How to prove $f:Z \to E$, given by $f(x) = 2x$ is bijective?

Asked Oct 22 '18 at 19:40

Active Oct 22 '18 at 20:44

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I know I must prove that $f$ is both injective and surjective. I've got the injective proof down. But I'm stuck on the surjective proof.

So far I have, for some $x,y ∈ Z$, $y = 2x \Rightarrow x = y/2$. Now does this prove that $f$ is surjective as well since $y$ would also $\in E$ (even numbers)? Or am I missing something?

edited Oct 22 '18 at 20:44

molarmass

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asked Oct 22 '18 at 19:40

johntc121

You're essentially correct. Be sure for the second part that you start from the assumption that $y$ is even. What does that tell you about $y/2$? – Ethan Bolker Oct 22 '18 at 19:43
Okay so just make sure to start with something like: for some x,y ∈ Z and y ∈ E? – johntc121 Oct 22 '18 at 19:44
You need to write clearer sentences. Start with $y\in E$. What precisely does this mean? – Ted Shifrin Oct 22 '18 at 19:44
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No. You start only with $y\in E$. – Ted Shifrin Oct 22 '18 at 19:44
Okay gotcha. Thanks.
Umm y∈E means that y is Even. I tried to keep the question short. But it describes the set E in our book, E being even integers.
– johntc121 Oct 22 '18 at 19:46
What is your definition of an even number? Start there. – Ethan Bolker Oct 22 '18 at 19:47

How to prove $f:Z \to E$, given by $f(x) = 2x$ is bijective?

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