Does the set, $S= \{x : x^2 >3\}$, have an upper or lower bound?

Question

I'm not sure whether or not the set, $S= \{x : x^2 >3\}$, has an upper or lower bound. At first, I was thinking that the set, $S= \{x : x^2 >3\}$, consists of all the real numbers greater than both $\sqrt3$ and $-\sqrt3$, so basically the real numbers greater than $\sqrt3$. And so I thought that $\sqrt3$ and even $-\sqrt3$ would be lower bounds, but the textbook's answer says that the set $S$ is actually unbounded.

Answer 1

You are almost there:

$x \in \mathbb{R}:$

$x^2 >3 $

$\iff $

$x<-√3$, or $x >√3.$

Hence

$S=${$x| x<-√3$}$\cup ${$ x| x>√3$} ,

not bounded.

Note: $x^2-3 > 0$

$\rightarrow:$

$(x-√3)(x+√3)>0.$

Both factors are positive or negative implies:

$x<-√3$ or $x >√3.$

Answer 2

Suppose $l$ is a lower bound for $S$. Then for all $x \in S$, we have that $l \leq x$. Now either $- \sqrt 3 < l$, or $l \leq -\sqrt3$. In the first case, any $y < -\sqrt 3 < l$ satisfies $y^2 > 3$. In the second case, any $y < l \leq - \sqrt 3$ also satisfies $y^2 > 3$. In either case, we have that there exists a $y \in S$ such that $y < l$. But this is a contradiction, as $l$ was a lower bound for $S$. Since this holds for an arbitrary lower bound for $S$, we conclude that $S$ can have no lower bound.

You can use the same logic to argue that $S$ cannot have an upper bound by considering the above inequalities with $\sqrt 3$ instead of $- \sqrt3$. Hence, $S$ is unbounded.

Answer 3

For all $n\in \Bbb Z\setminus \{-1,0,1\}$ we have $n^2>3.$