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I am trying to solve this exercise, but it is kind of confusing to me because the order relation involved "reverses" the standard order.

Let $\tau$ a binary relation over $\mathbb N$ defined as follows:

$$\begin{aligned} (\forall a,b \in \mathbb N)\;a\;\tau\; b \Leftrightarrow \left((b \leq a) \,\wedge\, (r(b,10)\leq r(a,10)) \right)\end{aligned}$$

where $r(n,10)$ is the remainder of the division of $n$ by $10$.

I have to find maximum, minimum, maximal and minimal elements in $(\mathbb N, \tau)$.

Due to the nature of this relation I think that the maximum element is $0$, which is also the only maximal element. If it weren't so then there should be $h \in \mathbb N$ such that:

$$\begin{aligned} h < 0 \wedge r(h,10) \leq r(0,10)\end{aligned}$$

which is absurd.

Using once again the definition of this relation I can say there isn't any minimum or minimal elements. If there were a minimum or minimal elements then let $h = 10\alpha,\; \alpha \in \mathbb N $, then $\nexists k = 10\beta,\; \beta \in \mathbb N$ where $\alpha < \beta$. This is absurd.

Is my reasoning correct or is there anything that does not hold?

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Your conclusions are correct, but I don’t follow your reasoning in showing that there is no $\tau$-minimal element. To do this, just note that for any $n\in\Bbb N$, $n+10\tau n$, since $r(n+10,10)=r(n,10)$. – Brian M. Scott Dec 27 '12 at 19:29
@BrianM.Scott I am sorry, I think I don't get what is your point, I'm afraid. Are you suggesting $\forall k \in \mathbb N : r(k, 10)=9$ are minimal elements? – haunted85 Dec 28 '12 at 9:48
Not at all: as I said, your conclusion that there are no $\tau$-minimal elements is correct. But I can’t make any sense of your argument attempting to justify that conclusion, so I made a comment showing you how to prove it correctly. – Brian M. Scott Dec 28 '12 at 9:50

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