This is my problem:
For every couple of integers $(a,b)\in\mathbb{Z}\times(\mathbb{Z}\backslash\{0\})$ we denote with $r(a,b)$ the remainder of the division between $a$ and $b$. Consider the application $$f:(a,b)\in\mathbb{Z}\times(\mathbb{Z}\backslash\{0\})\rightarrow r(a,b)\in\mathbb{N}$$
- Talk about injectivity and surjectivity
Considering in $\mathbb{Z}\times(\mathbb{Z}\backslash\{0\})$ the following relation $$(a,b)\varphi(c,t)\Leftrightarrow r(a,b)=r(c,t)$$ $$(a,b)\pi(c,t)\Leftrightarrow(a,b)=(c,t)\quad OR\quad r(a,b)<r(c,t)$$ 2.1 Prove that $\varphi$ is equivalence relation in $\mathbb{Z}\times(\mathbb{Z}\backslash\{0\})$, and show the equivalence class of $(6,3)$
2.2 Prove that $\pi$ is an order relation in $\mathbb{Z}\times(\mathbb{Z}\backslash\{0\})$. Is this total? In the ordered structure $(\mathbb{Z}\times(\mathbb{Z}\backslash\{0\}),\pi)$ find min, max, minimal and maximal element.
On first point, the function is not injective, cause exists at least two couple of integer that broke injection definition. (E.g. $10/3$ and $11/10$ has the same remainder). Is surjective 'cause every division between two integer number gave a remainder in $\mathbb{N}$.
For the point 2.1, a given binary relation on a set A is said to be an equivalence relation if and only if it is reflexive, symmetric and transitive. In our case:
- reflexive: cause if $r(a,b)$ is equal to $r(a,b)$ this will be true for every element $(a,b)\in\mathbb{Z}\times(\mathbb{Z}\backslash\{0\})$
- symmetric: if $r(a,b)$ is equal to $r(b,c)$ symmetricaly $r(b,c)$ will be equal to $r(a,b)$ and this will be true for every $(a,b)$ and $(b,c)$ in $\mathbb{Z}\times(\mathbb{Z}\backslash\{0\})$
- transitive: cause if $r(a,b)=r(c,t)$ and if $r(c,t)=r(p,q)$ then $r(a,b)=r(p,q)$ for every $(a,b),(c,t),(p,q)\in\mathbb{Z}\times(\mathbb{Z}\backslash\{0\})$
So this is an equivalence relation. About equivalence class of (6,3) it should be 0 cause, 6/3 get a remainder of 0. Is my resolution right???
How can I resolve 2.2????
Best Regards MC