Answer to Question #155601 in Calculus for nestar

Question

If x and y are arbitrary real numbers with x < y, prove that there exists at least one irrational number z satisfying x < z < y, and hence infinitely many.

Answer 1

Statement 1 (Density of rational numbers in $\R$ )

If $u\in\R$ and $v\in \R$ with $u<v,$ then there exists at least one rational number $w$ satisfying $u<w<v,$ and hence infinitely many.

Statement 2

$\sqrt{2}$ is not a rational number. In general, if $p$ is a prime number, then $\sqrt{p}$ is not a rational number.

Statement 3

If $a$ is an arbitrary rational number and $b$ is an arbitrary irrational number, then $ab$ is irrational provided $a\not=0.$

If we consider $u=x/\sqrt{2}$ and $v=y/\sqrt{2}$ then it follows from Statement 1 that there exists rational number $w\not=0$ such that 

$x/\sqrt{2}<w<y/\sqrt{2}$

Then

$x<w\sqrt{2}<y$

If we set $z=w\sqrt{2}$ then it follows from Statement 2 and Statement 3 that $z$ is irrational number.

Therefore

If $x\in \R$ and $y\in\R$ with $x<y,$ then there exists at least one irrational number $z$ satisfying $x<z<y,$ and hence infinitely many.