Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Please avoid using images in that fashion. How hard is it to type that in? – Aryabhata Sep 11 '11 at 14:50
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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.