If $I$ is an Ideal of $R$ (a ring) and $a$, $a'\in R$ why is the following true? $a + I = a' + I \iff a - a' \in I$

Question

My professor for Ring Theory uses this concept to prove certain theorems and corollaries in the class, so I asked him to explain but his explanation was not clear. What I can understand is that if $a + I = a' + I$ then $a = a'$ so $a - a' = 0 \in I$ (I'm not sure if I am correct here) If I'm correct, how about the proof for the converse? Please I need a clear explanation. Thank you.

Answer 1

$a+I=a'+I$ implies that $a'\in a+I, a'=a+i, i\in I, a-a'=-i\in I$.

$a-a'=i\in I$ implies that $a'+I=(a-i)+I=a+I$