(New page: <math>\text{Show: Given r } \in [-1,1] \text{, show there exist elements in the Cantor set } x,y \text{ such that } x-y=r.</math> <math>\text{Proof: Let } \mathcal{C} \text{ denote the Can...) |
|||
| (One intermediate revision by the same user not shown) | |||
| Line 1: | Line 1: | ||
| + | [[Category:MA598RSummer2009pweigel]] | ||
| + | [[Category:MA598]] | ||
| + | [[Category:math]] | ||
| + | [[Category:problem solving]] | ||
| + | [[Category:real analysis]] | ||
| + | |||
| + | =Solution for HW2.17, MA598, Weigel, Summer 2009= | ||
| + | |||
<math>\text{Show: Given r } \in [-1,1] \text{, show there exist elements in the Cantor set } x,y \text{ such that } x-y=r.</math> | <math>\text{Show: Given r } \in [-1,1] \text{, show there exist elements in the Cantor set } x,y \text{ such that } x-y=r.</math> | ||
<math>\text{Proof: Let } \mathcal{C} \text{ denote the Cantor set. Define } f: </math> <math> \mathcal{C} \times \mathcal{C} \rightarrow [0,1] \text{ by } (x,y) </math> <math> \mapsto \frac{x+y}{2}. </math> | <math>\text{Proof: Let } \mathcal{C} \text{ denote the Cantor set. Define } f: </math> <math> \mathcal{C} \times \mathcal{C} \rightarrow [0,1] \text{ by } (x,y) </math> <math> \mapsto \frac{x+y}{2}. </math> | ||
| Line 4: | Line 12: | ||
<math>\exists x, y \in \mathcal{C} \text{ s.t. } r+1 = x+y \Rightarrow r=x-(1-y). </math> | <math>\exists x, y \in \mathcal{C} \text{ s.t. } r+1 = x+y \Rightarrow r=x-(1-y). </math> | ||
<math>\text{ Since } 1-y \in \mathcal{C} \text{ by symmetry, } \square. </math> | <math>\text{ Since } 1-y \in \mathcal{C} \text{ by symmetry, } \square. </math> | ||
| + | ---- | ||
| + | ---- | ||
| + | [[MA_598R_pweigel_Summer_2009_Lecture_2|Back to Assignment 2, MA598, Summer 2009, Weigel]] | ||
| + | |||
| + | [[MA598R_%28WeigelSummer2009%29|Back to MA598R Summer 2009]] | ||
Latest revision as of 05:49, 11 June 2013
Solution for HW2.17, MA598, Weigel, Summer 2009
$ \text{Show: Given r } \in [-1,1] \text{, show there exist elements in the Cantor set } x,y \text{ such that } x-y=r. $ $ \text{Proof: Let } \mathcal{C} \text{ denote the Cantor set. Define } f: $ $ \mathcal{C} \times \mathcal{C} \rightarrow [0,1] \text{ by } (x,y) $ $ \mapsto \frac{x+y}{2}. $ $ \text{ Now f is clearly onto by examining the ternary representation of an element of } [0,1]. \text{ Given } r \in [-1,1], \frac{r+1}{2} \in [0,1] \Rightarrow $ $ \exists x, y \in \mathcal{C} \text{ s.t. } r+1 = x+y \Rightarrow r=x-(1-y). $ $ \text{ Since } 1-y \in \mathcal{C} \text{ by symmetry, } \square. $
