Question 1195956
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I'll use the tilde mark ~ in place of the symbol ≡n


So x ~ y means n | (x-y) and that is saying "n divides (x-y)"
i.e. "n is a factor of (x-y)"


Furthermore n|(x − y) is the same as nk = x-y for some integer k


Here's an article talking about Equivalence Relations
<a href = "https://www.cuemath.com/algebra/equivalence-relations/">https://www.cuemath.com/algebra/equivalence-relations/</a>


We need to prove three things:
Reflexive
Symmetric
Transitive



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Reflexive:
Let 'a' be an integer
The claim is to show that a ~ a is true


Well,
x ~ y
nk = x-y
nk = a-a
nk = 0
Since n is positive, this means k = 0 must be the case.
Therefore, a ~ a is true and the relation has been proven to be reflexive.


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Symmetric:
Define two integers. Call them a and b.
We need to show that if a ~ b, then b ~ a


Start with a ~ b and follow these steps
a ~ b
n | (a-b)
nk = a-b
-1*nk = -1(a-b)
n(-k) = -1(a-b)
n(-k) = -a+b
n(-k) = b-a
n | (b-a)
b ~ a
I multiplied both sides by -1 and did a bit of rearranging to get -nk into n(-k)
The k isn't that special since the nk is really n*(some integer)
k is an integer, and so is -k.


We have shown that a ~ b leads to b ~ a; hence proving the symmetric argument.


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Transitive:
We'll need three integers this time. Call them a,b,c
If a ~ b and b ~ c, then it should lead to a ~ c


a ~ b leads to nk = a-b for some integer k
while b ~ c leads to nm = b-c for some integer m


Add the equations and do a bit of algebra
(nk) + (nm) = (a-b) + (b-c)
n(k+m) = a-c
n*(some integer) = a - c
n | (a-c)
a ~ c


Therefore, a ~ b and b ~ c does indeed point to a ~ c
The transitive aspect has been proven.


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We have proven the three properties of reflexive, symmetric, and transitive. 


Therefore, the relation has been proven to be an equivalence relation.


Side note: For more information, check out the concept about congruence mod n.
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