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  1. #1
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    homework help

    For each ordered pair ( a, b) of integers define a mapping alpha (a, b) : Z . Z by alapha a, b( n) = an + b.

    ( a) For which pairs ( a, b) is alpha (a, b) onto?

    ( b) For which pairs ( a, b) is alpha (a, b) one- to- one?
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  2. #2
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    Is it \alpha(a,b): \mathbb{Z} \rightarrow \mathbb{Z} defined by \alpha(a,b)(n) = an+b ?

    For (a), consider |a|>1 ...can \alpha(a,b) be onto? Then examine remaining cases for a.

    In (b), again concentrate on the value of a (parameter b acts like a shift, it can't affect whether the function is one-to-one or not). There are not many cases in which the function is not one-to-one, can you see them?
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  3. #3
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    for part a is there a easier way to prove this proof instead looking at the problem from both sides of iff and only if

    I have this so far for part a

    right side: an+b= b+1
    an=1
    a=plus/minus 1
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  4. #4
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    Quote Originally Posted by ratterlgirl10 View Post
    For each ordered pair ( a, b) of integers define a mapping alpha (a, b) : Z . Z by alapha a, b( n) = an + b.

    ( a) For which pairs ( a, b) is alpha (a, b) onto?

    ( b) For which pairs ( a, b) is alpha (a, b) one- to- one?
    This notation is very hard to read.
    Here is a basic fact about linear functions.
    f(x) = ax + b\;\& \;a \ne 0 is both onto and one-to-one.

    But I don't understand your notation.
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  5. #5
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    Quote Originally Posted by ratterlgirl10 View Post
    I have this so far for part a

    right side: an+b= b+1
    an=1
    a=plus/minus 1
    That's actually a good idea! So now you know that for \alpha(a,b) to be onto, it is necessary that a=1 or a=-1. It is now trivial to verify that this is also sufficient.

    Is the rest clear?
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