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First, how do you define "exhibit"? To you, would this mean you need to prove that your bijection is 1-1 and onto? If so, I think I've got the 1-1 part, but am struggling with showing it is onto. The examples of a linear function being shown as onto just seem too elementary, and I'm not sure they work when mapping naturals to integers.

I'm saying $f(n)=2n+13$. To show 1-1, I've let $f(x)=f(y)$, substituted through, and came out with $x=y$. I'm sure there needs to be more detail given, but for now, I'm just trying to remember how to show a function is onto. From what I've been able to find examples of, I basically solve my proposed function for $y$ getting $(y-13)/2$, then plug that in? Does that really show anything?! Does this work when going from $\mathbb{N} \rightarrow\mathbb{Z}$?

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What exhibit means is probably a matter of context - if this is part of a course then you may be able to tell by how rigorous previous work was. –  Joe Tait Feb 8 '13 at 20:48
    
With regards to the actual maths; to show a function is on to you need to show that it maps to everything in the range. In this case that is odd numbers greater than 13. Given such a number, there should be an obvious way to write it (think about how you can write even numbers, and the fact that odd numbers are not even). –  Joe Tait Feb 8 '13 at 20:50
    
Do you mean odds represented as $2n+1$, and work backwards somehow? –  MrsMillz Feb 8 '13 at 20:52
    
That is exactly what I mean. Work backwards to find a natural number $m$ (Which will likely be in terms of $n$) such that $f(m) = 2n+1$. –  Joe Tait Feb 8 '13 at 20:55
    
Start with Let $f(m) \in \mathbb{Z}$ such that $f(m)=2n+1$ or should it be $2n+13$ in this case? The hardest part for me seems to be starting a proof. Once I get it rolling, I'm usually okay. –  MrsMillz Feb 8 '13 at 21:08
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2 Answers

Your $f$ is a bijection $\Bbb N\to \{1,3,5,7,\dots\}$. This should be shifted with $12$ (if $13$ is included in the target set, and by $14$ if that starts with $15$), and you're there: $$g(n):=f(n)+12=2n+13$$ will do it.

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I meant $2n + 13$. Shoot! –  MrsMillz Feb 8 '13 at 20:55
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Suppose that $n\in \mathbb Z$ is odd and that $n>13$.

Then we can write $n=2z+1$ for some $z \in \mathbb Z$.

Then $n= 2z+1 =2z+1 +(13-13) = 2z-12 +13 = 2(z-6)+13$.

Hence $f(z-6) = n$.

Now we need to show that $z-6 \in \mathbb N$. But since $2z+1 > 13$ then $2z >12$ and $z>6$ and finally $z-6>0$.

Obviously all the greater than's can be greater than or equal to, according to whether or not $0\in \mathbb N$ for you.

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Deinitely would not have thought of it this way. I need to let it sit for a bit, then come back to it. Thanks! –  MrsMillz Feb 9 '13 at 1:32
    
Let me know if you have any questions –  Joe Tait Feb 9 '13 at 9:11
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