Question 622759
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Let *[tex \LARGE \mathbb{Z}] be the set of all integers and let *[tex \LARGE m\ \in\ \mathbb{Z},\,m\ >\ 1], and accept the notation *[tex \LARGE \alpha\ \equiv\ \beta\,(\text{mod}\,n)] to mean that *[tex \LARGE \alpha\ -\ \beta] is divisible by *[tex \LARGE n].


Prove that the set *[tex \LARGE R\ =\ \{(x,y)\,|\,x,\,y\,\in\,\mathbb{Z},\,x\ \equiv\ y\,(\text{mod}\,m)\}] defines an equivalence relation:


1.  Since *[tex \LARGE x\ -\ x\ =\ 0], *[tex \LARGE m\ \Rightarrow\ x\ \equiv\ x\,(\text{mod}\,m)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ (x,x)\ \in\ R]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ R] is reflexive.


2.  Let *[tex \LARGE (x,y)\ \in\ R\ \Rightarrow\ x\ \equiv\ y\,(\text{mod}\,m)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ x\ -\ y\ =\ mq\ \exists\ q\ \in\ \mathbb{Z}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ y\ -\ x\ =\ m(-q)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ y\ \equiv\ x\,(\text{mod}\,m)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ (y,x)\ \in\ R]


Thus *[tex \LARGE (x,y)\ \in\ R\ \Rightarrow\ (y,x)\ \in\ R\ \Rightarrow R\ ]is symmetric.


3. Let *[tex \LARGE (x,y)\ \in\ R] and *[tex \LARGE (y,z)\ \in\ R]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ x\ \equiv\ y\,(\text{mod}\,m)\ \ ]and *[tex \LARGE\ y\ \equiv\ z\,(\text{mod}\,m)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ x\ -\ y\ =\ mq\ \ ]and*[tex \LARGE\ \ y\ -\ z\ =\ mq'\ \exists\ q,q'\ \in\ \mathbb{Z}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ (x\ -\ y)\ +\ (y\ -\ z)\ =\ m(q\ +\ q')]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Rightarrow\ (x\ -\ z)\ =\ m(q\ +\ q'),\ q\ +\ q'\ \in\ \mathbb{Z}]


Thus *[tex \LARGE (x,y)\ \in\ R\ \text{and}\ (y,z)\ in\ R\ \Rightarrow\ (x,z)\ \in\ R\ \Rightarrow R\ ]is transitive.


*[tex \LARGE R\ \ ] is reflexive, symmetric, and transitive *[tex \LARGE \Rightarrow\ R] is an equivalence relation.

 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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