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Math Help - Determining whether a function is 1-to-1 &/or onto

  1. #1
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    Exclamation Determining whether a function is 1-to-1 &/or onto

    For purposes of the problem, let Z=the set of all integers. Here goes:

    "Define f: Z x Z --> Z by f(n,m)=nm"

    I found f to NOT be 1-to-1. Now I need to determine whether or not it is onto (i.e. ran(f)=Z). If it is not onto I need to determine the range. I have an inkling it is onto but I have no idea to show that (nor how to determine the range!)

    Please help! And Thank You!
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  2. #2
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    Let's dissect this function really quickly, first we define the set of integers:

    \mathbb{Z} = \{..., -3, -2, -1, 0, 1, 2, 3, ...\}

    Ok, now we will define the Cartesian Product of itself, which I will call \mathbb{Z}^2:

    \mathbb{Z}^2 = \{..., (-3, -3), (-3, -2), (-3, -1), ..., (3, -3), (3, -2), ..., (3, 1), (3, 0), ...\}

    So, now we have what we need to define our function, which is:

    f : \mathbb{Z}^2 \to \mathbb{Z}

    This is a binary function defined by:

    f(n, m) = nm

    So, the range consists of all the products of the integers which is a subset of \mathbb{Z}, and the domain consists of any pair of integers.

    It can't be one-to-one because \mathbb{Z}^2 and \mathbb{Z} are equinumerous, so the range of the function is also a subset of \mathbb{Z}^2, and can therefore cover the entire range, and so it is onto.
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  3. #3
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    Quote Originally Posted by dolphinlover View Post
    For purposes of the problem, let Z=the set of all integers. Here goes:

    "Define f: Z x Z --> Z by f(n,m)=nm"

    I found f to NOT be 1-to-1. Now I need to determine whether or not it is onto (i.e. ran(f)=Z). If it is not onto I need to determine the range. I have an inkling it is onto but I have no idea to show that (nor how to determine the range!)
    z \in \mathbb{Z}\; \Rightarrow \;f(1,z) = z
    It is clearly onto. Then the range is \mathbb{Z}
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  4. #4
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    Quote Originally Posted by Plato View Post
    z \in \mathbb{Z}\; \Rightarrow \;f(1,z) = z
    It is clearly onto. Then the range is \mathbb{Z}

    It has been a while since I've been to the forum. What you stated was exactly what I used to show it was onto. Thanks!
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