm{Q}\left(\sqrt{{\mathrm{n}}_1}.......\sqrt{{\mathrm{n}}_{\mathrm{t}}}\right):\mathrm{Q}\left(\sqrt{{\mathrm{n}}_1}.......\sqrt{{\mathrm{n}}_{\mathrm{t}-1}}\right)\right]=2$

Here $\mathrm{x}\ne 0$because${\mathrm{n}}_1/{\mathrm{n}}_{\mathrm{t}+1}\mathrm{and}\sqrt{{\mathrm{n}}_{\mathrm{t}+1}}\in \left[\mathrm{Q}\left({\sqrt{\mathrm{n}}}_1............\sqrt{{\mathrm{n}}_{\mathrm{t}-1}}\right)\right]$

By this

$$\begin{gathered} {\mathrm{n}}_{\mathrm{t}+1}={\mathrm{x}}^2+{\mathrm{y}}^2{\mathrm{n}}_{\mathrm{t}}2\mathrm{xy}\sqrt{{\mathrm{n}}_{\mathrm{t}}} \\ 2\mathrm{xy}\sqrt{{\mathrm{n}}_{\mathrm{t}}}={\mathrm{n}}_{\mathrm{t}}-{\mathrm{x}}^2-{\mathrm{y}}^2{\mathrm{n}}_{\mathrm{t}} \end{gathered}$$

Since $2\mathrm{xy}\sqrt{{\mathrm{n}}_{\mathrm{t}}}\in \left[\mathrm{Q}\left({\sqrt{\mathrm{n}}}_1.........\sqrt{{\mathrm{n}}_{\mathrm{t}-1}}\right)\right]$

This is a contradiction. So,

$\sqrt{{\mathrm{n}}_{\mathrm{k}+1}}\in \left[\mathrm{Q}\left(\sqrt{{\mathrm{n}}_1}............\sqrt{{\mathrm{n}}_{\mathrm{t}}}\right)\right]$

This implies

$$\begin{gathered} \mathrm{x}\left[\mathrm{Q}\left(\sqrt{{\mathrm{n}}_1}............\sqrt{{\mathrm{n}}_{\mathrm{t}-1}}\right)\right]:\left[\mathrm{Q}\left(\sqrt{{\mathrm{n}}_1}............\sqrt{{\mathrm{n}}_{\mathrm{t}}}\right)\right]\le 2\mathrm{or} \\ \left[\mathrm{Q}\left(\sqrt{{\mathrm{n}}_1}............\sqrt{{\mathrm{n}}_{\mathrm{t}-1}}\right):\mathrm{Q}\right]\le 2 \end{gathered}$$

So the result is true for t=1 .Assume the result is true for t=k

That is

$$\begin{gathered} \left[\mathrm{Q}\left(\sqrt{{\mathrm{n}}_1}............\sqrt{{\mathrm{n}}_{\mathrm{t}}}\right):\mathrm{Q}\right]\le 2^{\mathrm{t}} \\ \mathrm{Thus}\left[\mathrm{Q}\left(\sqrt{{\mathrm{n}}_1}............\sqrt{{\mathrm{n}}_{\mathrm{t}}}\right):\mathrm{Q}\right]\le 2^{\mathrm{t}}. \end{gathered}$$