# Prove that if x and y

Prove that if $x$ and $y$ are irrational numbers, there exists an irrational number $z$ such that $y.
You can still ask an expert for help

## Want to know more about Irrational numbers?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

furniranizq
$x-y>0$ so $\left(x-y{\right)}^{2}>0$. By the Archimedean property, there is a positive integer $m$ such that $\frac{1}{m}<\left(x-y{\right)}^{2}$. By increasing $m$ if necessary, you can assume that $m=2{n}^{2}$ so that $\frac{1}{2{n}^{2}}<\left(x-y{\right)}^{2}$. This means
$\frac{1}{\sqrt{2}n}
Now let $k$ be the largest integer that is less than or equal to $\sqrt{2}ny$. This implies
$1+k\le 1+\sqrt{2}ny<\sqrt{2}nx.$
But we also have $\sqrt{2}ny, so
$\sqrt{2}ny