Question

Question:

(a) Show that$\mathbf{\{}\mathbf{1}\mathbf{,}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{3}}\mathbf{\}}$is linearly independent over
.

(b) show that $\left\{1,\sqrt{2},\sqrt{3},\sqrt{6}\right\}$is linearly independent over .

Short answer

Answer

1. Thus,$\left\{1,\sqrt{2},\sqrt{3}\right\}$ is linearly independent.
2. Thus,$\left\{1,\sqrt{2},\sqrt{3},\sqrt{6}\right\}$ is linearly independent.

Step-by-step solution

Explanation

First of all, we show that$\left\{1,\sqrt{2}\right\}⊸$is linearly independent over .

Now,

Let$a+b\sqrt{2}=0,a,b\in 0$

Suppose.$b\ne 0$

$\therefore \sqrt{2}=-\frac{a}{b}$

$\because -\frac{a}{b}\in Q$

Which is a contradiction,

$\therefore \left\{1,\sqrt{2}\right\}$Is linearly independent._____________(1)

Concept  

We have to show that, $\left\{1,\sqrt{2},\sqrt{3}\right\}$is linearly independent.

Let ,$a+b\sqrt{2}+c\sqrt{3}=0,a,b,c\in Q$

Let $c\ne 0$_________________(2)

$$\begin{gathered} \Rightarrow \sqrt{3}=-\frac{a}{c}-\frac{b}{c}\sqrt{2} \\ \Rightarrow \sqrt{3}=\alpha +\beta \sqrt{2},\alpha =-\frac{a}{c},\beta =-\frac{b}{c}\in Q \end{gathered}$$

,

Calculation

Squaring both sides,

$3={\alpha }^2+2{\beta }^2+2\alpha \beta \sqrt{2}$____________(3)

$$\begin{gathered} \Rightarrow \alpha \beta \sqrt{2}=\frac{3-{\alpha }^2+2{\beta }^2}{2}\in Q \\ \because \alpha ,\beta \in Q \\ =\frac{3-{\alpha }^2+2{\beta }^2}{2}\in Q \\ \Rightarrow \alpha \beta \sqrt{2}\in Q \\ \Rightarrow \alpha .\beta =0 \end{gathered}$$

Either$\alpha =0$ or$\beta =0$

If $\alpha =0$then $\beta =\sqrt{\frac{3}{2}}\notin Q$ (from (2))

which is a contradiction.

If, $\beta =0$$,{\alpha }^3=3$ ( from (2))

$\alpha =\pm \sqrt{3}\notin Q$

Here we get a contradiction.

Hence,c=0

We know that,

$a+b\sqrt{2}=0$

a=0,b=0$\left[\because \left\{1,\sqrt{2}\right\}linearly\text{independet over Q from (1)}\right]$

Thus, we get,

$a=0=b=c$

Hence,$\left\{1,\sqrt{2},\sqrt{3}\right\}$ is linearly independent.

Explanation

First of all, we show that$\left\{1,\sqrt{2}\right\}$ is linearly independent over .

Now,

Let$a+b\sqrt{2}=0,a,b\in 0$

Suppose.$b\ne 0$

role="math" localid="1658817191597" $$\begin{gathered} \therefore \sqrt{2}=-\frac{a}{b} \\ \because -\frac{a}{b}\in Q \end{gathered}$$

Which is a contradiction,

$\therefore \left\{1,\sqrt{2}\right\}$Is linearly independent

Concept

Here we have to show that,$\left\{1,\sqrt{2},\sqrt{3},\sqrt{6}\right\}$ is linearly independent.

Now, Let$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}=0$

$\Rightarrow \left(a+b\sqrt{2}\right)+\sqrt{3}\left(c+d\sqrt{6}\right)=0$_______(1)

Where.$a,b,c,d\in Q$

Let $c+d\sqrt{2}\ne 0$

$\therefore \sqrt{3}=\frac{a+b\sqrt{2}}{c+d\sqrt{2}}\times \frac{c-d\sqrt{2}}{c-d\sqrt{2}}$

$=\frac{ac+\left(bc-ad\right)\sqrt{2}}{c^2-2d^2}$

$\sqrt{3}=\alpha +\beta \sqrt{2},\alpha ,\beta \in Q$

Where,

,$$\begin{gathered} \alpha =\frac{ac}{c^2-2d^2},\beta =\frac{bc-ad}{c^2-2d^2} \\ \Rightarrow \alpha .1+\beta \sqrt{2}-1.\sqrt{3}=0 \end{gathered}$$

By (a) since$\left\{1,\sqrt{2},\sqrt{3}\right\}$ is linearly independent over .

$\therefore \alpha =0,\beta =0,-1=0$

Which is a contradiction.

$\therefore c+d\sqrt{2}=0$

as is linearly independent over .

Step:-3 calculation

We know that $a+b\sqrt{2}=0$from (1)

$\Rightarrow a=0,b=0$as$\left\{1,\sqrt{2}\right\}$ is linearly independent over Q.

Thus, we get

$a=0,b=0,c=0,d=0$

Which shows that $\left\{1,\sqrt{2},\sqrt{3},\sqrt{6}\right\}$is linearly independent.