Question

(a) Prove that the subset $\left\{1,\sqrt{2}\right\}$of $\mathit{R}$is linearly independent over $\mathit{Q}$.

(b)Prove that $\sqrt{\mathbf{3}}$is not linear combination of 1 and with coefficient in$\mathit{Q}$. Consclude that $\left\{1,\sqrt{2}\right\}$does not span $\mathit{R}$over$\mathit{Q}$.

Short answer

Answer:

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

(b)$\left\{1,\sqrt{2}\right\}$is not linear combination of 1 and$\sqrt{2}$.

Step-by-step solution

Definition of linearly independent.

A set of vector is said to be linear dependent if one of the vectors in the set can be written as a linear combination of others, and if no vector can be written as a linear combination of others then the vector is said to be linearly independent.

Showing that {1,2} is linearly independent.

(a)

Let $\alpha ,\beta \ne 0\in Q$are such that $\alpha 1+\beta \sqrt{2}=0$

This implies

$\beta \sqrt{2}=-\alpha$

If $\beta \ne 0$then $\sqrt{2}=\frac{-\alpha }{\beta }\in Q$

This is a contradiction as $\sqrt{2}\in Q$

Therefore $\beta =0$. Now $\beta =0$implies

$$\begin{gathered} \alpha 1+\beta \sqrt{2}=0 \\ \alpha =0 \end{gathered}$$

Since both the scalars $\alpha ,\beta \in Q$are zero.

Therefore, $\left\{1,\sqrt{2}\right\}$of $R$is linearly independent over $Q$.

Step-3: Showing that {1,2} not linearly combination of 1 and 2.

(b

Let if possible $\sqrt{3}$is a linear combination of 1 and $\sqrt{2}$with coefficient in Q.

Let $a,b,\in Q$are such that

$$\begin{gathered} \sqrt{3}=a1+b\sqrt{2} \\ {\left(\sqrt{3}\right)}^2={\left(a1+b\sqrt{2}\right)}^2 \\ 3=a^2+2b+2\sqrt{2}ab \end{gathered}$$

Now if $ab\ne 0than\sqrt{2}\in Q$. But this is a contradicitionas $\sqrt{2}\in Q$.

Therefore, $ab=0$. This implies either $a=0,or,b=o$.

Now if $a=0$then

$b=\frac{\sqrt{2}}{\sqrt{3}}\in Q$

Again a contradicition and if $b=0$then,

$\sqrt{3}=a\in Q$

This is contradicition as $\sqrt{3}\notin Q$

Therefore, no such $a,b\in Q$exists such that $\sqrt{3}=a+b\sqrt{2}$.

This implies $\sqrt{3}\notin span\left\{1,\sqrt{2}\right\}$and as $\sqrt{3}\notin R$but$\sqrt{3}\notin span\left\{1,\sqrt{2}\right\}$.

Therefore $\left\{1,\sqrt{2}\right\}$does not span R over Q.