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

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.

Solution 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.