Modular Arithmetic with a Negative Power Calculate $67^{-1} \pmod{119}$.
So I tried this and I got
\begin{align*}
x \equiv 67^{-1} \pmod{119} &\implies x \equiv \frac{1}{67} \pmod{119}\\ 
&\implies 67x \equiv 1 \pmod{119}\\ 
&\implies 67x = 1\\ 
&\implies x = \frac{1}{67}
\end{align*}
I just stopped after that because I knew I was going wrong can some one please help me with this one.
 A: In modular arithmetic (or more generally, in a group), $a^{-1}$ does not mean $\frac{1}{a}$ in the usual sense of $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, etc. (i.e. $67^{-1}$ does not mean $1 \div 67 \approx 0.0149$). In modular arithmetic, $a^{-1}$ means the multiplicative inverse of $a$. That is, it is the unique element in $\{1, \dots, 118\}$ such that $aa^{-1} = a^{-1}a \equiv 1 \pmod{119}$; note, the existence of an inverse for any non-zero element relies on the fact that the modulus is prime. If the modulus is not prime, the only non-zero elements which have multiplicative inverses are those which are coprime to the modulus.
Note: As Bill Dubuque outlines in his answer, provided $a$ is coprime to the modulus, you can treat $a^{-1}$ as $\frac{1}{a}$, but it represents a different value than it does in say $\mathbb{R}$.
Let $x$ be the multiplicative inverse of $67$ modulo $119$ (i.e. $x \equiv 67^{-1} \pmod{119}$). Then you are looking to solve $67x \equiv 1 \pmod{119}$. By definition, $67x \equiv 1 \pmod{119}$ means that $119 \mid (67x - 1)$ so there is some integer $y$ such that $67x - 1 = 119y$, or written differently $67x + 119y = 1$. Note, if $67$ and $119$ weren't coprime, this equation would have no integer solutions. This is a diophantine equation that you can solve by using the Euclidean algorithm and back substitution. Have you seen this before?
A: Think about this. What integer between 0 and 118 would, when multiplied by 67 produce an integer that is one more than some multiple of 119?
Hint: your division is flawed in your example. You are working exclusively with integers and therefore your intended meaning of 1/67 as it is in the real numbers is not allowed. 
A: Using Gauss's algorithm, $\rm\,mod\,\ 119\!:\ \dfrac{1}{67}\equiv\dfrac{2}{134}\equiv\dfrac{2}{15}\equiv\dfrac{16}{120}\equiv \dfrac{16}1$
Beware $\ $ The use of fractions in modular arithmetic is valid only when the denominator is invertible. Otherwise the quotient need not be unique, e.g. mod $\rm\:10,\:$ $\rm\:4\,x\equiv 2\:$ has solutions $\rm\:x\equiv 3,8,\:$ so the "fraction" $\rm\:x \equiv 2/4\pmod{10}\,$ cannot designate a unique solution of $\,4x\equiv 2.\,$ Indeed, the solution is $\rm\:x\equiv 1/2\equiv 3\pmod 5,\,$ which requires canceling $\,2\,$ from the modulus too, since $\rm\:10\:|\:4x-2\iff5\:|\:2x-1.\:$
Generally the grade-school rules of fraction arithmetic apply universally (i.e. in all rings) where the denominators are invertible. This fundamental property will be clarified conceptually when one learns in university algebra about the universal properties of fractions rings and localizations.
A: As $\displaystyle 119=7\cdot17$
$$67\equiv-1\pmod{17}\iff 67^{-1}\equiv-1\  \ \ \ (1)$$
$$67\equiv4\pmod7\equiv2^{-1}\text{ as }4\cdot2=8\equiv1\pmod7$$
$$\iff 67^{-1}\equiv2\  \ \ \ (2)$$
Apply CRT on $(1),(2)$
