Let $a,b \in R$ where $ a < b$. Prove that there exist a rational number $c$ and an irrational number $d$ such that $ a <c<b$ and $ a<d<b$. Question : Let $a,b \in R$ where $ a &lt; b$. Prove that there exist a rational number $c$ and an irrational number $d$ such that $ a &lt;c&lt;b$ and $ a&lt;d&lt;b$. Hint: consider decimal expansions of $a$ and $b$

Attempt: Theorem 2.7.5 states that a real number is rational if and only if its decimal expansion terminates or has an infinitely repeating sequence of digits. 

Set $I$ is irrational numbers 

$I =[x \in R: x \notin Q]$

Set $Q$ is rational numbers

$Q = [ \frac{a}{b}:a,b, \in Z$ and $b \neq 0]$

If the division process terminates, then we are done. Otherwise, since each digit of the quotient 

determines in turn its successor, and since there are at most $b-1$ possible remainders when dividing 

by $b$ (by the division algorithm), some digit of the remainder must show up again, forcing a sequence 

of digits to repeat forever. The length of the repeating cycle is at most $b-1$

Suppose the decimal expansion of r will terminate. Therefore, for $r=0$, we have $a_1,a_2,...a_k$ where 

each $a_i \in [0,1,2,3,4,5,6,7,8,9]$ and $a_k \neq 0$. Then 

$ r = \frac{a_110^{k-1}+a_210^{k-2}+...+a_k}{10^k}$

satisfies the definition of a rational number. Now, suppose the decimal expansion of r has an 

infinitely repeating sequence of digits that begins immediately after the decimal point:
$r=0.b_1b_2...b_k$

The sequence of digits are repeated forever. Therefore,

$10^kr=b_1b_2...b_k.b_1b_2...b_kb_1b_2...b_k$

so,

$$10^kr-r=b_1b_2...b_k$ 

giving

$r = \frac{b_1b_2...b_k}{10^k-1} \in Q$

Suppose $r$ has an initial sequence of digits before the repeating sequence: $r = 0.a_1a_2...a_lb_1b_2...b_k$. Then we let $r' = 0.b_1b_2...b_k$. By the previous case $r' \in Q$. It is 

easy to verify that 

$r=\frac{r'+a_1a_2...a_l}{10^l}$

As a result, $ r \in Q$

So my question is do I apply the decimal expansion to $c$ and $d$. Assuming I can, then the decimal 

expansion of $c$ will terminate since it's rational by theorem 2.7.5.

Then,  for $r=0.c_1c_2...c_k$ where each $ c_i \in [0,1,2,3,4,5,6,7,8,9]$ and $c_k \neq 0$. Then 

$ r = \frac{c_110^{k-1}+c_210^{k-2}+...+c_k}{10^k}$

Suppose $d$ is an irriational number, then the decimal expansion won't terminate. 

$10^kr=d_1d_2...d_k.d_1d_2...d_kd_1d_2...d_k$

so,

$10^kr-r=d_1d_2...d_k$ 

giving

$r = \frac{d_1d_2...d_k}{10^k-1} \in Q$.

If we let $r = 0.c_1,c_2...c_ld_1d_2...d_k$ and $r' =0.d_1d_2...d_k$, then $r' \in Q$ and $r=\frac{r'+c_1c_2...c_l}{10^l}$

I could've sworn that this is going to work for $c$ only, but not for $d$ because as I mentioned earlier $c$'s decimal expansion will stop and $d$ will go on forever.
Using the decimal expansion for this purpose is at least cumbersome, and while we can do it by observing that $a$ and $b$ must differ at some decimal for the first time, one has to be careful with lots of $9$s and $0$s causing trouble.
Anyway you have a lot of freedon with "late" decimals, so you can make things periodic or aperiodic at will.

It is much simpler to let $\epsilon=b-a&gt;0$ and then note that there is at least one onteger $n$ with $n&gt;\frac1\epsilon$ and then at least one integer $m$ with $na&lt;m&lt;nb$ (because $nb-na&gt;1$) and one integer $m'$ with $n(a-\sqrt 2)&lt;m'&lt;n(b-\sqrt 2)$. Then $c=\frac mn$ and $d=\frac {m'}n+\sqrt 2$ do the trick.
Consider the numbers $c=\dfrac{a+b}{2}$ and $d=(\sqrt{2}-1)(b-a)+a$.