elementary number theory - Calculating Modular Multiplicative Inverse for negative values of a.

If I'm calculating $a^{-1} \equiv 1 \pmod n$ where $a$ is negative. Do I simply add $kn$ to $a'$ until $0\lt a' \lt n$?

I'm currently using the extended euclidean algorithm to calculate my modular multiplicative inverses since I already have to make sure that $a$ and $n$ are coprime. From what little number theory I understand $a'=a+kn$ is going to give me the same result as $a \pmod n$. So that should mean that $a' \equiv 1 \pmod n$ is the same as $a \equiv 1 \pmod n$

I've confirmed this with a few values below but don't know if I'm understanding this properly.

$a=-36 \;\; a'=14$

$9 = -36^{-1} \pmod{25}$

$9 = 14^{-1} \pmod {25}$

$a=-11\;\; a'=15$

$7 = -11^{-1} \pmod{26}$

$7 = 15^{-1} \pmod{26}$

Here's a link to my python code.
https://paste.debian.net/1117624/

Answer

Hint: like sums & products, inverses too are preserved on replacing argument(s) by a congruent one

Congruence Inverse Law
$\ \color{#c00}{a'\equiv a}\,\Rightarrow\,{a'}^{-1}\equiv a^{-1}\,$ if $\ a\,$ is invertible, i.e. $\, ab\equiv 1\,$ for some $b$.

Proof $\ $ Notice $\,\ \color{c00}ab\equiv 1\ \Rightarrow\ \color{#c00}{a'} b\equiv \color{#c00}ab\equiv 1\,$ by the Congruence Product Rule, therefore we conclude $\, {a'}^{-1}\!\equiv b\equiv a^{-1}\,$ by Uniqueness of Inverses.