I am struggling with this question and was hoping somebody could help me, Thanks
Find a bijection between the intervals $(-1,1)$ and $(0,4)$ where $X \in R$
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2What is $X$? And what did you try? And why the "infinity" tag? – Ludolila May 01 '16 at 08:24
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You can consider the equation of a straight line. – Soby May 01 '16 at 08:25
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Are you shure about $(1,1)$? – T'x May 01 '16 at 08:25
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2Do you mean first interval $(0,1)$? Because $(1,1)$ is empty. – coffeemath May 01 '16 at 08:25
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4Once the intervals are corrected, try maps of the form $x\mapsto ax+b$ – Hagen von Eitzen May 01 '16 at 08:30
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Sorry it is between the interval (-1,1) and (0,4). I didn't know how to get the bijective so I tried getting the injective and the surjective but that confussed me even more – Emma May 01 '16 at 08:55
2 Answers
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$f:(\color\red{-1},\color\green{1})\rightarrow(\color\orange{0},\color\purple{4})$
$f(x)=(x-(\color\red{-1}))\cdot\frac{\color\purple{4}-\color\orange{0}}{\color\green{1}-(\color\red{-1})}+\color\orange{0}=2x+2$
$g:(\color\orange{0},\color\purple{4})\rightarrow(\color\red{-1},\color\green{1})$
$g(x)=(x-\color\orange{0})\cdot\frac{\color\green{1}-(\color\red{-1})}{\color\purple{4}-\color\orange{0}}+(\color\red{-1})=\frac12x-1$
barak manos
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Thanks a lot. Is that a formula that you used to get the answer? Sorry, just wondering for in future if i get a similar question again. Really appreciate your help. – Emma May 01 '16 at 09:50
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@Emma: Yes, I've used four different colors in order to make it clear what the formula is. – barak manos May 01 '16 at 09:52
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the bijection between two open intervals is simple the line joining those two points where "y" is f(x)
