Understanding Equivalence Classes: Even and Odd Numbers in Relation to 0 and 1

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In the discussion on equivalence classes, even numbers are associated with 0 and odd numbers with 1, represented as [0] and [1]. It is noted that any even number can serve as a representative of its equivalence class, not just 0; for example, [2] or [4] could also be used. The key point is that the addition of equivalence classes remains consistent regardless of the chosen representative. Specifically, the results of adding these classes—such as [even] + [even] = [even] and [odd] + [odd] = [even]—are invariant. This highlights the fundamental properties of equivalence classes in relation to even and odd numbers.
rajeshmarndi
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Why in equivalence class of N of even number and odd number, the even number are taken as related to 0 and odd number are related as 1 i.e [0] and [1].

Instead of [0], even number can also be related to [2] or [4]? Or [2] or [4] could also be taken, as it is just an convention.

Thanks.
 
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Yes, any even number can be taken as a "representative" of its equivalence class. The important thing is that addition of equivalence classes doesn't depend on the choice of representative, i.e.
[even]+[even]=[even], [odd]+[even]=[even]+[odd]=[odd], [odd]+[odd]=[even].
 
thanks a lot!
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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