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

• I
• JVEK7713
In summary, the author claims that ##\sqrt{3}## is the least upper bound of the set ##A##, and that it is defined using the completeness argument.
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?

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

Last edited:
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
##\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.

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.

### Similar threads

• Precalculus Mathematics Homework Help
Replies
4
Views
1K
• Linear and Abstract Algebra
Replies
11
Views
1K
• Linear and Abstract Algebra
Replies
4
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
1K
• Topology and Analysis
Replies
7
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
905
• Calculus and Beyond Homework Help
Replies
7
Views
750
• Linear and Abstract Algebra
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
828
• Topology and Analysis
Replies
4
Views
446