Question

Let $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)=\left\{r+s\sqrt{2}\overline{)r,s\in \mathrm{\mathbb{Q}}}\right\}$. Show that$\mathrm{\mathbb{Q}}\sqrt{2}$ is a subfield of$\mathrm{ℝ}$ .

Short answer

It is proved that $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)=\left\{r+s\sqrt{2}\overline{)r,s\in \mathrm{\mathbb{Q}}}\right\}$ is a subfield of$\mathrm{ℝ}$ .

Step-by-step solution

Subfield 

It is given that $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)=\left\{r+s\sqrt{2}\overline{)r,s\in \mathrm{\mathbb{Q}}}\right\}$ and we have to show $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)=\left\{r+s\sqrt{2}\overline{)r,s\in \mathrm{\mathbb{Q}}}\right\}$ is a subfield of $\mathrm{ℝ}$.

If $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)=\left\{r+s\sqrt{2}\overline{)r,s\in \mathrm{\mathbb{Q}}}\right\}$ satisfies the following conditions for a field then,$\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)=\left\{r+s\sqrt{2}\overline{)r,s\in \mathrm{\mathbb{Q}}}\right\}$ is subset of $\mathrm{ℝ}$ then, we can say$\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)=\left\{r+s\sqrt{2}\overline{)r,s\in \mathrm{\mathbb{Q}}}\right\}$ is a subfield of $\mathrm{ℝ}$.

Conditions of field

(1). Let $a=r_1+s_1\sqrt{2},b=r_2+s_2\sqrt{2}\in \mathrm{\mathbb{Q}}\sqrt{2}$.

$\begin{array}{rcl}& \Rightarrow & a+b=\left(r_1+s_1\sqrt{2}\right)+\left(r_2+s_2\sqrt{2}\right)\\ & =& \left(r_1+r_2\right)+\left(s_1+s_2\right)\sqrt{2}\\ & =& r+s\sqrt{2}\end{array}$

(2). Let $a=r_1+s_1\sqrt{2},b=r_2+s_2\sqrt{2}\in \mathrm{\mathbb{Q}}\sqrt{2}$.

$\begin{array}{rcl}& \Rightarrow & ab=\left(r_1+s_1\sqrt{2}\right)\left(r_2+s_2\sqrt{2}\right)\\ & =& \left(r_1{r}_2+2s_1{s}_2\right)+\left(s_1{r}_2+r_1{s}_2\right)\sqrt{2}\\ & & \end{array}$

It implies that $ab\in \mathrm{\mathbb{Q}}\sqrt{2}$.

(3). For zero vector, $0\in \mathrm{\mathbb{Q}}$ .

$\begin{array}{rcl}0& =& 0+0\sqrt{2}\\ & =& 0\end{array}$

It implies that$0\in \mathrm{\mathbb{Q}}\sqrt{2}$ .

(4). From the first condition, we can write $a+b\in \mathrm{\mathbb{Q}}\sqrt{2}$ where, $b=r_2+s_2\sqrt{2};\forall r_2,s_2\in \mathrm{\mathbb{Q}}$ . Let $b=-r_1+\left(-s_1\right)\sqrt{2};\forall -r_1,-s_1\in \mathrm{\mathbb{Q}}$ then it gives,

$\begin{array}{rcl}& \Rightarrow & a+b=\left(r_1+s_1\sqrt{2}\right)+\left(r_2+s_2\sqrt{2}\right)\\ & =& \left(r_1-r_1\right)+\left(s_1-s_1\right)\sqrt{2}\\ & =& 0\end{array}$

Hence, the set$\mathrm{\mathbb{Q}}\sqrt{2}$ is a subring of $\mathrm{ℝ}$.

(5). From the second condition, we can write $ab\in \mathrm{\mathbb{Q}}\sqrt{2}$ where, $b=r_2+s_2\sqrt{2};\forall r_2,s_2\in \mathrm{\mathbb{Q}}$. Let $b=\frac{1}{r_1}+\frac{1}{s_1}\sqrt{2};\forall \frac{1}{r_1},\frac{1}{s_1}\in \mathrm{\mathbb{Q}}$then it gives,

$\begin{array}{rcl}ab& =& \left(r_1+s_1\sqrt{2}\right)\left(r_2+s_2\sqrt{2}\right)\\ & =& \left(r_1+s_1\sqrt{2}\right)\frac{1}{\left(r_1+s_1\sqrt{2}\right)}\\ & =& 1\end{array}$

Thus, it implies that $a=\frac{1}{b}$.

Now, it can also be written as:

$\begin{array}{rcl}a& =& \frac{1}{b}\\ & =& \frac{1}{r_1+s_1\sqrt{2}}\end{array}$

After rationalizing, it gives,

$\begin{array}{rcl}a& =& \frac{\left(r_1-s_1\sqrt{2}\right)}{\left(r_1+s_1\sqrt{2}\right)\left(r_1-s_1\sqrt{2}\right)}\\ & =& \frac{r_1-s_1\sqrt{2}}{r_1^2-2s_1^2}\\ & =& \frac{r_1}{r_1^2-2s_1^2}+\frac{-s_1\sqrt{2}}{r_1^2-2s_1^2}\end{array}$

It implies that $\frac{r_1}{r_1^2-2s_1^2}+\frac{-s_1\sqrt{2}}{r_1^2-2s_1^2}=a\in \mathrm{\mathbb{Q}}\sqrt{2}$ where$\frac{r_1}{r_1^2-2s_1^2},\frac{-s_1}{r_1^2-2s_1^2}\in \mathrm{\mathbb{Q}}$ .

Hence, $\mathrm{\mathbb{Q}}\left(\sqrt{2}\right)=\left\{r+s\sqrt{2}\overline{)r,s\in \mathrm{\mathbb{Q}}}\right\}$ is a subfieldof$\mathrm{ℝ}$ .