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I have a function $f_2: \mathbb Z \times \mathbb Z \to \mathbb Z $ defined by $f_2(m,n)=m^2+n$.

How do I know if it is one-to-one, onto or both?

What I am most confused about is what $\mathbb Z \times \mathbb Z \to \mathbb Z$ means and how that is different from just $\mathbb Z \to \mathbb Z$.

I know one-to-one means every $x$ has a unique $y$ and onto means for all $y$, there exists an $x$ such that $f(x)=y$.

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    For one-to-one, informally, you could say that for any $y$, there is **at most** one $x$.2011-10-21

2 Answers 2

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This notation means that the "x" in your function is a pair of integers $(m,n)$.

So the question is: Do you know two pairs $(m_1,n_1)$ and $(m_2,n_2)$ of integers that give the same $m_1^2+n_1=m_2^2+n_2$?

The two pairs count as distinct if at least one element changes.

  1. one-to-one?

Choose two different $m$s and try to find $n$s such that the image of the function is the same for the two pairs.

  1. onto?

Try $m=0$.

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    Okay, that makes sense. So does that mean it will be onto? I'm not sure how to actually prove which one it is.2011-10-21
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For "onto" you need to answer this: Given any integer $y$, can you find a pair $(m,n)$ so that $m^2+n = y$?

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    `m = 0`, `n = y` -- so yes. therefore it is onto.2011-10-21