Let a be a positive rational number. Let $A = \{ x\in\Bbb Q \mid x^2 < a \}$ . Show that $A$ is bounded in $\Bbb Q$ I need to prove the following:


  Let $a$ be a positive rational number. 
  
  Let $A = \{ x\in\Bbb Q \mid x^2 &lt; a \}$ . 
  
  Show that $A$ is bounded in $\Bbb Q$


So I know I have to show that $A$ is bounded above and below in $\Bbb Q$. I have figured out that A is bounded above by $\sqrt(a)$ but I am unsure to prove that it is bounded below.
Because a square root can be either positive or negative, when you have $x^2&lt;a$ and you take the square root of both sides, you actually get two inequalities out.

$$\begin{align}
x^2 &amp;&lt; a \\
\implies\quad x&lt;\sqrt{a} \;\;\;&amp;\text{and}\;\;\; x&gt;-\sqrt{a}\\
\implies\quad -\sqrt{a}&lt; \;&amp; x &lt; \sqrt{a}
\end{align}$$

That gives you a lower bound.
$A=[-\sqrt 2,\sqrt 2]\cap \Bbb Q$