Talk:Division Theorem/Proof 1
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Prime.mover, although I was able to figure out what the syntax intended, it is by no means perfectly good. $x = y+2, y \in \Z$ (or whatever) is really not a substitute for $\exists y \in \Z: x = y + 2$. --Dfeuer (talk) 05:16, 10 April 2013 (UTC)
- It's good enough. --prime mover (talk) 05:24, 10 April 2013 (UTC)
- It's perfectly good enough. --prime mover (talk) 20:38, 10 April 2013 (UTC)
- Compromise. --prime mover (talk) 07:46, 11 April 2013 (UTC)