# Prove r is one-to-one and onto.

$\mathbb N_k$ = { n $\in$ $\mathbb N$: n $\le$ k}, let k $\ge$ 2. Define r: $\mathbb N_k$ $\to$ $\mathbb N_k$ by r(k)=1 and for all x $\in$ $\mathbb N_k$ , if x < k, then r(x)=x+1. Prove r is one-to-one and onto.

I'm sorry if this looks weird. This is my first time posting here and I couldn't find a guide on how to properly post mathematical symbols, etc.

I don't really know where to even begin proving this. Upon looking at it, all I can tell is that it doesn't appear to be one-to-one, but I'm not sure about onto. I don't have a complete understanding of that term.

Any help would be appreciated, thanks!

Note: 0 $\notin$ $\mathbb N$

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There's some links in the FAQ on how to typeset equations. A basic tutorial is here. –  Douglas S. Stones Oct 26 '12 at 1:27
Please, learn some TeX;) –  nikita2 Oct 26 '12 at 1:28
Thanks! I'll edit it as soon as I figure it out. –  Christina Oct 26 '12 at 1:34
Under your definition, $0 \not \in \mathbb N?$ –  Ross Millikan Oct 26 '12 at 1:34
Correct. I forgot to add that. –  Christina Oct 26 '12 at 1:47

A function $f \colon A \to B$ is called one-to-one (or 1-1, or injective) if different elements of $A$ get mapped to different elements of $B$. That is, for any $x,y \in A$, $x \neq y$, it must hold that $f(x) \neq f(y)$. Does your $r$ satisfy this? (possible hint: check for the cases where each of $x$ and $y$ are less than $k$ and equal to $k$.)
A function $f \colon A \to B$ is called onto (or surjective) if every element of $B$ is mapped by some element of $A$. That is, for any $x \in B$, there is some $y \in A$ with $f(y) \neq x$. Does your $r$ satisfy this? (possible hint: check for the cases $x=1$ and $x>1$.)