Let O be the set of odd integers and E be the set of even integers.
Prove that |Z| = |E|
Any advice?
I did get that part then.
Is this right? To show it is onto I said that any x plugged into the function would come out even. Then the image of f is E and thus onto.
To show 1-1, I said for every x plugged into the function, there would be exactly one output, thus 1-1.
How does that sound?
1-1 means, $\displaystyle f(x_1) = f(x_2) \implies x_1 = x_2$ for all $\displaystyle x_1,x_2 \in \mathbb{Z}$.
So assume $\displaystyle f(x_1) = f(x_2) \implies 2x_1 = 2x_2 \implies x_1 = x_2$ by dividing through by 2. hence $\displaystyle f(x)$ is one-to-one
Follow Isomorphisms directions to show $\displaystyle f(x)$ is onto