If $a$ and $n$ are relatively prime, there is a unique natural number $b < n$ such that $ab \equiv_n 1$ I really don't know how to tackle this proof because it has mod in it. There's 3 parts to the question. You don't have to answer all three parts (would be cool to check answer with though), I just need a starting point so get this proof rolling. 
Question: 
Relatively Prime: Means if two numbers' greatest common divisor is $1$.
Let $a$ and $n$ be two natural numbers. Prove that if $a$ and $n$ are relatively prime, there exists a unique natural number $b < n$ such that $ab \equiv_n 1$ by doing the following:
a) Prove that 
$$\exists b \in \mathbb N, b < n \land ab\equiv_n 1. \tag1$$
b) Add to equation ($1$) in part (a) to express that $b$ is unique.
c) Now prove $b$ is unique.
(If you need a picture of the question Question picture
 A: if $\gcd (a,n) = 1$
then by the Euclidean algorithm there exists integer $x$ and $y$ such that $ax + ny = 1$
or $ax \equiv 1 \pmod n$
We have not shown that $0<x<n$
if $ax \equiv 1\pmod n \implies a(x-n) \equiv 1\pmod n$
$x$ is not in the desired interval, there exists a $z$ such that $x-zn = b$ with $b$ in the the interval.
$ab \equiv 1\pmod n, 0<b<n$ 
Is b unique?  Suppose there were another, $ac \equiv 1\pmod n$ 
$a(c-b)\equiv 0\pmod n$
But that is only possible if $n|(c-b)$
A: $(1.)$ There exist $x$ and $y$ such that $ax+ny = 1$. Then $ax \equiv_n 1$.
$(2.)$There exists integers $q$ and $b$ such that $x = qn + b$ where $0 \le b < n$. We see, in this case, that $0 < b < n$.
$(3.)$ Suppose $0<b,b'<n$ and $ab \equiv_n ab' \equiv_n 1$.
Then $a(b-b') \equiv_n 0$.
What follows is a proof that, if $a$ and $n$ are relatively prime, then $b-b' =0$.

So $n$ divides $a(b-b')$.
Since $ax+ny=1$, then $a(b-b')x + n(b-b')y = b-b'$.
Since $n$ divides $a(b-b')$, it divides $a(b-b')x$.
Clearly $n$ divides $n(b-b')y$.
So $n$ must divide $b-b'$.
Since $0 < b < n$ and $-n < -b' < 0$, then $-n < b-b' < n$.
The only multiple of $n$ between $-n$ and $n$ is $0$. So b-b' = 0$.

Hence $b=b'$. So $b$ is unique. 
