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Math Help - [SOLVED] proving bijectivity

  1. #1
    Junior Member scottie.mcdonald's Avatar
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    [SOLVED] proving bijectivity

    Define g: Integers to Integers by g(n)=4n-5. Is g injective? surjective? justify your answer

    injectivity
    suppose g(n)=g(m) for some m element of integers
     \Rightarrow 4n-5=4m-5

     \Rightarrow 4n=4m

     \Rightarrow n=m
    so therefor this statement is injective since each element of g(n goes to an element of g(m)

    for surjectivity i know i have to let b E integers and let a E integers since g:integers to integers and i know i have to show that f(a)=b. so:

    f(a)=b
    4a-5=b
    a=  \frac{b+5}{4}

     f(a)= 4 \frac{b+5}{4} - 5
     = \frac{4b+20}{4} - 5
    =b+5-5
    =b

    this is wrong, but i don't understand why. i tried following the books example, but i guess it only works for their specific example.

    how am i to show a generic proof for surjectivity in this case? i know that f(a)=2 has no solution in integers (from what my professor wrote on my paper). as well, how do i write the integers symbol using latex?

    thank you,

    Scott
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  2. #2
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    Your mistake is in not realizing that a must be an integer.
    If b=1 can you find such an a? NO!
    Is the function surjective?
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  3. #3
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    Quote Originally Posted by scottie.mcdonald View Post
    Define g: Integers to Integers by g(n)=4n-5. Is g injective? surjective? justify your answer

    injectivity
    suppose g(n)=g(m) for some m element of integers
     \Rightarrow 4n-5=4m-5

     \Rightarrow 4n=4m

     \Rightarrow n=m
    so therefor this statement is injective since each element of g(n goes to an element of g(m)

    for surjectivity i know i have to let b E integers and let a E integers since g:integers to integers and i know i have to show that f(a)=b. so:

    f(a)=b
    4a-5=b
    a=  \frac{b+5}{4}

     f(a)= 4 \frac{b+5}{4} - 5
     = \frac{4b+20}{4} - 5
    =b+5-5
    =b

    this is wrong, but i don't understand why. i tried following the books example, but i guess it only works for their specific example.

    how am i to show a generic proof for surjectivity in this case? i know that f(a)=2 has no solution in integers (from what my professor wrote on my paper). as well, how do i write the integers symbol using latex?

    thank you,

    Scott
    The definition for surjectivity is :

    FOR all bεZ there exists an aεD(g) ( = domain of g = Z) such that g(a) =b

    Now the negation of the above statement is:

    There exists a bεZ SUCH that for all aεD(g) g(a)=/=b.

    So if you take b=2 you get that,according to the formula g(a) = 4a-5,a=7/4.

    That means that if you try ALL the members of your domain none of them will give you the No 2.Only the No 7/4 will give you 2 ,which does not belong to D(g)

    Hence the function is not surjective.

    Note for surjectivity you must have :

    Range of g = Z
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  4. #4
    Junior Member scottie.mcdonald's Avatar
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    ahh yes, a must be an integer. *sobs*, i should have noticed that.

    thank you for your help.

    Scott
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