1. ## infimum proof question

prove that:

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

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

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

3. Originally Posted by Abu-Khalil
Let $\displaystyle A=\inf\{x\in \mathbb{Q}:0<x\land x^2>3\}$ and suposse $\displaystyle A\neq \sqrt{3}$. Then $\displaystyle A>3$ or $\displaystyle A<3$.

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

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

Tonio

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

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

5. Originally Posted by Abu-Khalil
No, i mean closure in terms of limit points. You usually read as $\displaystyle \mathbb{Q}$ is dense in $\displaystyle \mathbb{R}$.

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

Tonio