Multiplicative Inverse? The book I'm reading defines the multiplicative inverse of $a\pmod N$ as $x$, such that $ax \equiv 1\pmod N$. It then states not all numbers have a multiplicative inverse, such as $2 \pmod 6$. It states that for a multiplicative inverse to exist, $N$ and $a$ have to be co-prime.
But wouldn't the multiplicative inverse just be the reciprocal? And since every number has a reciprocal, wouldn't every number have a multiplicative inverse?
I.e It states $2 \pmod 6$ does not have a multiplicative inverse. But what about $\frac{1}{2}\pmod 6$? Doesn't this qualify as the multiplicative inverse since $\frac{1}{2}\pmod 6 \times 2\pmod 6 = 1 \pmod 6$?
What am I misunderstanding here?
 A: There are two problems, one each depending on what $\frac{1}{2}$ means.


*

*We can define $\frac{1}{a}$ to be "the [unique] number such that $a\times\frac{1}{a}=1$, if it exists" in whatever system we are working on (that is, "number" here would mean "number modulo $N$"). But then you cannot assume that such a thing as $\frac{1}{2}\pmod{N}$ exists in the first place. You must prove it exists. 
Note that they don't always exist: for example, $2$ has no multiplicative inverse in the integers either.
In this situation, it is true that $\frac{1}{2}\pmod{6}$ is a multiplicative inverse of $2\pmod{6}$, if it exists. But in fact, no such thing exists. Just like an even prime number greater than $2$ would be congruent to $0$ modulo $2$ if it existed, but no such thing exists. 

*If by $\frac{1}{2}$ you mean the rational number $\frac{1}{2}$, then $\frac{1}{2}\pmod{6}$ makes no sense in the integers modulo $6$: we only allow integers! That is, when we write things like $a\pmod{N}$, we are implicitly asserting that $a$ is an integer. We cannot do that with $\frac{1}{2}$.
To see that there cannot exist an integer $x$ such that $2x\equiv 1\pmod{6}$, note that $2x-1$ is always odd, so it is never a multiple of $6$; hence, $2x$ can never be congruent to $1$ modulo $6$, no matter what integer $x$ is.
A: One thing you're missing is that you shouldn't be referring to such alleged objects as $2\bmod 6$, etc.
$$
\text{Right:}\quad (x \equiv y)\pmod n
$$
$$
\text{Wrong:}\quad x \equiv \Big( y \bmod n \Big)
$$
To say that $x$ and $y$ are mod-$n$ congruent to each other means their difference is a multiple of $n$.  Thus for example, $(69\equiv 62) \pmod 7$.
Now observe that $(3\cdot5 \equiv 1)\pmod 7$, so $3$ and $5$ are each other's mod-$7$ reciprocals.
A: The rational number $1/2$ is not allowed as a value of $x$. 
Only integers are allowed for modular arithmetic (in the context of the question).

But wouldn't the multiplicative inverse just be the reciprocal? And since every number has a reciprocal, wouldn't every number have a multiplicative inverse?

This is only true for fields like the real numbers $\mathbb{R}$ or complex numbers $\mathbb{C}$, where every non-zero "number" has a "reciprocal".
A: I guess we need layman's terms here for OP's benefit.
What you are missing here: is that in arithmetic modulo 6, the only set of numbers you have is $\{0, 1, 2, 3, 4, 5\}.$ There is no $\frac{1}{2}.$ You are no longer working with all numbers. You're only working with a restricted subset, namely: $0, 1, 2, 3, 4, 5.$ (more formally $\{[0], [1], \ldots [5]\}$ where $[x] = \{ y : y \equiv x \pmod{6} \}$)
I will steal these pictures from Wolfram|Alpha: that's arithmetic $\pmod{6}.$
 
A: The use of $a^{-1}$ instead of $\dfrac 1a$ (or $\div a$) is often but not always required. Basically, it's complicated. 
With a little numerical manipulation, both ways will make sense; but, you still need to be careful.
For example, we find $3^{-1} \equiv 9 \pmod{13}$ (Note that $3 \times 9 \equiv 27 \equiv 1 \pmod{13}$). In the following cases, I will show that some times multiplying by $3^{-1} \equiv 9 \pmod{13}$ is appropriate, and some times dividing by $3$ is appropriate.

(1.) Consider the equivalence, $$3x \equiv 12 \pmod{13}.$$ 

So we can compute $ x \equiv 9 \times 12 \equiv 108 \equiv 4 \pmod{13}$. But it would be silly to do all of that work when we can see that $12 \div 3 = 4$.

(2.) Consider the equivalence, $$3x \equiv 10 \pmod{13}.$$

Since $10 \div 3$ is not an integer; we can't use that method. By repeatedly adding $13$, we see that $10 \equiv 23 \equiv 36 \pmod{13}$ and $36 \div 3 =12.$ So $x \equiv 12 \pmod{13}$. This method is only practical though for very small moduli. It would be much more efficient, usually, to just compute 
$x = 3^{-1}\times 10 \equiv 9 \times 10 \equiv 90 \equiv 12 \pmod{13}$.
There is a theorem that states that the equivalence $ax \equiv b \pmod c$ if and only if $\gcd(a,c) \mid b$. If there is a solution, then the equivalence reduce to 
$\dfrac ag x \equiv \dfrac bg \pmod{\dfrac cg}$ where $g = \gcd(a,c)$.

(3.) Consider the equivalence, $$3x \equiv 12 \pmod{15}.$$ 

In this case, $\gcd(3,15)=3 \mid 12$, so there is a solution and we find $x \equiv 4 \pmod{5}.$

(4.) Consider the equivalence, $$3x \equiv 10 \pmod{15}.$$ 

In this case, $\gcd(3,15)=3 \not \mid 10$, so there is no solution.
A: The multiplicative inverse, if it exists, would be $\frac{1}{2}$.
The problem is, definitions do not imply existence. The definition describes properties that would make it fulfil these definitions, but then you still need to find such an element that actually exists.
Axioms can be used to put these new elements into existence. The rational numbers will let these multiplicative inverses exist, but the integers won't. This is all based on the axioms underlying these systems.
When working with modular arithmetic here, it's likely that you are working with the integers, where the existence of a multiplicative inverse has not been demanded/assumed.
