# What rational number is between these two real numbers?

According to several texts and professors, there exists a rational number between any two real numbers. But suppose you had two real numbers which had the same digits in the same places up to some place, where they differed by one digit - say the smaller one had a $4$ and the bigger one had a $5$. The smaller number's digits following the four are all $9$s, and the bigger number's digits after the five are all $0$s. For example: $$\beta=1.235\overline{0} \\ \alpha=1.234\overline{9}$$

What rational number is between $\alpha$ and $\beta$?

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There is a rational number between any two distinct real numbers. But what you have is two equal real numbers. –  Zhen Lin Mar 21 '12 at 14:02
It had to be done. I drew the line. –  Asaf Karagila Mar 21 '12 at 15:15
By the way $\beta$ ($=\alpha$) is rational. –  UncleZeiv Mar 21 '12 at 17:28
$1.234\overline{9} \le \dfrac{247}{200} \le 1.235\overline{0}$ –  Henry Mar 21 '12 at 17:55
Hey OP, did you ever notice that there is no fraction that generates $.99\overline{9}$? There is $\frac 19 = .11\overline{1}$, and $\frac 29 = .22\overline{2}$, ... and $\frac 89 = .88\overline{8}$, and, finally $\frac 99 = 1$. I have no idea why (or if) that's significant. –  Jeff Dec 16 '12 at 4:03

I think the most helpful thing you can do here is to think about what decimal expansions really are.

What do we mean when we write $x=1.234999\ldots$? As you probably know, finite decimal expansions are just sums of certain decimal fractions, for example: $1.234$ means is by definition $1+\frac2{10}+\frac3{100}+\frac4{1000}$. So infinite decimal expansions are in fact just another way to write an infinite series, in your case: $$1.234999\ldots = 1+\frac2{10}+\frac3{10^2}+\frac4{10^3}+\frac9{10^4}+\frac9{10^5}+\frac9{10^6}+\cdots$$ which we may write more concisely as: $$\pm a_0.a_1a_2a_3a_4\ldots = \pm\sum_{n=0}^\infty\frac{a_n}{10^n},$$ where $a_0\in{\mathbb N_0}$ and $a_i\in\lbrace0,1,2,3,4,5,6,7,8,9\rbrace$ for $i\in\mathbb N$.

In your case, $a_0=1,a_1=2,a_2=3,a_3=4$ and $a_i=9$ for all $i>3$. So we are going to solve the problem if we determine the value of this series: $$1.234999\ldots =1+\frac2{10}+\frac3{10^2}+\frac4{10^3}+\frac9{10^4}+\frac9{10^5}+\frac9{10^6}+\cdots =\\=1.234+\frac9{10^4}(1+\frac1{10}+\frac1{10^2}+\frac1{10^3}+\cdots).$$ Now the expression in parentheses is a geometric series, i.e. a series of the form $1+x+x^2+x^3+\cdots$ (or if you prefer the more concise notation, $\sum\limits_{n=0}^\infty x^n)$. As you probably already know (otherwise you will most likely learn this soon), the geometric series converges to $\frac1{1-x}$ for $-1<x<1$. So in our case where $x=\frac1{10}$, we have: $$\sum_{n=0}^\infty(\frac1{10})^n =1+\frac1{10}+\frac1{10^2}+\frac1{10^3}+\cdots = \frac1{1-\frac1{10}}=\frac{10}9.$$ Now we just plug this into the expression above and we get $$1.234999\ldots = 1.234 + \frac9{10^4}\frac{10}9 = 1.234+\frac1{10^3} = 1.234+0.001=1.235$$ Thus the two numbers are equal.

To see that for two distinct real numbers $a,b$ we indeed have a rational between them, we note that $c=\frac{a+b}2$ is a number strictly between them. But as every rational number has a decimal expansion, we get a sequence of truncated decimal expansions, $\sum\limits_{n=0}^N\frac{c_n}{10^n} = c_0+\frac{c_1}{10}+\frac{c_2}{10^2}+\cdots+\frac{c_N}{10^N}$, that converges (gets arbitrarily close) to this number. These are rational numbers. Now, as $a$ and $b$ are both a positive distance away from $c$, this means there will be a number in this sequence that is closer to $c$ than $a$ and $b$ are. This means we have found a rational number strictly between $a$ and $b$, so we are done.

Added: The thing to remember here is that some real numbers have two distinct decimal expansions. If you think about it a bit, you will see that these are precisely the numbers that have a terminating decimal expansion, i.e. the numbers which have $0$ from some place on. (These are the rational numbers which can be written in the form $\frac{m}{10^k}$ for some $m\in\mathbb Z$ and $k\in\mathbb N_0$.) These numbers can be written in two ways because we can always replace the last digit before the zeroes with a digit that is one smaller and add nines after it. Note that a funny thing happens with $0$. It may still be regarded as having two decimal expansions in a sense: $0=+0.000\ldots=-0.000\ldots$ This is because we have defined the decimal expansion using the formula $$\pm a_0.a_1a_2a_3a_4\ldots = \pm\sum_{n=0}^\infty\frac{a_n}{10^n},$$ for $a_0\in{\mathbb N_0}$ and $a_i\in\lbrace0,1,2,3,4,5,6,7,8,9\rbrace$ for $i\in\mathbb N$. The strange behaviour of $0$ arises because this way we treat positive and negative numbers differently. It goes away if we use a different formula: $$a_0.a_1a_2a_3a_4\ldots = \sum_{n=0}^\infty\frac{a_n}{10^n}$$ where $a_0\in{\mathbb Z}$ and $a_i\in\lbrace0,1,2,3,4,5,6,7,8,9\rbrace$ for $i\in\mathbb N$. I don't believe this is completely standard, however, probably because in this case the other decimal expansion looks a bit funny: $(-1).999\ldots=0.000\ldots=0$. Note, though, that using this (possibly impractical) notation every number that can be written as $\frac{m}{10^k}$ behaves the same. In both cases every number that is not of this form has a unique decimal expansion.

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Have you seen the fact that $$0.99999\bar{9}=1.$$ These two numbers are literally equal. Take a look at this question to see more in depth answers regarding this fact: Does .99999... = 1?

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This may be one of the more maddeningly subtle aspects of the real numbers and it's the hardest to really grasp.I know experienced mathematicians that still have trouble "getting" it. –  Mathemagician1234 Mar 21 '12 at 16:25
@Mathemagician1234 : If someone does not "get" this (to the point where it is obvious to them), then they are not an experienced mathematician. –  Adam Smith Mar 21 '12 at 16:56
A twin. Your point is made. (alternate mixedmath) –  mixedmath Mar 21 '12 at 18:05
@mixedmath: Not a bad answer for a throwaway account! –  The Chaz 2.0 Mar 23 '12 at 3:21

I can give you a clear picture of why these two numbers are the same real number. It is NOT an "algebraic" or something like that proof, which I found a lot in the other question, but it is based on the very definition of the decimal expansion.

For now consider a real number $x=1/5$. What is this number? It is a number such that $x+x+x+x+x=1$. It takes many more steps starting from the very axioms of the real numbers to show that it is well-defined, i.e. there exists only one such real number, but, of course, we are going to skip all those discussions as not relevant to our question.

When it comes to decimal representation, one defines it as follows. Let us define it for a number $y$ in $[0,1]$. First, we divide the interval into 10 equal subintervals ($[0,1/10][1/10,2/10]...$ when each $k/10$ is well-defined as before). Now, we assign a number from $0$ to $9$ to each interval and find the one containing $y$. So, suppose it is interval #2. Then, we start $y$'s decimal representation with 2: $y=0.2\dots$. Then, we divide the second interval (where we found $y$) into another 10 subintervals, and find the one containing $y$. Suppose, it is interval #3, then $y=0.23\ldots$. In fact, we proceed this way infinite number of steps, and construct the sequence of digital numbers representing $y$. It is just a common convention not to write down all the tail zeros if the sequence are all zeros starting from some point.

Now, let's go back to $x=1/5$. At the very first step we can choose either 1st or 2nd interval, because $1/5=2/10$, and our $x$ is right at the border of the two intervals. If we choose the interval #1, then after that we will divide the first interval into subintervals, and our $x$ will always be at the very right end of the last subinterval, i.e. it will always be in interval #9: $x=0.1999\ldots$. On the other hand, if we choose interval #2, then we will proceed dividing this interval into subintervals, and $x$ will always be in interval #0: $x=0.2000...$.

I hope, this gives you a clear picture of what is going on here. So, it is NOT the case that those are two different numbers. In fact, the reason is that the very definition of the decimal expansion allows for two different representations of same numbers.

So, vice versa: when you see $0.19999\ldots$, you know that it is THE number that lies in the first interval $[1/10,2/10]$ and always in the last interval #9 in further divisions, but there is only one such point (by the intersection theorem), which is $1/5$.

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