Prove that between any two different real numbers there is a rational number and an irrational number.

My attempt at a proof: (Throughout this proof it is assumed without any loss of generality that $x < y$)

1.) $x$ is irrational and $y$ is irrational.

In this case the arithmetic mean $\frac{x+y}{2}$ is always an irrational number between $x$ and $y$. Also, if $y$ is irrational then it has a non-terminating decimal expansion and so we can always find a positive integer $n$ such that $x < y - 10^{-n}y < y$. The middle of the three numbers here will have a terminating decimal expansion, and hence it is rational. It also clearly lies between $x$ and $y$. These examples of rational and irrational numbers should also work in the cases:

2.) $x$ is irrational and $y$ is rational.

3.) $x$ is rational and $y$ is irrational.

There is one last case to consider.

4.) $x$ is rational and $y$ is rational.

In this case the arithmetic mean of $x$ and $y$ is rational. An irrational number for example could be the ratio $\frac{ex + \pi y}{e + \pi}$

I'm not sure if this proof is a good method or if it is even valid, so some feedback would be appreciated.

  • 2
    $\begingroup$ The arithmetic mean of irrational numbers need not be irrational; consider $-\sqrt2,\sqrt2$. $\endgroup$ Sep 6, 2016 at 22:35

5 Answers 5


Nice attempt, but unfortunately your proof is wrong. $y-10^{-n}y=y(1-10^{-n})$. Since $1-10^{-n}$ is rational and $y$ is irrational, $y(1-10^{-n})$ is irrational. Also, as pointed out by Mees de Vries in comments, $\frac{x+y}{2}$ may be rational.

In this link, you can find a proof by joeA that there is a rational between two real numbers. He uses the Archemidean Property of the real numbers, which can be stated as follows:

For every number $x\in\mathbb{R}$, there exists a natural number $n$ such as $x<n$.

for this purpose (even the intuitive fact that there is an integer between two numbers $x,y$ satisfying $y-x>1$ can be proved using this property) . Then you can conclude the result for irrationals; indeed if $x,y\in\mathbb{R}$, take any irrational number of your choice, say $\sqrt{2}$. Suppose that $x<y$. Then $x-\sqrt{2}<y-\sqrt{2}$. Thus there exists a rational number $q$ such as $x-\sqrt{2}<q<y-\sqrt{2}$, that it $x<q+\sqrt{2}<y$ and $q+\sqrt{2}$ is irrational.

  • $\begingroup$ This proof can actually be taken one step deeper if you want as well by proving that $q + \sqrt{2}$ is indeed irrational. $\endgroup$ Apr 16, 2021 at 16:01
  • $\begingroup$ @GrayLiterature Indeed; I assumed it is clear enough. $\endgroup$ Apr 19, 2021 at 9:19
  • $\begingroup$ Yes. My extra step is not necessary, but for someone new to these proofs it could be helpful in their journey to prove that if X rational and Y irrational that X+Y is irrational. It’s basic, but a good exercise nevertheless. $\endgroup$ Apr 19, 2021 at 13:56
  • $\begingroup$ @GrayLiterature You're right, I agree. Even for the most basic properties, it is a good practice to learn to prove them. If one thinks it's too obvious to prove it, but then fails to, then it means it was not obvious at all for them, but rather simply "made sense" to them. $\endgroup$ May 3, 2021 at 0:14

Given two real numbers $a,b$ with $a<b$ we have that $n(b-a)>1$ for some $n\in\mathbb{N}^+$.
It follows that there is some integer $m$ in the interval $(an,bn)$ and $\frac{m}{n}$ is a rational number belonging to the interval $(a,b)$. In a similar way, for some $N\in\mathbb{N}^+$ we have that $N\sqrt{2}(a-b)>1$, hence there is some integer $M$ in the interval $(aN\sqrt{2},bN\sqrt{2})$ and $\frac{M}{N\sqrt{2}}$ is an irrational number belonging to the interval $(a,b)$.


Once you've got a rational number between any two reals, and you've also proved that some irrational number exists, it's easy to prove there is an irrational number between to reals, as follows.

Suppose $x<y$. Some rational number $a$ is between them. And then some rational number $b$ is between $a$ and $y$. So you have $x<a<b<y$. Now let $\alpha$ be some irrational number. If $\alpha<0$, then multiply $\alpha$ by $-1$ and let that be the new $\alpha$. If not $\alpha>1$, then let $1/\alpha$ be the new $\alpha$. Now you have an irrational number $\alpha$ between $0$ and $1$. Now let $$ c = \alpha a + (1-\alpha)b. \tag 1 $$ To prove $c$ is irrational, assume $c={}$an integer over an integer, and solve $(1)$ for $\alpha$, and recalling that $a$ and $b$ are rational, show that $\alpha$ would then have to be rational.

Now the harder part. There is no real number greater than all of the finite integers $1,2,3,4,\ldots$. (If there were, then the integers would have a least upper bound $m$, and then $m-1$ would have to be less some integer, so so would $m$, so it wouldn't be an upper bound.)

Therefore $1/(y-x)$ cannot be greater than all integers. Let $n$ be an integer greater than $1/(y-x)$. Then $1/n < y-x$, so one of the rational numbers $k/n$, $k\in \mathbb Z$, must be between $x$ and $y$.


Proof that between any two real numbers there exists a rational:

Let $x,y \in \mathbb R$, and assume without loss of generality that $x \lt y$. Then $y-x \gt 0$, and by the Archimedean property $\exists n \in \mathbb N$ such that $n(y-x) \gt 1$ and hence $ny \gt nx + 1$. Again by the Archimedean property, $\exists m \in \mathbb N$ such that $m \gt -nx$ and $\exists k \in \mathbb N$ such that $k \gt nx$. Then $-m \lt nx \lt k$ and $\exists j \in \mathbb N$ with $-m \le j \le k$ such that $j-1 \le nx \lt j$. Combining inequalities we have $nx \lt j \le nx + 1 \lt ny$, hence $x \lt \frac jn \lt y$, where $j,n \in \mathbb Z$, hence $\frac jn \in \mathbb Q$, as desired.

Note: we have to account for the possibility that $x<0$, which is why we bring in the $k$ and $j$.

Now to prove that between any two reals there exists an irrational.

Let $x,y \in \mathbb R$ and assume without loss of generality that $x \lt y$. Then $x - \sqrt{2} \lt y - \sqrt{2}$. By the density of the rationals (shown above), $p \in \mathbb Q$ with $x - \sqrt{2} \lt p \lt y - \sqrt{2}$. Then $x \lt p + \sqrt{2} \lt y$, where $p + \sqrt{2} \notin \mathbb Q$, as desired.

I assume you are familiar with the irrationality of $\sqrt{2}$.

I suggest you take as an exercise the proof that if $p \in \mathbb Q, r \notin \mathbb Q$, then $p+r \notin \mathbb Q$, and look at my proof afterwards.

Since $p \in \mathbb Q$, $\exists a,b \in \mathbb Z$ such that $p = \frac ab$. Suppose $p+r \in \mathbb Q$. Then $\exists m,n \in \mathbb Z$ such that $p +r = \frac mn$. Then $r = \frac mn - p = \frac mn - \frac ab = \frac {mb -an}{nb}$, where $mb-an, nb \in \mathbb Z$, contradicting the irrationality of $r$. Thus $p+r$ cannot be rational.

Let me know if you have any questions!


Case 1: x,y $\in\ \mathbb {Q}$

  • $\cfrac {x+y}{2} \in\ \mathbb {Q}$
  • $x < \cfrac {x+y}{2} < y$
  • Thus $\exists\ a \in\ \mathbb {Q} , a \in\ (x,y)$
  • y-x > 0 & y-x $\in\ \mathbb {Q}$
  • $\cfrac {y-x}{\sqrt 2} \in\ \mathbb {Q^c}$
  • $x < x+\cfrac {y-x}{\sqrt 2} < y$
  • Thus $\exists\ b \in\ \mathbb {Q^c} , b \in\ (x,y)$
  • True for case 1

Case 2: x,y $\in\ \mathbb{Q^c}$

  • x < $\cfrac {x+y}{2}$ < y

  • r = $\cfrac {\lfloor 10^n(\cfrac {x+y}{2}) \rfloor}{10^n} ; n \in\ \mathbb {N}$

  • r $\in\ \mathbb {Q}$

  • $\exists\ n \in\ \mathbb {N}$ , x < r < y

  • Thus $\exists\ a \in\ \mathbb {Q} , a \in\ (x,y)$

  • $\cfrac {x+r}{2} \in\ \mathbb {Q^c}$

  • $x < \cfrac {x+r}{2}$ < r < y

  • Thus $\exists\ b \in\ \mathbb {Q^c} , b \in\ (x,y)$

  • True for case 2

Case 3: x $\in\ \mathbb {Q} , y \in\ \mathbb {Q^c}$

  • i = $\cfrac {x+y}{2} \in\ \mathbb {Q^c}$

  • x < i < y

  • Thus $\exists\ b \in\ \mathbb {Q^c} , b \in\ (x,y)$

  • $\exists\ r \in\ \mathbb {Q}$ , i < r < y ; (Proved in case 2 ; i,y $\in\ \mathbb {Q^c}$)

  • Thus $\exists\ a \in\ \mathbb {Q} , a \in\ (x,y)$

  • True for case 3

There exists a rational number and irrational number between any two real numbers

Is my answer correct?


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