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I am stuck with,

Give a one to one correspondence between Z+ and positive even integers.

Now, I don't have an idea how to show that there is a one to one correspondence between the two. I would be thankful for some hints.

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Please avoid using images in that fashion. How hard is it to type that in? –  Aryabhata Sep 11 '11 at 14:50
btw, these (this one and your other recent question) look like homework. If so, I would encourage you to tag them as such. –  Aryabhata Sep 11 '11 at 14:52

2 Answers 2

up vote 2 down vote accepted

When asked to prove that "$\exists$ a ..."; then what you are really doing is actually finding whatever is that you need to prove exists.

For example in your question you must prove that there exists a one-to-one correspondance between $\mathbb{Z}^{+}$ and positive even integers, which I will now denote $\mathbb{Z}_e$ so we should attempt to find a map which takes $\mathbb{Z}^{+} \rightarrow \mathbb{Z}_e$.

So how should we go about finding one? well first lets think what is the formal definition of even? I would say an integer $x$ is even if $x = 2k$ for some $k\in \mathbb{Z}$ so the set of positive even integers is $\mathbb{Z}_e = \{x = 2k : k\in\mathbb{Z}^{+}\}$.

Now once we have actually formalized what a positive even integer is it is not hard to think of a map, for example take: $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}_e$ defined by : $k \mapsto 2k$

Now we've got a map we think we will work, and we just need to check if it is one-to-one.

Suppose $f(r) = f(s)$ Then $2r = 2s$, but this quickly implies that $r = s$ so the map is one-to-one, as desired.

Furthermore the map is also onto, because $\mathbb{Z}_e = \{x = 2k : k\in \mathbb{Z}^{+}\}$ is the set of integers of the form $2k$ by definition.

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Consider the f(x)=2x. Now if f(a)=f(b) then 2a=2b then a=b ( because $2\neq0$) this proves the function is injective, to prove that is surjective a positive even integer is of the form 2k with k a positive integer, then f(k)=2k.

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