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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 rules2012-01-13
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    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.2012-01-13
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    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 $f(a,b)=x$.2012-01-13
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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

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