2 irrational numbers for same Dedekind cut I was reading a book on history of numbers and came across Dedekind cut. I understand that 2 different cuts represent 2 unique numbers. But one thing that is not clear to me is if it is possible that the same Dedekind cut represent 2 different irrational numbers, i.e. 2 adjacent irrational numbers sandwiched between same cut. Can someone give proof for this?
 A: If two irrational numbers $a$, $b$ are "adjacent", then where is the number $\dfrac{a+b}{2}$?
The fact that between any two irrational numbers there is a rational number is essentially the same thing as saying the real line has no nonzero infinitesimals.  A number $\varepsilon>0$ would be "infinitesimal" if
$$
\underbrace{\varepsilon+\cdots\cdots+\varepsilon}_{n \text{ terms}} < 1
$$
no matter how large the finite number $n$ of terms is.  But if $n$ can be made big enough so that the sum exceeds $n$, then the rational number $1/n$ is less than $\varepsilon$.  If two irrational numbers have no rational number between them, then the difference between them would be a nonzero infinitesimal.
In a sense, this is all just a matter of what the conventional definition of "real number" is.  But that conventional definition can be shown to be the only one that has certain nice properties, e.g. an ordered field in which sups and infs exist.
A: First, a few definitions that I'll be using (they may not correspond precisely with what you've encountered, so let me know if yours are different, and I'll see what I can do to adapt).
We say that an ordered pair $\langle A,B\rangle$ with $A,B$ sets of rationals is a Dedekind cut if the following hold:

(i) $A,B$ are non-empty, disjoint sets with $A\cup B=\Bbb Q$,
(ii) $A$ has no greatest element ($B$ may or may not have a least element), and
(iii) every element of $A$ is less than every element of $B$.

We define $\Bbb R$ to be the set of left sides of Dedekind cuts--that is, the set of all $A\subseteq\Bbb Q$ such that $\langle A,\Bbb Q\smallsetminus A\rangle$ is a Dedekind cut. We will say that some $A\in \Bbb R$ corresponds to $a\in \Bbb Q$ if and only if $A=\{x\in\Bbb Q:x<a\}$--that is, if and only if $\Bbb Q\smallsetminus A$ has a least element, which will necessarily be $a$.
It's a nice exercise to show that $\subset$ is a total order relation on $\Bbb R$, and that the map $\Bbb Q\to\Bbb R$ given by $a\mapsto\{x\in\Bbb Q:q<a\}$ is an order-preserving injection--that is, $a<b$ if and only if $\{q\in\Bbb Q:x<a\}\subset\{x\in\Bbb Q:x<b\}$.
Given $X\in\Bbb R$, define $-X:=\{-q:q\in\Bbb Q\smallsetminus X, q\,\text{ not the least element of }\,\Bbb Q\smallsetminus X\}$. It can be seen that $-X\in\Bbb R$ for any $X\in\Bbb R$, that $-(-X)=X$. Also, given $X,Y\in\Bbb R$, we have $X\subset Y$ if and only if $-Y\subset-X$. Furthermore, $X$ corresponds to a rational if and only if $-X$ does.

Proposition: Given any $X,Y\in\Bbb R$ with $X\subset Y$, there is some $A\in\Bbb R$ corresponding to a rational such that $X\subset A\subset Y$.
Proof: If $X,Y$ correspond to rationals $x,y$, then by definition, we have $x<y$ since $X\subset Y$. Letting $a:=\frac{x+y}{2}$, it can be shown that $a\in\Bbb Q$, $x<a<y$, and letting $A:=\{q\in\Bbb Q:q<a\}$, $A$ corresponds to a rational, and we clearly have $X\subseteq A\subseteq Y$. Since $x\in A\smallsetminus X$ and $a\in Y\smallsetminus A$, then $X\subset A\subset Y$.
Now, suppose $X,Y$ don't both correspond to rationals. Without loss of generality, we may suppose $X$ doesn't correspond to a rational, for if $Y$ doesn't, then neither does $-Y$, and an identical proof will show that there is some $A\in\Bbb R$ corresponding to a rational with $-Y\subset A\subset-X$ (so $-A$ corresponds to a rational and $X\subset-A\subset Y$, as desired). Since $X$ doesn't correspond to a rational, then $\Bbb Q\smallsetminus X$ has no least element. Take any $a\in Y\smallsetminus X=(\Bbb Q\smallsetminus X)\cap Y$, so by definition of Dedekind cut, $a$ is greater than every element of $X$ and less than every element of $\Bbb Q\smallsetminus Y$. Since $a\in\Bbb Q\smallsetminus X$ and $\Bbb Q\smallsetminus X$ has no least element, then there is some $p\in\Bbb Q\smallsetminus X$ such that $p<a$. Similarly, there is some $r\in Y$ such that $a<r$. Put $A:=\{q\in\Bbb Q:q<a\}$, so that $A\in\Bbb R$ corresponds to the rational $a$. Since $a$ is greater than every element of $X$, then $X\subseteq A$, and since $a\in\Bbb Q\smallsetminus X$, then $X\subset A$. Since every element of $\Bbb Q\smallsetminus Y$ is greater than $a$ and $A=\{q\in\Bbb Q:q<a\}$, then $\Bbb Q\smallsetminus A=\{q\in\Bbb Q:q\geq a\}\supset\{q\in\Bbb Q:q>a\}\supseteq\Bbb Q\smallsetminus Y$, and so $A=\Bbb Q\smallsetminus(\Bbb Q\smallsetminus A)\subset\Bbb Q\smallsetminus(\Bbb Q\smallsetminus Y)=Y$. $\Box$

It follows from the above proposition that the rationals are densely embedded in the reals. This is how we (may) define the reals, and those that don't correspond to a rational are those that we define to be the irrationals. Thus, there aren't any distinct reals (in particular, distinct irrationals) without a rational between them, meaning that they must correspond to distinct Dedekind cuts.
