Timeline for Guidelines for finding maximum, minimum, maximal and minimal elements of a poset
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 28, 2012 at 17:23 | vote | accept | haunted85 | ||
| Aug 28, 2012 at 10:05 | answer | added | Brian M. Scott | timeline score: 1 | |
| Aug 28, 2012 at 9:50 | comment | added | Brian M. Scott | @Tobias: It depends on the definition that you’re using: it doesn’t work if you define $\gcd(a,b)$ to be, literally, the greatest (largest) common divisor. | |
| Aug 28, 2012 at 9:42 | comment | added | haunted85 | @BrianM.Scott I'm very sorry while copying my homework on screen, I must have messed up the definition of $\pi$, now it's fixed. | |
| Aug 28, 2012 at 9:40 | comment | added | Tobias Kildetoft | @BrianM.Scott What is wrong with defining $gcd(0,0) = 0$? $0$ is the unique common divisor of $0$ and $0$ such that any common divisor of $0$ and $0$ divides it. | |
| Aug 28, 2012 at 9:39 | history | edited | haunted85 | CC BY-SA 3.0 | edited body |
| Aug 28, 2012 at 9:38 | comment | added | Brian M. Scott | Because a partial order is antisymmetric: if $a\,R\,b$ and $b\,R\,a$, then $a=b$. Your $\pi$, however, is not antisymmetric, as was already noted: $\langle 1,2\rangle\,\pi\,\langle 1,3\rangle$ and $\langle 1,3\rangle\,\pi\,\langle 1,2\rangle$, but $\langle 1,2\rangle\ne\langle 1,3\rangle$. (If your text actually defines $\gcd(0,0$, that’s fine, but it’s by no means a universal convention, so I didn’t expect it.) | |
| Aug 28, 2012 at 9:32 | comment | added | haunted85 | @BrianM.Scott why is $\pi$ not a partial order? And my textbook clearly states if $a = b = 0$ then the (only) $\gcd(a,b)=0$. | |
| Aug 28, 2012 at 9:26 | comment | added | Brian M. Scott | You still have the problem that $\pi$ isn’t a partial order, and in any case $\gcd(0,0)$ is undefined. | |
| Aug 28, 2012 at 9:23 | history | edited | haunted85 | CC BY-SA 3.0 | deleted 8 characters in body |
| Aug 28, 2012 at 9:21 | comment | added | haunted85 | Yes, you are totally right. Fixing that up. | |
| Aug 28, 2012 at 9:08 | history | asked | haunted85 | CC BY-SA 3.0 |