Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Mathematics
Linear and Abstract Algebra
Show ##sup\{a \in \mathbb{Q}: a^2 \leq 3\} = \sqrt{3}##
Reply to thread
Message
[QUOTE="nuuskur, post: 6848829, member: 519618"] 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. [/QUOTE]
Insert quotes…
Post reply
Forums
Mathematics
Linear and Abstract Algebra
Show ##sup\{a \in \mathbb{Q}: a^2 \leq 3\} = \sqrt{3}##
Back
Top