- #1

JVEK7713

- 1

- 0

- TL;DR Summary
- Proof verification of ##sup\{a \in \mathbb{Q}: a^2 \leq 3\} = \sqrt{3}##

I would wish to receive verification for my proof that ##sup\{a \in \mathbb{Q}: a^2 \leq 3\} = \sqrt{3}##.

• It is easy to verify that ##A = \{a \in \mathbb{Q}: a^2 \leq 3\} \neq \varnothing##. For instance, ##1 \in \mathbb{Q}, 1^2 \leq 3## whence ##1 \in A##.

• We claim that ##\sqrt{3}## is an upper bound of ##A##: to see why, let ##a \in A##. Then, ##a^2 \leq 3 \Rightarrow |a| \leq \sqrt{3} \Rightarrow a \leq \sqrt{3}##.

• We claim ##\sqrt{3}## is the least upper bound of ##A##: to see why, let ##x \in \mathbb{R}## be an upper bound of ##A##. Then, for any ##a \in A##, ##a \leq x \Rightarrow a^2 \leq x^2##. As ##a^2 \leq 3##, it must be the case that ##a^2 \leq \text{min}\{x^2, 3\}.## (*) We claim that ##x^2 \geq 3##. To prove this, suppose, upon the contrary, that ##x^2 < 3##. Then by definition of ##A## and the density of ##\mathbb{Q}##, there exists ##a \in \mathbb{Q}## s.t. ##x^2 \leq a^2 < 3##, which implies that ##x## is not an upper bound for ##A##–– a contradiction! Thus, ##\sqrt{3}## must be the least upper bound of ##A##, as desired.

Note:

• It is easy to verify that ##A = \{a \in \mathbb{Q}: a^2 \leq 3\} \neq \varnothing##. For instance, ##1 \in \mathbb{Q}, 1^2 \leq 3## whence ##1 \in A##.

• We claim that ##\sqrt{3}## is an upper bound of ##A##: to see why, let ##a \in A##. Then, ##a^2 \leq 3 \Rightarrow |a| \leq \sqrt{3} \Rightarrow a \leq \sqrt{3}##.

• We claim ##\sqrt{3}## is the least upper bound of ##A##: to see why, let ##x \in \mathbb{R}## be an upper bound of ##A##. Then, for any ##a \in A##, ##a \leq x \Rightarrow a^2 \leq x^2##. As ##a^2 \leq 3##, it must be the case that ##a^2 \leq \text{min}\{x^2, 3\}.## (*) We claim that ##x^2 \geq 3##. To prove this, suppose, upon the contrary, that ##x^2 < 3##. Then by definition of ##A## and the density of ##\mathbb{Q}##, there exists ##a \in \mathbb{Q}## s.t. ##x^2 \leq a^2 < 3##, which implies that ##x## is not an upper bound for ##A##–– a contradiction! Thus, ##\sqrt{3}## must be the least upper bound of ##A##, as desired.

Note:

**Is it simply obvious from this point (*) that ## 3 \leq x^2##, so that ##\sqrt{3} \leq x##, QED? Or is this elaboration needed?**
Last edited: