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 $\beta$?

  • 38
    $\begingroup$ There is a rational number between any two distinct real numbers. But what you have is two equal real numbers. $\endgroup$
    – Zhen Lin
    Mar 21, 2012 at 14:02
  • 15
    $\begingroup$ It had to be done. I drew the line. $\endgroup$
    – Asaf Karagila
    Mar 21, 2012 at 15:15
  • 8
    $\begingroup$ By the way $\beta$ ($=\alpha$) is rational. $\endgroup$
    – UncleZeiv
    Mar 21, 2012 at 17:28
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    $\begingroup$ $1.234\overline{9} \le \dfrac{247}{200} \le 1.235\overline{0}$ $\endgroup$
    – Henry
    Mar 21, 2012 at 17:55
  • 1
    $\begingroup$ 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. $\endgroup$
    – Jeff
    Dec 16, 2012 at 4:03

7 Answers 7


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: Is it true that $0.999999999\ldots = 1$?

  • $\begingroup$ 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. $\endgroup$ Mar 21, 2012 at 16:25
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    $\begingroup$ @Mathemagician1234 : If someone does not "get" this (to the point where it is obvious to them), then they are not an experienced mathematician. $\endgroup$
    – Adam Smith
    Mar 21, 2012 at 16:56
  • $\begingroup$ A twin. Your point is made. (alternate mixedmath) $\endgroup$
    – davidlowryduda
    Mar 21, 2012 at 18:05
  • 5
    $\begingroup$ @mixedmath: Not a bad answer for a throwaway account! $\endgroup$ Mar 23, 2012 at 3:21

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.


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 $1/5$.


As was pointed out, your "two" real numbers both represent the same quantity so there is nothing "between" them. The problem is NOT with our intuition concerning repeating decimals or limits as some people claim. The problem is with the notion "equals." A repeating decimal that ends in endless nines is NOT the "SAME" as its limit that terminates. They are obviously different since the former requires a (countable) infinity of numerals to write down and the latter does NOT. The quantity REPRESENTED is the same but obviously its CHARACTERIZATIONS are NOT. The problem is that it is conventional to speak of "equal" when we mean "equivalent." The two forms REPRESENT the same QUANTITY but they are clearly two different FORMS of that quantity. The closely connected concepts of FORMS and EQUIVALENCE of different forms arise throughout mathematics and should have been made clear way back when we learned about decimal numbers. I wish the equals sign would be limited to CONDITIONAL EQUALITY and NOT be used for quantitative equivalence. So, for example, 1+1≡2 would mean that 1+1 is quantitatively EQUIVALENT to 2, and y≡f(x) would mean y is quantitatively EQUIVALENT to f(x), i.e., another form or representation of f(x), etc. People wrongly blame lack of intuition for what is really lack of clarity. If I may I'd like to make two more points. Proofs do not explain; they assert the authority of logical consistency. And so-called "intuition" is required for all understanding. The idea that formality is not based on intuition is absurd. "Intuition" does not mean guessing or making up. It means SEEING in some sense.


Think of each positive real number as representing a length, or a distance along a line. If you mark each of those numbers two numbers you have there on that line you will mark the same position on the line twice. If you subtract those two numbers you have you will get 0. The point is, they are two different ways of writing the same number. Two numbers which represent the same distance along the line are the same number.


Whenever you see that 'bar stuff' at the end of a decimal expansion, you know that is shorthand for a repeating decimal and that the number is rational.

So the two numbers

$\beta=1.235\overline{0} \\ \alpha=1.234\overline{9}$

are both rational numbers.

The average of two rational numbers is also a rational number, and unless you are taking the average of two equal numbers, you'll be getting a new number between them.

So, just take the average and see what happens,

$\quad (.5) \times \bigr(1.235\overline{0} + 1.234\overline{9}\bigr) = $
$\quad (.5) (2.46\overline{9}) =$
$\quad 1.23 + (5) (0.\overline{9})\cdot 10^{-3} =$
$\quad 1.23 + 5 \cdot 10^{-3} =$
$\quad 1.235 = \beta$

So we are starting with two equal numbers.


Yet another perspective that I think is worthwhile.

$\alpha$ and $\beta$ are two real numbers that are arbitrarily close together. Precisely, for any $\epsilon > 0$, $|\alpha - \beta| < \epsilon$. To see this, just continue the repeating decimal until it's clear that the difference between $\alpha$ and $\beta$ is surely less than $\epsilon$.

For example, suppose $\epsilon = .000001$. We know $$ 1.2349999 < 1.234\overline{9} = \alpha \leq \beta = 1.2345\overline{0} = 1.2345 $$ which means $|\alpha - \beta| < |1.2349999 - 1.2345| = .0000001 < .000001 = \epsilon$. Hopefully it is clear that we could play this same game with any choice of $\epsilon > 0$.

The claim is that if for all $\epsilon > 0$ we have $| x - y | < \epsilon$, then $x = y$. Suppose not. Then $x \neq y$, so $| x - y | > 0$. But, by hypothesis, this means $| x - y | < |x - y|$, which is absurd: no number is strictly less than itself.

Therefore, since $|\alpha - \beta| < \epsilon$ for any $\epsilon > 0$, we conclude $\alpha = \beta$.


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