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This is not a homework, I'm just doing some revision and I saw this exercise:

Consider the function $f :$ $N× N$ $\rightarrow$ $N× N$ given by: $f(m, n) = (3m+n, n^2)$

(a) Is $f$ one-to-one?
(b) Is $f$ onto?

What should I do in this case $(3m+n, n^2)$ ? I usually assume $f(m)=f(n)$ and so on but here it's different.

Community
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user113311
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    Because $f(m,n)=(3m+n,n^2)$, the second coordinate of $f(m,n)$ is a perfect square. If follows that $f(m,n)$ can never be $(17,2)$, since $2$ is not a perfect square. It follows that $f$ is not onto. – André Nicolas Dec 04 '13 at 04:22
  • In this case (two-variable function) you would start with $f(m,n)=f(m',n')$ and continue as usual. – Doc Dec 04 '13 at 04:26

2 Answers2

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Do the same. Assume that $f(a,b) = f(m,n)$. What does this tell you?

Calvin Lin
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    Would you mind explain more carefully your answer? It is kinda of disrespectful to answer a quarter of the question given that the asker is just learning about it, and you know it. Be more humane to the asker. If you want to asnwer, try to give a complete solution. – ILoveMath Dec 04 '13 at 04:43
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It's not really different. Here as well you start by assuming that $f$ takes the same value at two places and show that then the places must be the same. However, the "places" are not simple numbers ($m,n$ if you start with $f(m)=f(n)$ as you say), but pairs of numbers (and so are the values). That is: You start with the assumption $$\tag1 f(m,n)=f(m'n')$$ and try to show from this that $$\tag2(m,n)=(m',n'),$$ which is just the same as: $m=m'$ and $n=n'$ (this is the definition of equality of pairs). So let's see: From $(1)$ we have $(3m+n,n^2)=(3m'+n',n'^2)$, i.e. $3m+n=3m'+n'$ and $n^2=n'^2$. From the latter we find $n=n'$ (as there are no negatives in $N$) and plugging this into the firstgives $3m=3m'$, i.e. $m=m'$. So we have indeed shown $(2)$.

Hagen von Eitzen
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