Does a Dedekind cut define a real number into existence? Suppose I cut the rationals greater than $-q$ into two sets, $L=(-q,q)$ and $R=[q,\infty )$. 
Take $L$ to be the set of all rationals whose squares are less than $2$, and $R$ to be the set of all  rationals whose squares are greater than $2$.

There are certainly no rational numbers between $L$ and $R$. Is $ \sqrt 2$ the only number between $L$ and $R$? How do I know this? 
How do I know that anything exists between these two sets?
 A: The point is that we have $\mathbb{Q}$ in our hands and want to "construct" real numbers.
To define this we define Dedekind cuts as all pairs $(L,R)$ where $L,R \subseteq \mathbb{Q}$ are non-empty, disjoint, $L \cup R = \mathbb{Q}$ and such that 


*

*$\forall x \in L: \forall x' \in R: x < x'$ i.e. $L$ "lies left of" $R$.

*$\forall x \in L: \exists x' \in L: x < x'$, i.e. $L$ has no maximum.


One easily shows that $L$ is closed under taking smaller points, and $R$ under taking larger points.
For every $q \in \mathbb{Q}$ we can define $L_q = \{x \in \mathbb{Q}: x < q\}$ and $R_q = \{x \in Q: x \ge q\}$. One easily shows that this is always a Dedekind cut. So we have plenty of cuts, and we can define an order $(L,R) \le (L',R')$ (where both are cuts) iff $L \subseteq L'$.
This defines an order on $\mathbb{R}:= \{(L,R): (L,R) \text { a Dedekind cut}\}$
and $(L_q, R_q) \le (L_{q'}, R_{q'})$ iff $q \le q'$. All this can be shown just reasoning inside $\mathbb{Q}$ and its subsets. So the new set (which is formally a set if $\mathbb{Q}$ is) contains $\mathbb{Q}$ as an order-embedded subset.
But cuts like 
$$L = \{x \in \mathbb{Q}: x < 0 \text{ or } x^2 < 2\} ,R = \{x \in \mathbb{Q}: x >0 \text{ and } x^2 > 2\}$$ 
are not of the form $(L_q, R_q)$ so give a new point that we can call $\sqrt{2}$. To justify this we first have to define $+$ and $\times$ on the set of cuts, using the already (hopefully) present $+$ and $\times$ on $\mathbb{Q}$ (this can be done, but it's a tedious checking) and show that the "square"  of the above cut really equals $2$, so that $\sqrt{2}:= (L_2, R_2)$.
So $\sqrt{2}$ exists because we can define this cut, and show it has the required properties we expect of it. Every distinct cut by definition is a different "real" number and in this construction the real numbers just are these cuts, by definition. It's not like we already have real numbers and then reason about cuts. The cuts are there only to define the real numbers, so we can reason about them. But we come preloaded with conceptions and intuitions so it can get confusing. 
In my old university we had a short course in our first year, "Foundations of Calculus" , where we did all of the above (including field operations etc.) and all proofs of them in detail, basically as a training in formalisation: show how we can prove all we "think we know" and use in Calculus (like order completeness of the reals and denseness of $\mathbb{Q}$) and other courses starting from just Peano's axioms (constructing and proving all about $\mathbb{Z}$ and $\mathbb{Q}$ as well) and some basic set theory. IMHO every wannabe mathematician should go through it once, just to see it is possible and to enhance your faith in your intuitions.
There are other ways to construct the real numbers based on $\mathbb{Q}$, namely identifying a real with an equivalence class of Cauchy sequences in $\mathbb{Q}$. This is more like viewing the reals as decimal expansions e.g. ( which are special Cauchy sequences based on powers of $\frac{1}{10}$ really). Here also we can define field operations and a linear order etc. all in other ways. This way of doing it is due to Cantor I believe (and it's a model for constructing completions for all metric spaces), while the cuts method is due to Dedekind and completes the order (not the metric), to get all suprema of bounded above sets. It turns out that the resulting ordered, metric fields are isomorphic and homeomorphic, so both can be identified and called $\mathbb{R}$. 
A: Short answer: yes, a cut defines a new kind of number into existence.
The square root of two is not between L and R. Instead, as your question suggests, we identify "the square root of two" as this cut.
There's no natural number less than 0 (or 1 if you are so inclined). But we can identify a new beast, called an integer, out of naturals. And there is no integer between 1 and 2, but we can identify a new beast, called a rational, which lives between 1 and 2 (actually many such critters). It makes sense to ask, "Is there a number between 1 and 2?", and definitively the answer is no—until we define the rationals, and call them "numbers." For people to be happy with your definition, they'll have to act like numbers in the usual ways, such as being amendable to some new kind of "addition". (Defining such things gets progressively more complicated.)
So it makes sense to ask whether the square root of two is rational, and then to prove that there is no number "the square root of 2". But, happily, we can make new numbers. This cut is one such method.
A: Does a Dedekind cut define a real number into existence?
No, not really in this case.
The irrational number $\sqrt 2$ can be defined using a triangle, 
the property of unique factorization, and a few other basic concepts
such as perpendicularity, ratio, measurement etc.
Real world right triangles can easily be constructed. 
Granted, the lines are not perfectly thin or accurate, but most non-mathematicians 
can easily grasp what is meant by a "mathematically perfect" version of the construction. 
A: You don't.  And nothing does exist between them. (Nothing rational at any rate).
But the set exists.  And in the set of sets we  can prove the set of all cuts forms an ordered  field.  And it has a proper subfield of cuts at rational points.  And that subfield is isomorphic and equivalent to the rationals.
And I guess this is the point where numbers "will themselves into existance".  We have an isomorphism of the Rationals to a Field equivalent to the rationals.  This equivalent field is a proper subfield of another.  So if we invert the isomophorism as applied to the superfield, we get an equivalent super field of the rationals.
Now if this were the real and practical world, it might be that no inverse of this real field of cuts exists really.  But this is the abstract and ideal world.  If the isomorphism exists than the image of the inverse isomorphism must as well.
I think.  At least it can be talked about abstractly.  I think that is the same thing.
