# 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 $a < b$. Prove that there exist a rational number $c$ and an irrational number $d$ such that $a <c<b$ and $a<d<b$. 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,

$$10^kr-r=b_1b_2...b_k giving r = \frac{b_1b_2...b_k}{10^k-1} \in Q Suppose r has an initial sequence of digits before the repeating sequence: r = 0.a_1a_2...a_lb_1b_2...b_k. Then we let r' = 0.b_1b_2...b_k. By the previous case r' \in Q. It is easy to verify that r=\frac{r'+a_1a_2...a_l}{10^l} As a result,  r \in Q So my question is do I apply the decimal expansion to c and d. Assuming I can, then the decimal expansion of c will terminate since it's rational by theorem 2.7.5. Then, for r=0.c_1c_2...c_k where each  c_i \in [0,1,2,3,4,5,6,7,8,9] and c_k \neq 0. Then  r = \frac{c_110^{k-1}+c_210^{k-2}+...+c_k}{10^k} Suppose d is an irriational number, then the decimal expansion won't terminate. 10^kr=d_1d_2...d_k.d_1d_2...d_kd_1d_2...d_k so, 10^kr-r=d_1d_2...d_k giving r = \frac{d_1d_2...d_k}{10^k-1} \in Q. If we let r = 0.c_1,c_2...c_ld_1d_2...d_k and r' =0.d_1d_2...d_k, then r' \in Q and r=\frac{r'+c_1c_2...c_l}{10^l} I could've sworn that this is going to work for c only, but not for d because as I mentioned earlier c's decimal expansion will stop and d will go on forever. ## 3 Answers Using the decimal expansion for this purpose is at least cumbersome, and while we can do it by observing that a and b must differ at some decimal for the first time, one has to be careful with lots of 9s and 0s causing trouble. Anyway you have a lot of freedon with "late" decimals, so you can make things periodic or aperiodic at will. It is much simpler to let \epsilon=b-a>0 and then note that there is at least one onteger n with n>\frac1\epsilon and then at least one integer m with na<m<nb (because nb-na>1) and one integer m' with n(a-\sqrt 2)<m'<n(b-\sqrt 2). Then c=\frac mn and d=\frac {m'}n+\sqrt 2 do the trick. • I've read about this somewhere...this doesn't use the decimal expansion...something about showing that it's rational, but it's irrational instead. x_X Commented Apr 24, 2014 at 9:55 Hint: Given a\lt b in \mathbb{R}, find any positive integer n so that$$ n(b-a)\gt1 $$This assures that$$ \frac1n\lt b-a $$Find the smallest integer k so that$$ \frac kn\ge b $$Show that$$ a\lt\frac{k-1}{n}\lt b $$Find the smallest integer m so that$$ \frac mn+\sqrt2\ge b $$Show that$$ a\lt\frac{m-1}{n}+\sqrt2\lt b 

• so we need to find two smallest integers for k and m. So do we let $c = \frac{m}{n}$? and then I'm kind of stuck afterwards except n must be positive. Commented Apr 24, 2014 at 10:47
• @usukidoll: I changed the multiplication to addition to overcome a problem of multiplication by $0$. $c$ would be $\frac{k-1}{n}$ and $d$ would be $\frac{m-1}{n}+\sqrt2$.
– robjohn
Commented Apr 24, 2014 at 10:56

Consider the numbers $c=\dfrac{a+b}{2}$ and $d=(\sqrt{2}-1)(b-a)+a$.

• ok? So what do I do with that? I was reading the hint and wanted to approach that way unless it's not a good idea because of all the numbers. Commented Apr 24, 2014 at 9:41
• Prove they are between $a$ and $b$, and in the case of $d$, that it is irrational. Commented Apr 24, 2014 at 9:42
• @FireGarden: $a$ and $b$ are not necessarily rational. Commented Apr 24, 2014 at 9:44
• @FireGarden what the?! The question said that a rational number c exists for a<c<b and an irrational number d exists for a<d<b. Hint: consider decimal expressions of a and b which I've provided. Commented Apr 24, 2014 at 9:45
• @TonyK Once you have the first part, it doesn't matter. If you look at $(a,b)$ with $a,b$ irrational, then you can by density of rationals consider $(a',b')\subset (a,b)$ with $a',b'$ rational.. Commented Apr 24, 2014 at 9:46