Is it true: 2 does not divide 1? Is this true? 2 does not divide 1? Does a proof exist for this? Is it because the outcome will be a fraction and that is not allowed in discrete mathematics? I am new to this. 
 A: If you carefully write out the definition, to say that $2$ divides $1$ means that there is an integer $m$ for which
$$1 = 2m.$$
However, there is no solution to this equation in the integers (e.g. division by $2$ requires that $m = 1/2$, which is non-integral). The key point in the above definition is the bolded part.
A: How deep do you want to go?  First we need to define integers.  And then we need to define $1$ and $2$ and what "$a$ divides $b$" mean.
"$a$ divides $b$" means that there is a number $k$ so that $a*k = b$. This is usually taken in context and it is usually assumed that $k$ is an integer. (It's trivial if $a$ and $b$ are elements of a field, such as the rational, real, or complex numbers and $a \ne 0$ and $k$ can be any element of the field-- the $k = \frac ba$ will always be the case that $a*k = b.)
So "$2$ divides $1$" means that there is an integer $k$ so that $2k = 1$.  There is not.
Why not?  And is there a proof?  Well, that really depends an the axioms and definitions for your system. 
If it were up to me.... well, I'd do something of the nature of taking it as axiomatic that our numbers form an Ordered Field of which the Integers are a subset.  This means among other things that $0 < 1$ and that $2 = 1+ 1 > 1$, and whenever $k > 0; a<b$ we have an axiom that $ak > bk$.  This means that if $2k = 1$ then $0 < k < 1$. (if $k < 0$ then $2>0$ would mean $2k < 2*0 = 0 < 1$.  If $k > 1$ then $2k > 2*1 =2 > 1$).  Part of the definitions of Integers would be an axiom that $1$ is the smallest integer that is greater than $0$.  So $k$ being between $0$ and $1$ can not be an integer.
That's a little breezy but I hope it gives the idea.
A: For $a$ to divide $b$ the following must hold (by definition of "divides"):
$$b = a \cdot m, m \in \mathbb{Z}$$
When $b=1$ and $a=2$ there is no such m.
