Proof that $a\equiv b \pmod n \iff a \pmod n = b\pmod n$

Question

Proof that for every $a,b \in \mathbb Z,\ n \in \mathbb N$, that

$$a\equiv b \pmod n \iff a \pmod n = b \pmod n.$$

My approach is:

$n\mid a$ and $n\mid b$

$a\equiv b \pmod n \iff \exists x,y: a\cdot x+n\cdot y=b$

Then I'm stuck.

Answer 1

When you take a number mod $n$, you are taking its representative in $\{0,1,\dots, n-1\}$

Use the fact that $a$ (mod n) - $b$ (mod n) $\equiv a-b$ (mod n).

(To see this, say $a=m_1n +r_1$ and $b=m_2n+r_2$. Then $r_1+r_2=(a-m_1n)+(b-m_2n)$ so that $r_1+r_2\equiv a+b$ (mod n) )

Hence

$a$ (mod n) $= b$ (mod n) $\Longrightarrow$ $a$ (mod n) $-b$ (mod $n$) $=0 \Longrightarrow a-b\equiv 0$ (mod n) $\Longrightarrow a\equiv b$

Conversely, prove the contrapositive.

$a$ (mod $n$) $\not= b$ (mod $n$) $\Longrightarrow a$ (mod $n) - b$ (mod $n)\not=0 \Longrightarrow a$ (mod $n) -b$ (mod $n)\not\equiv 0$ (mod $n \Longrightarrow a-b\not\equiv 0$ (mod $n$)