$\sqrt{180\mathbb Z} = 30\mathbb Z$, $(180\mathbb Z:700\mathbb Z)= 9\mathbb Z$ (First time studying rings, and I need some help on this example about radical ideal and fraction ideal)
Let $\sqrt{I}$ be the radical ideal on the commutative ring $R$, defined as $\sqrt{I}=\{r\in R: r^n\in I\ \mathrm{for\ some\ } n\in\mathbb{Z}^{+}\}$.
Then, for example, $\sqrt{180\mathbb Z} = 30\mathbb Z$, since $180=2^23^25$, and $30 = 2*3*5$.
Using this example, I can now solve any similar problem, but I do not understand why this method of prime factorization works.
I was also confused about the example regarding a fraction ideal.
$(I:J)=\{x\in R:xJ\subseteq I\}$, where $I, J$ are ideals in $R$. 
The example that was given to me was this:
$(180\mathbb Z:700\mathbb Z)= 9\mathbb Z$.
The method to obtain this is as follows: $180=2^23^25, 700 = 2^25^27$.
And we see that
$180/gcd(180, 700) = 9$.
I feel like once I see the answer it'll be clear, but as of now I do not see it.
 A: The key fact for the first question is this:

If $p$ is prime and $p$ divides $a^n$, then $p$ divides $a$.

If $a^n \in m\mathbb Z$, then all prime factors of $m$ divide $a^n$ and so divide $a$. Therefore, $a$ is multiple of $r=rad(m)$, and so $\sqrt{m\mathbb Z} \subseteq r\mathbb Z$.
Conversely, $r^n \in m\mathbb Z$, for every $n$ at least equal to the largest exponent in the prime factorization of $m$. Therefore, $r \in \sqrt{m\mathbb Z}$ and so $r\mathbb Z \subseteq \sqrt{m\mathbb Z}$.
For the second question, note that $700a \in 180 \mathbb Z$ implies $700a=180b$ and so $35a=9b$. Since $9$ and $35$ are coprime, $9$ must divide $a$, that is, $a \in 9 \mathbb Z$. The reverse inclusion is easy.
A: We wish to show that $(m\mathbb{Z} : n\mathbb{Z}) = \frac{m}{(m,n)} \mathbb{Z}$. 
Courtesy of Elegant Proof that $m | xn \implies \frac{m}{(m,n)} | x$, I can now give a much simpler proof. 
Take $x \in (m\mathbb{Z} : n\mathbb{Z})$. Then, by definition, $xn\mathbb{Z} \subset m\mathbb{Z}$. In particular, this means that there exists $k \in \mathbb{Z}$ such that $xn  = mk$. Hence, $m | xn$. 
Let $d = (m,n)$. By Bezout's identity, there exist $a,b \in \mathbb{Z}$ such that $am + bn = d$. Thus $amx + bnx = dx$. Since $m | xn$, $m | bnx$. Clearly $m | amx$. Thus, $m | amx + bnx$, i.e. $m | dx$. Thus, there exists $r \in \mathbb{Z}$ such that $mr = dx$. Thus, $x = \frac{m}{d} r$, which shows $x \in \frac{m}{d} \mathbb{Z}$. That is, $x \in \frac{m}{(m,n)} \mathbb{Z}$. Thus, we have shown $(m\mathbb{Z} : n\mathbb{Z}) \subset \frac{m}{(m,n)} \mathbb{Z}$. 
For the other containment, take $x \in \frac{m}{(m,n)} \mathbb{Z}$. Then there exists $s \in \mathbb{Z}$ such that $x = \frac{m}{(m,n)} s$. By definition, $\frac{n}{(m,n)} \in \mathbb{Z}$. Thus $t =\frac{ns}{(m,n)} \in \mathbb{Z}$. We compute
$$
mt = m \frac{ns}{(m,n)} = n \frac{m}{(m,n)} s = nx = xn
$$
Thus, there exists $t \in \mathbb{Z}$ such that $mt = xn$. Hence, for any $j \in \mathbb{Z}$ there exists $k = jt \in \mathbb{Z}$ such that $mk = mjt = mtj = xnj$. That is, for any $j \in \mathbb{Z}$ there exists $k \in \mathbb{Z}$ such that $xnj = mk$. This is precisely the statement that $xn\mathbb{Z} \subset m\mathbb{Z}$, which itself is precisely the statement that $x \in (m\mathbb{Z} : n\mathbb{Z})$. Thus, we have shown $\frac{m}{(m,n)} \mathbb{Z} \subset (m\mathbb{Z} : n\mathbb{Z})$. 
We conclude $(m\mathbb{Z} : n\mathbb{Z}) = \frac{m}{(m,n)} \mathbb{Z}$. 
