tag:blogger.com,1999:blog-4264810422877338082.post7670805101862450034..comments2023-10-03T22:19:01.016-07:00Comments on MetaMathological: Maths on Steroids: Tute 5Werewolframhttp://www.blogger.com/profile/11388384764792312184noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-4264810422877338082.post-13087977100599523212013-04-22T14:07:08.431-07:002013-04-22T14:07:08.431-07:00Hey Anonymous! It's great to finally meet you ...Hey Anonymous! It's great to finally meet you - you seem to be omnipresent on the interwebs, how are you so prolific? :P<br /><br />I don't really know of a cooler way to do this actually - I mean, you could do it more directly with any bijection \(\mathbb{R}\rightarrow(0,1)\) and do the same trick but using the usual addition and scalar multiplication on R. And for a function like \(\mathrm{Arctan}\) there might be a semi-nice interpretation of the operations in terms of trigonometric identities and geometry...=/<br /><br />Btw: you can TeX using backslash+open-bracket, backslash-close bracket in MathJax. Werewolframhttps://www.blogger.com/profile/11388384764792312184noreply@blogger.comtag:blogger.com,1999:blog-4264810422877338082.post-37648342457520410522013-04-22T07:59:21.094-07:002013-04-22T07:59:21.094-07:00There's a bijection
$\varphi: \mathbb{R}^+ \ri...There's a bijection<br />$\varphi: \mathbb{R}^+ \rightarrow (0,1)$ given by $x \mapsto \frac{x}{x+1}$.<br /><br />Define vector addition by $x + y = \varphi( \varphi^{-1}(x) \oplus \varphi^{-1}(y))$ where $\oplus$ is the sum you defined in your post. <br /><br />Scalar multiplication is defined similarly. You've checked all the axioms for me!<br /><br />Was there a cooler way to do this? <br /><br />Oh you should probably delete / not post this comment if you prefer to leave the question open for your students.Anonymousnoreply@blogger.com