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

Question

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.

Answer 1

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.

Answer 2

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 $(a,b)$.

Answer 3

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 $y$.

Answer 4

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!

Answer 5

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?