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Thread: Onto Proof

  1. #1
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    Exclamation Onto Proof

    Can someone give me some feedback, please?
    ********
    f: Ζ⇒Ζ

    f(x) = { x/2 if x is even, 0 if x is odd

    Prove whether f is onto and one-to-one.
    Here's my proof.
    ******
    Proof. To show that f is onto, let b∈Ζ. Consider the 2 following cases:

    1. If b=0, f(x)=0 implies x=z.
    2. If b≠0, x/2=b gives x=2b which implies f(2b)=2b/b=b.

    This shows that f has 2 pre-images which proves that f is onto. Since f(0)=0=f(1),
    we have also proved that f is not one-to-one. ■
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  2. #2
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    Onto and one-to-one function

    Hello yvonnehr
    Quote Originally Posted by yvonnehr View Post
    Can someone give me some feedback, please?
    ********
    f: Ζ⇒Ζ

    f(x) = { x/2 if x is even, 0 if x is odd

    Prove whether f is onto and one-to-one.
    Here's my proof.
    ******
    Proof. To show that f is onto, let b∈Ζ. Consider the 2 following cases:

    1. If b=0, f(x)=0 implies x=z.
    2. If b≠0, x/2=b gives x=2b which implies f(2b)=2b/b=b.

    This shows that f has 2 pre-images which proves that f is onto. Since f(0)=0=f(1),
    we have also proved that f is not one-to-one. ■
    Your proof is fine (I take it $\displaystyle x =z$ should be $\displaystyle x=0$), but I think you can simplify it a bit.

    You don't really need to say anything about $\displaystyle b = 0$ at all for the 'onto' proof. (You're not trying to divide anything by zero, are you? And $\displaystyle 0 \div 2 = 0$, which is OK.)

    So I think all you really need to say is:

    $\displaystyle \forall b \in \mathbb{Z}, 2b$ is even and $\displaystyle 2b \in \mathbb{Z}$

    $\displaystyle \Rightarrow f(2b) =\frac{2b}{2}=b$

    $\displaystyle \Rightarrow f$ is onto.

    And your counterexample ($\displaystyle f(0) =0= f(1)$) is all you need to prove that $\displaystyle f$ is not one-to-one.

    Grandad
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