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Math Help - providing example of one to one ,onto functions......

  1. #1
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    providing example of one to one ,onto functions......

    I have problem with following....
    I was given constrain that function must have input between (0,1) and output (1,3)-open intervals
    I want to prove or disprove that example function I have given f(x)=1/2*x +1 is one to one or onto.
    Can someone give me example how to prove or disprove this.
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  2. #2
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    Your function is not correct. It needs to be f(x)=2x+1. That is 1-1 and onto (i.e., a bijection).

    To show this is 1-1, look at the definition of 1-1. A function is 1-1 if and only if for every x,y\in\mathcal{D}(f) such that x\not=y, it follows that f(x)\not=f(y). Or, equivalently, f(x)=f(y) implies x=y. So, assume x,y\in(0,1) and f(x)=f(y). Can you prove that x=y?

    For onto, you have to show that the function "hits" every value in the range. That is, for every y\in\mathcal{R}(f), there exists x\in\mathcal{D}(f) such that y=f(x). So, pick a y\in(1,3). Now show that you can find an x\in(0,1) such that y=f(x).

    Can you finish from here?
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  3. #3
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    Problem finding example

    I have trouble finding actual example of function. Is there any rule you can use
    to find example.
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  4. #4
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    Any linear function is a bijection.

    Consider the function: f(x) = \left( {\frac{{d - c}}{{b - a}}} \right)\left[ {x - a} \right] + c

    Does that map \left( {a,b} \right) \leftrightarrow \left( {c,d} \right)?
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  5. #5
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    thank you

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  6. #6
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    Reason I asked this is because my teacher during test can come up with some other constrain and ask if we can produce functions that follows the constrain. He might ask to produce one function for 1-1 ,onto.
    I was just curious if there is rule you can use that no matter what constrain he puts on I can produce function examples.
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  7. #7
    A Plied Mathematician
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    It is not true that for any two sets A and B, there exists a bijection between them. In particular, if the two sets have different cardinality, you won't be able to find a bijection. Make sense?
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