infimum proof question

**pseudonym** · November 7th 2009, 08:37 AM

prove that:

√3 = inf{ x in rationals: x > 0 and x^2 > 3}

thank you for your time!

**Abu-Khalil** · November 7th 2009, 11:48 AM

Let $A=\inf\{x\in \mathbb{Q}:0<x\land x^2>3\}$ and suposse $A\neq \sqrt{3}$ . Then $A>3$ or $A<3$ .

Now you can use the fact that $\mathbb{Q}\subset \mathbb{R}\subset \overline{\mathbb{Q}}$ .

**tonio** · November 7th 2009, 01:05 PM

Originally Posted by Abu-Khalil

Let $A=\inf\{x\in \mathbb{Q}:0<x\land x^2>3\}$ and suposse $A\neq \sqrt{3}$ . Then $A>3$ or $A<3$ .

Now you can use the fact that $\mathbb{Q}\subset \mathbb{R}\subset \overline{\mathbb{Q}}$ .

What do you mean by $\overline{\mathbb{Q}}$ ? Because if this is, as usually denoted, the algebraic closure of Q then definitely $\mathbb{R} \nsubseteq\overline{\mathbb{Q}}$ . For example, $\pi\notin\overline{\mathbb{Q}}$

Tonio

**Abu-Khalil** · November 7th 2009, 02:55 PM

Originally Posted by tonio

What do you mean by $\overline{\mathbb{Q}}$ ? Because if this is, as usually denoted, the algebraic closure of Q then definitely $\mathbb{R} \nsubseteq\overline{\mathbb{Q}}$ . For example, $\pi\notin\overline{\mathbb{Q}}$

Tonio

No, i mean closure in terms of limit points. You usually read as $\mathbb{Q}$ is dense in $\mathbb{R}$ .

**tonio** · November 7th 2009, 03:13 PM

Originally Posted by Abu-Khalil

No, i mean closure in terms of limit points. You usually read as $\mathbb{Q}$ is dense in $\mathbb{R}$ .

Oh, I see...but then in fact $\mathbb{R}=\overline{\mathbb{Q}}$

Tonio

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