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

[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?

 

[anuttarasammyak]

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 &lt; a &lt; \sqrt{3}

 

[Office_Shredder]

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.

 

[Erland]

$\sqrt 3$ is here defined as the supremum of the set $A$. It is not a priori given that the square of this number is 3, nor that it cannot be greater than any number whose square is less than or equal to 3.

 

[nuuskur]

The set in question is nonempty and bounded from above. Hence it has a supremum, call it $s$ and we can define $\sqrt{3}:=s$. One can't prove anything about something that is undefined. The symbol $\sqrt{3}$ has no meaning beforehand.

It is reasonable to ask, whether $\sqrt{}$ defined on the nonnegative rationals using the completeness argument is well defined. And it is.