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Math Help - injectivity and surjectivity (onto and one-one)

  1. #1
    Junior Member scottie.mcdonald's Avatar
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    injectivity and surjectivity (onto and one-one)

    I was going over my assignment to study for the upcoming midterm, and I thought I had a grasp on surjectivity and injectivity...I don't. Can someone tell me where I went wrong in proving these to 'jectives'.

    define g:integers -> integers be defined by g(n)=4n-5. is it injective? surjective?

    I wrote:

    Injective:
    Suppose g(n)=g(m)
    4n-5=4m-5
    4n=4m
    n=m therefor it is injective

    Surjective:
    let f(a)=b
    4a -5=b
    a=(b+5)/4

    sub back into the equation g(a)=4a-5 to get 4[(b+5)/4]-5
    =b+5-5
    =b, so therefor it is surjective

    I think this is correct, but he wrote something along the lines of f(x)=2 has no solution in integers. This is true, so why does my way of doing surjectivity not work. What am I doing or assuming wrong?
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by scottie.mcdonald View Post
    I was going over my assignment to study for the upcoming midterm, and I thought I had a grasp on surjectivity and injectivity...I don't. Can someone tell me where I went wrong in proving these to 'jectives'.

    define g:integers -> integers be defined by g(n)=4n-5. is it injective? surjective?

    I wrote:

    Injective:
    Suppose g(n)=g(m)
    4n-5=4m-5
    4n=4m
    n=m therefor it is injective

    Surjective:
    let f(a)=b
    4a -5=b
    a=(b+5)/4

    sub back into the equation g(a)=4a-5 to get 4[(b+5)/4]-5
    =b+5-5
    =b, so therefor it is surjective

    I think this is correct, but he wrote something along the lines of f(x)=2 has no solution in integers. This is true, so why does my way of doing surjectivity not work. What am I doing or assuming wrong?
    The Key is what set is your domain.

    Your function is going from the integers to the integers

    The statement \frac{b+5}{4} the input is NOT an integer for all b.

    Using your example

    suppose 4n-5=2 \implies n=\frac{7}{4} but this is a rational number Not an integer.

    So the function would be surjective over the rationals, but it is not over the integers.

    I hope this helps.

    Good luck.
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  3. #3
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    Quote Originally Posted by scottie.mcdonald View Post
    I was going over my assignment to study for the upcoming midterm, and I thought I had a grasp on surjectivity and injectivity...I don't. Can someone tell me where I went wrong in proving these to 'jectives'.

    define g:integers -> integers be defined by g(n)=4n-5. is it injective? surjective?

    I wrote:

    Injective:
    Suppose g(n)=g(m)
    4n-5=4m-5
    4n=4m
    n=m therefor it is injective

    Surjective:
    let f(a)=b
    4a -5=b
    a=(b+5)/4

    sub back into the equation g(a)=4a-5 to get 4[(b+5)/4]-5
    =b+5-5
    =b, so therefor it is surjective

    I think this is correct, but he wrote something along the lines of f(x)=2 has no solution in integers. This is true, so why does my way of doing surjectivity not work. What am I doing or assuming wrong?
    Proving surjectivity ( onto) for a function some times is very difficult if not impossible.

    Remember the definition that the function g is surjective over integers is:

    for all integers m we must find an integer n such that 4n-5=m

    But for each definition you must consider its negation ,and the negation of the above definition is:

    there exists an integer m such that for all integers n ,  4n-5\neq m.

    If we put now m=1,we have: 4n-5=1 or n=6/4 .That result indicates that even if we try all the integers n we will not get 4n-5 equal to 1.

    Hence the negation of the definition for surjectivity is satisfied,because:

    there exists an integer m=1 such that for all n integers  4n-5\neq m.

    Thus g is not surjective over all integers ,but definitely is surjective over a
    subset A of Z : g: Z\rightarrow A\subset Z

    A good problem is to find that A
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