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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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If I understand your first question correctly, to prove surjectivity you can, for each element of the codomain, give a precise element in the domain that is mapped to it. This is quite easy to do in your case, and once you have done that you can be sure there are no elements that break the rules – Daniel Freedman Jan 13 '12 at 16:07
For the second question, you should show that the set is not totally ordered. You can do that by finding two elements $(a,b)$ and $(c,d)$ such that neither $(a,b)\Sigma(c,d)$ nor $(c,d)\Sigma(a,b)$ holds. – Michael Greinecker Jan 13 '12 at 16:13
Note that the statement "This is surjective [be]cause, every couple of numers $(a,b)\in \mathbb{Z}$ have a correspective inside the codomain" is not very well-phrased (if not outright incorrect). To me, it reads like you are saying that the image of every element in the domain is an element of the codomain, but that is not surjectivity. To prove surjectivity, you need to show that for every $x\in\mathbb{Z}$ there exists an $(a,b)\in\mathbb{Z}\times\mathbb{Z}$ with the property that $f(a,b)=x$. – Arturo Magidin Jan 13 '12 at 16:51
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)?" – David Mitra Jan 13 '12 at 17:06

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. – Seamus Jan 13 '12 at 16:33

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