Proof that for any interval (a,b) with aBackground: We are assuming that the elements of $\mathbb{R}\setminus\mathbb{Q}$ are irrational number. If $x$ is irrational and $r$ is rational then $y=x+r$ is irrational. Also, if $r\neq 0$ then $rx$ is irrational as well. Likewise, if a number is irrational then its reciprocal is irrational as well.
Theorem: Every interval $(a,b)$, no matter how small, contains both rational and irrational numbers.
Proof. First we can see that the interval $(0,1)$
  contains both rational and irrational, just select the numbers $1/2$
  and $1/\sqrt{2}$. For the general interval $(a,b)$, think of $a$
  and $b$
  as cuts, that is $a=A\mid A^{'}$
  and $b=B\mid B^{'}$
 , such that $a<b$
 . The fact that $a<b$
  implies that $B\setminus A\neq\emptyset$
 , moreover since $B$
  has no greatest value if we select an a rational element $r\in B\setminus A$
  then we can find another rational element $s\in B\setminus A$
  such that $r<s$
 . Thus $a\leqslant r<s\leqslant b$
 . The transformation
$$T:t\mapsto r+(s-r)t$$
sends the interval $(0,1)$
  to the interval $(r,s)$
 . Since $r,s,$
  and $s-r$
  are all rational, the transformation $T$
  sends rationals to rationals and irrationals to irrationals. That is, $(r,s)$
  contains and irrationals, and so does the larger interval $(a,b)$.$\space\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \square$
My Confusion: My first confusion is what does the transformation  $T$ map from and to? If $t\in \mathbb R$ then for any value of  $t$  we should have $r\leqslant r+(s-r)t \leqslant s$ right? However, if  $t<0$, $r<0$, and $s<0$ then the preceding inequality shouldn't hold (unless I am missing something); and if the preceding inequality doesn't hold then what good is the transformation $T$ for  showing that there is an irrational in the interval $[r,s]$?
 A: As the proof says, we actually define $T$ only on the open unit interval $(0,1)$, which it maps onto $(r,s)$. If we were to use the same formula to define $T$ on all of $\Bbb R$, we’d find that $T$ maps $\Bbb R$ bijectively onto $\Bbb R$.


*

*$T(0)=r$, $T(1)=s$.  

*$T$ maps $(0,1)$ onto $(r,s)$ in order-preserving fashion.  

*$T$ maps $(\leftarrow,0)$ onto $(\leftarrow,r)$ in order-preserving fashion: as $t$ moves leftward from $0$, $T(t)$ moves leftward from $r$.  

*$T$ maps $(1,\to)$ onto $(s,\to)$ in order-preserving fashion: as $t$ moves rightward from $1$, $T(t)$ moves rightward from $s$.


But we don’t actually care about the first, third, and fourth of these points.
A: $T$ maps from $(0,1)$ to $(r,s)$, as the proof says! The point is that $$a < r < r + (s-r)\frac{1}{\sqrt{2}} < r + (s-r)1 = s < b$$ and the third number in that inequality is irrational (and lies in $(a,b)$).
This only works because $0 < \frac{1}{\sqrt{2}} < 1$ — it doesn't work for all $t\in\mathbb{R}$.
A: Another method is 
to use the Archimedean axiom
for the reals:
For any two positive reals $x$ and $y$,
there is a positive integer $n$
such that
$nx > y$.
We want to find a rational number
$m/n$ such that
$a < \frac{m}{n} < b$.
To do this,
choose $n$ such that
$n(b-a) > 1$,
so that
$\frac1{n} < b-a$.
Let $j = \lfloor na \rfloor$.
Then $j \le na < j+1$
so
$\frac{j}{n} \le a < \frac{j+1}{n}
=\frac{j}{n}+\frac{1}{n}
<a+(b-a)
= b
$.
Therefore the rational
$\frac{j+1}{n}$
is between $a$ and $b$.
To find an irrational between $a$ and $b$,
do the same as before,
but choose $n$ such that
$n(b-a) > 2$
and
$j = \lfloor na \rfloor$.
By the same reasoning,
both 
$\frac{j+1}{n}$
and
$\frac{j+2}{n}$
are between $a$ and $b$.
Then
$\frac{j+\sqrt{2}}{n}$
is an irrational that is between $a$ and $b$
since 
$a < \frac{j+1}{n} < \frac{j+\sqrt{2}}{n}
< \frac{j+2}{n} < b$.
