Are $r + \frac{1}{3}(s-r)$ and $r + \frac{2}{3}(s-r)$ irrational, given that $r < s$ and both are irrational? Are $r + \frac{1}{3}(s-r)$ and $r + \frac{2}{3}(s-r)$ irrationals, given that $r < s$ and both are irrational?
The context is the following:
There is an interval $E$ with irrational endpoints $r,s$. Do I get two other intervals if remove the middle third of $E$?
The doubt came from below [Cooke-1976]:
Q.: Is there a non-empty perfect set in $R^1$ which contains no rational number?
A.: Yes. Let $\{r_1, r_2,...,r_n,...\}$ be rational numbers in the interval $[-\pi, \pi]$. Let $E_0 = [-\pi, \pi]$. Now assume that $E_k$ has been chosen for $k < n$ in such a way that $E_k$ is a pairwise disjoint union of at most $2^{k+1}-1$ closed intervals with irrational endpoints, each of positive length at most $(\frac{2}{3})^k\pi$ and that $E_k$ does not contain $r_j$ if $j \le k$. (All of these conditions hold trivially for $k=0$.) Define a set $F_{k+1}$, which is obtained from $E_k$ by removing first the middle third of each of the intervals that constitute $E_k$. The result is a set of at most $2^{k+2}-2$ pairwise disjoint intervals having irrational endpoints, each interval being of length at most $(\frac{2}{3})^{k+1}\pi.$ If $r_{k+1} \notin F_{k+1}$, let $E_{k+1} = F_{k+1}$. If $r_{k+1} \in F_{k+1}$, then $r_{k+1}$ is not the endpoint of the interval $I = [a,b]$ of $F_{k+1}$ that it belongs to. Hence let $\delta$ be an irrational positive number less than the minimum of $r_{k+1}-a$ and $b-r_{k+1}$, and let $E_{k+1}$ be obtained from $F_{k+1}$ by removing the interval $(r_{k+1}-\delta, r_{k+1}+\delta)$ (which has irrational endpoints). Then $E_{k+1}$ consists of at most $2^{k+2}-1$ pairwise disjoint closes intervals, each of positive length at most $(\frac{2}{3})^{k+1}\pi$, and each having irrational endpoints. ... (and the proof goes on)
[Cooke-1976] - Solutions Manual to Walter Rudin's Principles of Mathematical Analysis
 A: That is not necessarily true. For example, with $r=-\sqrt{2}, s=2\sqrt{2}$, we get $r+\frac13(s-r)=0$. 
What you can show is that you cannot have both of your considered points rational, so at least one of your resulting 'remaining intervals' has both end points irrational.
A: They can't both be rational because if they were, then $2(r + \frac 13(s-r))-(r+\frac 23(s-r)) = r$ would also be rational, which $r$ is not.
But one or  or the other may be rational.  Just set one of them to any rational you want and then solve for $r$ in terms of $s$.
e.g.  $r + \frac 13(s-r) = 0\implies s = -2r$.
But this would mean $r + \frac 23(s-r)= -r$ which, of course, is not rational.
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Actually it looks like Cooke maybe made an oversight.
If you take an interval with irrational endpoints $r,s$ where $r$ and $s$ having been derived by linear combinations of $\pi$ and arbitrary irrational $\delta$s, and finding $(r + \frac 13(s-r))$ and $(r + \frac 23(s-r))$ which he assumed to be irrational, which they are unless by some coincidence it just happened that $\frac {(2|1)}3r = q - \frac {(1|2)}3s$ for some rational $q$. 
That would be a very strange coincidence but it could happen.
But... we can avoid it.
As the irrationals are dense in the reals we don't need an algorithm to find them.  We just provided one for the sake of confidence it gives us.
Do this instead:
Take an interval $[A,B] = E_0$ where $A$ and $B$ are irrational.  Take an enumerated list $(r_1,......)$ of the rationals in $[A,B]$.
Now for each $k$th step of the procedure do the following:  If $r_k$ is one of the intervals, $[a,b]$ (i.e. $a < r_k < b$, composing the set $E_{k-1}$,  remove a small chunk of the interval containing $r_k$ so that $[a,b]$ has been broken down into $[a,b']\cup  [a', b]$ where $a < b' < r_k < a' < b$.
Then for all the other intervals $[a_\alpha, b_\alpha]$ that compose $E_{k-1}$, remove about one third from the center.  Do it so that you are left with $[a_\alpha, b_\beta] \cup [a_\beta, b_\alpha]$ where $a_\alpha < b_\beta < a_\beta < a_\alpha$ and $b_\beta, a_\alpha$ are irrational.
Call the resulting union of closed intervals with irrational endpoints $E_k$.
Do this a countably infinite number of times.
The result is a cantor like set with only irrational points.
A: Put $r+(s-r)/3=100$, $r=\sqrt{2}$. Solve this linear equation for $s$, that will give you a counter-example. 
