# Show ##sup\{a \in \mathbb{Q}: a^2 \leq 3\} = \sqrt{3}##

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JVEK7713
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: Is it simply obvious from this point (*) that ## 3 \leq x^2##, so that ##\sqrt{3} \leq x##, QED? Or is this elaboration needed?

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## Answers and Replies

Gold Member
I think the point of proof is:
For any ##\epsilon##>0 there exists {##a |a=m/n, a^2<3## }such that
$$\sqrt{3}-\epsilon < a < \sqrt{3}$$

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Staff Emeritus
Gold Member
I suspect the real point here is that you're supposed to prove ##\sqrt{3}## even exists - you assume it does abs show it's the supremum, but existence as a number is not obvious, and is generally done by showing the Supremum of this set, when squared, must equal 3.

Hall