Help with understanding Dedekind Cuts definition I am trying to understand the definition of what a Dedekind Cut is and I get that these are subsets of the rationals that are non-empty not not the whole of the rationals. I also sort of understand what it meant by downwards closed, but I do not understand the third part of the definition that states there is no largest element. Any suggestions? 
A Dedekind Cut is a subset $x\subset\mathbb{Q}$ such that:
$\emptyset\neq\ x\neq\mathbb{Q}$
$x$ is downwards closed, i.e. if $q\in\ x$ and $r<q$, then $r\in\ x$
$x$ has no largest element.
 A: If we restrict our space to only rational numbers (no irrationals) then the open interval $(-\infty, 1)$ is an example of a dedekind cut.
1) It isn't empty and it is not all of $\mathbb Q$.
2) If $r < q$ and $q \in (-\infty, 1)$ then $r\in (\infty, 1)$
3) If $q \in (-\infty, 1)$ then there is an $r > q$ so that $r \in (\infty, 1)$.  
To find such an $r$ in property 3):  We know that $q < 1$ so $1 - q > 0$.  And $1-q > \frac {1-q}2 > 0$.  So let $r = q + \frac {1-q}2 < q + (1-q) = 1$.  As $r$ is rational, $r \in (-\infty, 1)$ and $r >q$.  So $q$ is not a maximal element.  
No element in $(-\infty, 1)$ is maximal because for any rational number less than $1$, one can always find a larger rational number between it and $1$. 
Reading the comments you seem be confusing "maximal" elements with "numbers that are as large or larger than all elements in the set".  The difference being that a maximal element has to actually be an element in the set.  In the comments you asked "Why is $1$ not the maximal element of $(-\infty, 1)$".  The answer to that is because $1$ isn't IN $(-\infty, 1)$.
All $(-\infty, q)$ for rational $q$ are dedikind cuts.  In fact the only difference between an open interval $(-\infty, q)$ and a dedekind cut is that a dedekind cut might not have a well defined endpoint.  A dedekind cut will be something like $(-\infty, ????)$ but that $????$ might not be describable.
For example:.  $D = \{q \in \mathbb Q|q < 0$ or if $q\ge 0$ then $q^2< 2\}$ is of the form $(-\infty, ???)$ but $???$ would have to be some sort of square root of $2$ and there's no such thing in the rational numbers so there is no way to write that.
Which gives a hint to the punchline of Dedekind's joke.  Once we define the real numbers, the dedekind cuts are nothing more or less than $(-\infty, x)\cap \mathbb Q$ where $x$ is a real number.  But have to define the dedekind cuts before we define the real numbers in order to define the real numbers in the first place.
A: So informally the Dedekind Cut of the real number $x$ is $\{r \in \Bbb Q: r < x\}$. Note the sign of $<$ instead of $\le$, such that the set has no maximum element (but does have a $\sup$remum, namely $x$).
A: For example, $x = \mathbb{Q} \cap (-\infty,1]$ satisfies all the requirements in the definition except that it has a largest element $1$, so this $x$ is not a Dedekind cut.
But $x = \mathbb{Q} \cap (-\infty,1)$ has no largest element. Notice that this $x$ does have a least upper bound, namely the number $1$, but $1$ is not an element of $x$. And since $x$ is downward closed and $\emptyset \ne x \ne \mathbb{Q}$, it follows that this $x$ is a Dedekind cut.
A: For any rational number $p$, there is a corresponding subset of rational numbers defined by
$\quad I_p = \{q \in \mathbb Q \; | \; q \lt p \}$
So we have a function 
$\tag 1 \mathcal R: \mathbb{Q} \to \mathcal{P}(\mathbb{Q})$ 
with domain $\mathbb Q$ and target the power set of $\mathbb Q$; it is defined by
$\tag 2 p \mapsto I_p = \mathcal R(p)$
Proposition 1: The function $\mathcal R$ is an injection.
Proof: Exercise.
Proposition 2: Each subset $I_p$ of $\mathbb Q$ satisfies the following conditions:
$\tag 3 I_p \ne \emptyset$
$\tag 4 I_p \ne \mathbb Q$
$\tag 5 \text{IF } q \in I_p \text{ And } r \lt q \text{ THEN } r \in  I_p$
$\tag 6 \text{IF } s \in I_p \text{ THEN There Exists } t \in I_p \text{ such that  } t \gt s$
Proof: Exercise.
Define $D_{x^2 = 2\text{?}} = \{q \in \mathbb Q \; | \; q^2 \lt 2 \} \; \bigcup \; I_0$.
Proposition 3: The set $D_{x^2 = 2\text{?}} \subset \mathbb Q$ in not in the range of $\mathcal R$. Moreover, this set satisfies the conditions (3) thru (6).
Proof: Exercise.
So the set of all Dedekind Cuts $x\subset\mathbb{Q}$ contains all the $I_p$ but also 'other stuff'. In fact, it contains a 'whole bunch of other stuff'.
A: A Dedekind cut on $\Bbb Q$ can also be defined as a pair $(A,B)$ of subsets of $\Bbb Q$ such that 
(i).  $A\ne \phi \ne B$ and $A\cap B=\phi$  and $A\cup B=\Bbb Q $.
(ii).  Every $a\in A$ is less than every $b\in B.$ 
(iii).  $A$ has no largest member.
For $q\in \Bbb Q$ let $A_q=\{r\in \Bbb Q: r<q\}$ and $B_q=\{r\in \Bbb Q:r\geq q\}.$ Then $(A_q,B_q)$ is a Dedekind cut on $\Bbb Q.$ In this case the second member $B_q$ of the Dedekind cut has a least member ($q$). 
There are Dedekind cuts $(A,B)$ on $\Bbb Q$ for which $B$ has no least member. For example let $A=\{q\in \Bbb Q:q\leq 0\lor (q>0\land q^2<2)\}$ and $B=\Bbb Q \setminus A.$ Such Dedekind cuts correspond to members of $\Bbb R \setminus \Bbb Q.$ 
In fact we $define$ $\Bbb R$ in such  a way that 
(i'). If $r\in \Bbb R\setminus \Bbb Q$   then there is a unique Dedekind cut $(A_r,B_r)$ on $\Bbb Q$ such  that $a<r<b$ for all $a\in A_r$ and $b\in B_r$, and such that $B_r$ has no least member .
(ii'). If $(A,B)$  is a Dedekind cut on $\Bbb Q$ such that $B$ has no least member then there is a unique $r\in \Bbb R\setminus \Bbb Q$ such that $a<r<b$ for every $a\in A$ and every $b\in B$.
