Bijective function

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Albi

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Construct a bijective function f : (0,1) => (2,3) u {5} u (10,100)

Can someone help me solve this problem, because I'm having difficulties.
 
Albi said:
Construct a bijective function f : (0,1) => (2,3) u {5} u (10,100)

Can someone help me solve this problem, because I'm having difficulties.
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Can you provide the definition and an example of a bijective function?
 
Albi said:
We pick a point in (0,1), say 0.5 and I map it to 2.5 for example
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This is not a function. You need a function, e.g. `y=x^2`. Can you think of one that maps (0,1) to (2,3)?
 
Albi said:
Is it f(x) = x + 2
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Yes. For a complete answer you should probably demonstrate that it is, in fact bijective for the specified intervals.

How about mapping (0,1) to (0,3)? And (0,1) to (2,4)?
 
lev888 said:
Yes. For a complete answer you should probably demonstrate that it is, in fact bijective for the specified intervals.

How about mapping (0,1) to (0,3)? And (0,1) to (2,4)?
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This is the point that I don't understand, why should we map (0,1) to (0,3) and (0,1) to (2,4). How should we "group" these
 
Albi said:
A bijective function is a function that is onto and one-to-one.
Ex: f(x) = x² from R to R
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X^2 is not 1-1. I can get 4 from -2 or 2. It is not onto since f(x)>=0. For example -9 is in R but nothing from R maps to it. In the end, f(x): R -> R is neither 1-1 nor onto
 
Albi said:
Anyways, we can map (0,1) to (0,3) using g(x) = 3x, and for the second part we can map (0,1) to (2,4) using h(x) = 2x +2
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Last one: (2,3) to (5,20). Feel free to skip it if you already know how to map any open interval into another open interval.
Now to the original problem: can you do it or something is not clear?
 
Jomo said:
X^2 is not 1-1. I can get 4 from -2 or 2. It is not onto since f(x)>=0. For example -9 is in R but nothing from R maps to it. In the end, f(x): R -> R is neither 1-1 nor onto
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Then I can change that and say f(x) = x² from N => N
 
Albi said:
Construct a bijective function \(\Large f : (0,1) \to (2,3) \cup \{5\} \cup (10,100)\)
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\(f(t)=2+2t,\text{ for }0<t<0.5\\\;\;\;~~~~5.\text{ for }t=0.5\\~~~{\Large~?}\text{ for }0.5<t<1\)
 
Find the mapping (a linear one would be perfect) that maps (0, .5) to (2,3) and a 2nd mapping that maps (.5, 0) to (10, 100) (think linear again. Now guess what maps to 5.

By the way, in (0, .5), that .5 could be replaced with any number in (0,1). Do you see that?
 
Jomo said:
Find the mapping (a linear one would be perfect) that maps (0, .5) to (2,3) and a 2nd mapping that maps (.5, 0) to (10, 100) (think linear again. Now guess what maps to 5.

By the way, in (0, .5), that .5 could be replaced with any number in (0,1). Do you see that?.5
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For the first mapping we can say f(x) = 2x +2
but I'm not sure about the second mapping, by the way, why (.5, 0 ), I mean why are we going backwards?
 
\(f(t) =\)\( \left\{ {\begin{array}{*{20}{c}} {2 + 2t}&,&{0 < t < 0.5} \\ 5&,&{t = 0.5} \\ {180t-80}&,&{0.5 < t < 1} \end{array}} \right.\)
 
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