# Proving Set bounded vs unbounded

I know what a given set means to be bounded or unbounded, I'm a bit confused about how I would prove the set is bounded or unbounded. For bounded do I show an example of an upper bound and lower bound and have that be enough to prove it?

Two problems involving these: Prove that the set $\{x \in \mathbb{R}: 10(x)^{1/2} - x > 0 \}$ is bounded.

Prove that the set $\{x \in \mathbb{R}: x^2 - 25x > 0 \}$ is unbounded.

• Perhaps you could dominate MathJax before the world? – copper.hat Feb 5 '15 at 6:07
• I'm sorry I'm not familiar with how it works. – WorldDominator Feb 5 '15 at 6:09
• To show a set is bounded, you need to show that it is bounded above and below. To show a set $A$ is unbounded. You may show that $\forall b\in \mathbb{R}^+,\exists a\in A, s.t. \; |a|>|b|$. – Brian Ding Feb 5 '15 at 6:11

To answer the first problem and prove that $\{x \in \mathbb{R}: 10(x)^{1/2} - x > 0 \}$ bounded, look at the condition for $x$ being in the set: \begin{align} && 10(x)^{1/2} - x &> 0 \\ &\implies& 10(x)^{1/2} &> x \\ &\implies& 100x &> x^2 \\ &\implies& 100 &> x \end{align} So $x$ is strictly bounded above by $100$. As WorldDominator points out in the comments, because we had to divide by zero above, we should do a separate check to see if zero is in our set. Since $$10(0)^{1/2} - (0) = 0 \ngtr 0$$ zero cannot be in our set. To get a lower bound on our set, just notice that $x$ cannot be negative because there is a $x^{1/2} = \sqrt{x}$ in our condition. So $$0 < x < 100$$ The second one can be done similarly.
• What do you mean by "proper"? I don't quite see the motivation for expressing it as $x(100-x) > 0$. – Mike Pierce Feb 5 '15 at 7:27
Problem 1 : $0 \lt x \lt 100$ Problem 2 : $x \lt 0 \lor x \gt 25$