Modulo Concept

Here is a proof by example:
Let p be an odd prime and let S = ([1], [2 ],...,[ p-1]) the set of equivalence classes of nonzero modulo p.
Show that for any integer a such that [a] εS, [-a εS] and [-a] is not [a].

The congruence classes mod n are the remnants left
when n divides all the integers, for example, if n = 5,
the remains will be 0,1,2,3,4 and all numbers
in the [class 0] are multiples of five, they leave a
remainder of 0 when divided by 5, and similarly for
Other classes

[0] = {...,- 25, -20, -15, -10, -5,0,5,10,15,20,25 ...
[1] = {..., -24, -19, -14, -9, -4, 1,6,11,16,21,26 ,...}
[2] = {...,- 23, -18, -13, -8, -3, 2,7,12,17,22,27 ,...}
[3] = {...,- 22, -17, -12, -7, -2, 3,8,13,18,23,28 ,...}
[4] = {...,- 21, -16, -11, -6, -1, 4,9,14,19,24,29 ,...}

Thus, in classroom management, just take one representative each (equivalence) and form
the set S *= ([0] [1], [2 ],...,[ p-1]) from which your set S
is obtained.
Suppose that p is an odd prime and
S = ([1], [2], [3 ],...,[ p-1]) is the set of nonzero equivalence
classes modulo p. Then we interpret the class [-a], for
positive p <, as equivalent to the Palestinian Authority] [class because
Pa-A and leave the rest even when divided by
P. Is it possible that [-a] = [a]? If yes, then] a year [=] [a
and since pa is less than p and positive, they
leave the same remainder and are the same
remainder when divided by p which means then pa =
as PP = 2a, which even instead of odd. It is a
contradiction, so [-a] does [not equal to].
For the second part, [2 a ^] = [b ^] 2 means that
a ^ 2 = b ^ p 2mod then a ^ 2-b ^ 2 = 0 mod p, (a + b) (ab) = 0
mod p, then p divides (a + b) (ab), then p divides (a + b)
or p divides (ab) and resulting in a + b = p 0mod
or AB = 0mod p then a = bmod p-or a = b mod p and
back translation: [a] = [-b] or [a] = [b].