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$$\forall x, y, n \in \mathbb{Z}, x=y \ \implies x \equiv y\ (mod\ n)\ $$

I am not sure if I am doing this properly and I don't know if I can show my proof like this:

Suppose x = y which means x = x

x ≡ x mod n

n | x - x

kn = x - x

kn = 0 and since

x ≡ y mod n

jn + y = x

jn = x - y and since x = y, y=y and x=x

jn = 0

kn = jn

RobPratt
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Fred Vann
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  • Consider the following: Let $c\equiv x\mod n.$ Now you know that for some $k\in\mathbb{Z},\ x=k\cdot n+c=y$ by the assumption that $x=y.$ Can you formalize the rest? –  Sep 28 '20 at 20:58
  • @DavidKipper I'm not really sure where to go from there, other than that would mean c = x mod n and c = y mod n, which seems redundant? – Fred Vann Sep 28 '20 at 21:51
  • This is the same as showing that congruence is reflexive relation, see the linked dupe. – Bill Dubuque Sep 29 '20 at 08:04

2 Answers2

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First let's prove this: $\forall n,x \in \mathbb{Z}, x\equiv x \hspace{0.1cm} \mod n$.

Let $n \in \mathbb{Z}$. Then: $$0 = 0*n$$ $$ n | 0 $$ Let $x \in \mathbb{Z}$. Then $ x - x = 0$, so: $$ n | x -x $$ $$ x\equiv x \hspace{0.1cm} \mod n $$ Proof ended.

Now the little proof you have to do is: Let $x,y,z \in \mathbb{Z}$ such that $x=y$. We now know that $ x\equiv x \hspace{0.1cm} mod(n) $, but since $x=y$ then $ x\equiv y \hspace{0.1cm} \mod n $. Proof ended.

user577215664
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franncapurro
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By definition, $x \equiv y \pmod n \iff n|x-y$.

Suppose $x = y$. Then $x-y = 0$, and since $n|0$, we have $n|x-y$, therefore $x \equiv y \pmod n$.

Ovi
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