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Math Help - One-to-one and onto

  1. #1
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    One-to-one and onto

    Hey everyone, for homework we have to have one example of each under the condition that A = {odd numbers} and B = {even numbers}, thus:

    give examples of functions from A to B which are:

    a) One-to-one but not onto
    b) Onto but not one-to-one
    c) Neither one-to-one nor onto
    d) One-to-one and onto

    I've been working at this and I can't seem to get even the simplest of examples, ugh. Thanks for the help!
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  2. #2
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    Quote Originally Posted by teacast View Post
    Hey everyone, for homework we have to have one example of each under the condition that A = {odd numbers} and B = {even numbers}, thus: give examples of functions from A to B which are:
    a) One-to-one but not onto
    b) Onto but not one-to-one
    c) Neither one-to-one nor onto
    d) One-to-one and onto
    Here is part a) g(n)=4n.
    Reply by showing that the function works.

    Also reply with your attempt at b).
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  3. #3
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    For a) g(n) = 4n,

    Anything can be put into n, so when you apply g(n), then the number will always be even, thus always being in B? So 4 times any odd number will yield an even number, is that right?


    As for b)

    I would say that g(A) = B because there is at least one a in A such that g(a) = b

    Let me know if/what I am doing or saying wrong, ugh 1-1 and onto escape me
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  4. #4
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    1-to-1 means no element in your domain A is simultaneously sent/mapped to two different images in your Range B. Nor are two different elements in your domain A mapped to the same image in the range B.

    Onto means every element in your codomain B is 'hit'. That is, your map/function will map (some or all of) the odd numbers to all of the even numbers.

    Your function needs to specify what it does to an element in A

    so g(A) is ambiguous, as A is the whole set.

    g(input) = output

    your input is a member of A, your output is a member of B, g "maps" or "sends" your input to your output. You get to define the rule for this mapping.
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  5. #5
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    Update:

    So I looked it over, and read about onto and one to one and I think I get it.

    for a)

    g(x) = 4x because whatever 4 is multiplied by, it will always yield an even number, thus satisfying what I want (input is odd, output is even).

    for b)

    f(x) = x^2+1 where x is in A

    because Onto means that for every y in B, there is an x in A such that f(x)=y. So x^2 + 1 when the odd number is squared and one is added will always yield an even number


    I'm still working on c) and d)
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  6. #6
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    Quote Originally Posted by teacast View Post
    Hey everyone, for homework we have to have one example of each under the condition that A = {odd numbers} and B = {even numbers}, thus:

    give examples of functions from A to B which are:

    a) One-to-one but not onto
    b) Onto but not one-to-one
    c) Neither one-to-one nor onto
    d) One-to-one and onto

    I've been working at this and I can't seem to get even the simplest of examples, ugh. Thanks for the help!
    Hmmm...
    the best notation to use would probably be the:

     f: \{ \begin{array}{l} \ A \rightarrow B\\ n \mapsto f(n) \end{array}

    c) might be:
     f: \{ \begin{array}{l} \ A \rightarrow B\\ n \mapsto f(0) \end{array}
    d) is probably pretty trivial
     f: \{ \begin{array}{l} \ A \rightarrow B\\ n \mapsto f(2n) \end{array}
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  7. #7
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    Quote Originally Posted by teacast View Post
    For a) g(n) = 4n,

    Anything can be put into n, so when you apply g(n), then the number will always be even, thus always being in B? So 4 times any odd number will yield an even number, is that right?
    Yes, that is true. But now you need to show g is "one to one but not onto".


    As for b)

    I would say that g(A) = B because there is at least one a in A such that g(a) = b

    Let me know if/what I am doing or saying wrong, ugh 1-1 and onto escape me
    That makes no sense at all. What "g" are you talking about? Not g(n)= 4n, certainly, that was given to you as a function that is "one to one but not onto". You can't also say that it is "onto but not one to one"!
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