Timeline for If $0 \le a \lt b$ prove that $a^2 < b^2$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 17, 2015 at 16:38 | history | edited | kingW3 | Added tags | |
| Feb 17, 2015 at 16:34 | answer | added | kahen | timeline score: 0 | |
| Feb 17, 2015 at 16:33 | answer | added | Quickbeam2k1 | timeline score: 0 | |
| Feb 17, 2015 at 16:32 | answer | added | user207710 | timeline score: 5 | |
| Feb 17, 2015 at 16:31 | answer | added | Uncountable | timeline score: 0 | |
| Feb 17, 2015 at 16:31 | answer | added | Aaron Maroja | timeline score: 0 | |
| Feb 17, 2015 at 16:29 | answer | added | Ivo Terek | timeline score: 3 | |
| Feb 17, 2015 at 16:27 | answer | added | graydad | timeline score: 1 | |
| Feb 17, 2015 at 16:26 | comment | added | Bill Dubuque | See here for a simpler proof of a slightly stronger result. | |
| Feb 17, 2015 at 16:26 | comment | added | Nicolas | You could just say that $x\mapsto x^{2}$ is a non-decreasing function on $\mathbb{R}_{+}$, so preserves the inequality. | |
| Feb 17, 2015 at 16:23 | history | asked | idonno | CC BY-SA 3.0 |