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Given the function: $f:(a,b)\in\mathbb{Z}\times\mathbb{Z}\longrightarrow ab^2\in\mathbb{Z}$ What can you say about injectivity and surjectivity?

This is not injective. I have easily find two elements that broke injectivity definition. For instance $(1,2)$ is different from $(1,-2)$ but their images is 4 in both case. This is surjective cause, every couple of number $(a,b)\in\mathbb{Z}$ have a correspective inside the codomain. But, what if I can't easily find the couple of element that broke the rules? What mathematic process can I use to confirm my results?

Considering in $\mathbb{Z}\times\mathbb{Z}$ the following relation: $(a,b)\Sigma(c,d)\Leftrightarrow a\leq c\quad\text{AND}\quad b\leq d$ Prove that $\Sigma$ is not a total order relation and determine min, Max, minimal and maximal element. I'm able to responde this question when I can sketch Hasse digram, so when I have finite set of element, but, what in this case?

Help me, best regards MC

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    I believe he may be asking in his first question "what if I can't obviously find two distinct elements that have the same image under the function? How would I show one-to-oneness (or not)?"2012-01-13

2 Answers 2

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For every $a$, $f(a,1)=a$ so the function is surjective.

The relation is obviously transitive, so for it to fail to be a total order, it must fail to be total. As the comments suggest, you must find $a,b,c,d$ such that neither $(a,b)\Sigma(c,d)$ nor $(c,d)\Sigma(a,b)$. Think of $(a,b)$ as a line segment with ends at $a$ and at $b$, and think about what the relation says about line segments.

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Seamus’s suggestion for the second problem may work very well for you, but there’s another picture that I find more helpful, so I’ll offer it as well. If you think of $(a,b)$ and $(c,d)$ as points in the plane, the relation $(a,b)\;\Sigma\;(c,d)$ has a simple geometric interpretation. If you figure out what that interpretation is, you shouldn’t have much trouble finding pairs $(a,b)$ and $(c,d)$ such that neither $(a,b)\;\Sigma\;(c,d)$ nor $(c,d)\;\Sigma\;(a,b)$.

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    That's a good point! I think that's a much more natural interpretation.2012-01-13