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**Deveno** suppose i have the entire set of integers:

Z = {.........-4,-3,-2,-1,0,1,2,3,4,5........}

what set do i get if i add 1 to every integer in this set?

one of the peculiarities of the integers is that there is no "smallest integer" nor no "largest integer" so there is no "bottom" that gets bigger when we add 1 to everything, nor no "top" that keeps everything under a certain amount.

EDIT: here is a "semi-formal" proof that Z = Z+1.

suppose k is in Z. then k-1 is in Z as well, so k = (k-1) + 1 is in Z+1.

on the other hand, suppose m is in Z+1. then m = n+1, for some integer n. hence m = n+1 is also an integer, whence m is in Z.

thus Z+1 = Z, since the two sets contain each other.

EDIT #2:

let's look at a particular m, say m = 5.

now 5Z = {........-25,-20,-15,-10,-5,0,5,10,15,20,25,30.......}.

1+5Z = {.......-24,-19,-14,-9,-4,1,6,11,16,21,26,31......}

2+5Z = {......-23,-18,-13,-8,-3,2,7,12,17,22,27,32.......}

3+5Z = {......-22,-17,-12,-7,-2,3,8,13,18,23,28,33......}

4+5Z = {.....-21,-16,-11,-6,-1,4,9,14,19,24,29,34......}

so far, we haven't had any "overlap" at all, every integer only occurs in ONE of these sets.

what happens when we have:

5+5Z = {.......-20,-15,-10,-5,0,5,10,15,20,25,30,35.....}

all we have done is "shifted 5Z over 5". but 5Z has no "start" and no "end" (it's infinite both ways), so we don't get any "extras" at the "end" or any "new ones" at the "start".

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personally, i don't like this way of looking at Z/5Z. i prefer to think of it as a PARTITION of Z into 5 subsets:

5Z = all integers of the form 5k, for some integer k.

1+5Z = all integers of the form 5k+1

2+5Z = all integers of the form 5k+2

3+5Z = all integers of the form 5k+3

4+5Z = all integers of the form 5k+4

if you like you can think of this partition as being the equivalence classes of the following equivalence relation:

m~n if m-n is divisible by 5.

so now your question becomes: why are the integers of the form 5k+5 of the form 5k?

because 5k+5 = 5(k+1) (we just change k's).