Squaring Both Sides Of An Inequality (With Proof Using The Axioms Of Ordered Fields)

Опубликовано: 27 Апрель 2026
на канале: Amour Learning
654
11

In this video, I prove that if 0 is less than x and x is less than y, then x^2 is less than y^2 (where x and y are elements of an ordered field F).

This problem can be found in Dr. Jirí Lebl's free open-access textbook: "Basic Analysis I: Introduction to Real Analysis, Volume I"
https://www.jirka.org/ra/

Any feedback on how these videos can be improved would be great. I will be constantly updating these videos to ensure that everything is accurate. Thank you all for being such a great community. All content is released under Creative Commons "No Rights Reserved", so please feel free to use this video however you want.

My name is Kody Amour, and I teach free undergraduate mathematics courses on YouTube. I'm currently developing three YouTube courses simultaneously: Discrete Math, Linear Algebra, Real Analysis, Topology, Number Theory and Complex Analysis. These courses correspond to playlists that I am constructing indefinitely.

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Credit to all of the authors of all of the resources below that greatly helped influence this video.

Free open-access online math textbooks (I only use textbooks from this resource, since everything is free for my students):
https://aimath.org/textbooks/approved...

A free in-depth tutorial for SageMath: https://doc.sagemath.org/html/en/tuto...

Learn LaTeX in 30 minutes: https://www.overleaf.com/learn/latex/...
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My favorite math-oriented YouTube channels (in no particular order):
Aleph 0:    / @aleph0  
The Bright Side of Mathematics:    / @brightsideofmaths  
Math Visualized:    / @mathvisualized355  
Insights into Mathematics:    / @njwildberger  
ScienceClic English:    / @scienceclicen  
Stand-up Maths:    / @standupmaths  
Zach Star:    / @zachstar  
Numberphile:    / @numberphile  
3Blue1Brown:    / @3blue1brown  
MIT OpenCourseWare:    / @mitocw  
Khan Academy:    / @khanacademy  
PBS Infinite Series:    / @pbsinfiniteseries  
Lex Fridman:    / lexfridman