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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.

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    $\begingroup$ Perhaps you could dominate MathJax before the world? $\endgroup$ – copper.hat Feb 5 '15 at 6:07
  • $\begingroup$ I'm sorry I'm not familiar with how it works. $\endgroup$ – WorldDominator Feb 5 '15 at 6:09
  • $\begingroup$ 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 $\forall b\in \mathbb{R}^+,\exists a\in A, s.t. \; |a|>|b|$. $\endgroup$ – Brian Ding Feb 5 '15 at 6:11
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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.

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    $\begingroup$ In other words: first figure out a more convenient expression for each set (a union of intervals, in this case); then determine whether they're bounded. $\endgroup$ – Greg Martin Feb 5 '15 at 7:03
  • $\begingroup$ Isn't it proper to do x(100-x) > 0 though? Thanks for your answer, I thought this was the way to do it but I thought I was missing something. $\endgroup$ – WorldDominator Feb 5 '15 at 7:21
  • $\begingroup$ What do you mean by "proper"? I don't quite see the motivation for expressing it as $x(100-x) > 0$. $\endgroup$ – Mike Pierce Feb 5 '15 at 7:27
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    $\begingroup$ Well as a part of factorization you get x = 0, which you would lose if you divide by x. $\endgroup$ – WorldDominator Feb 5 '15 at 7:42
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Problem 1 : $0 \lt x \lt 100$ Problem 2 : $x \lt 0 \lor x \gt 25$

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