Answer to Question #155680 in Calculus for Vishal

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

Here, we have $x',y'\in\R,$ such that $x'<y'.$

Now, we already know that there exists a rational number $q$ such that $x'<q<y'.$

Now, let $n\in\R,$ so $\sqrt{n}\in\R$ .

Dividing all sides by $\sqrt{2}$ :-

$$\frac{x'}{\sqrt{2}}<\frac{q}{\sqrt{2}}<\frac{y'}{\sqrt{2}}$$

Now, as $q\in\mathbb{Q},$ so $\frac{q}{\sqrt{2}}=z$ $\in\mathbb{Q^c}$ .

And $x',y'\in\R$, so we have $\frac{x'}{\sqrt{2}},\frac{y'}{\sqrt{2}}$ $\in\R$

So,

$$x<z<y$$

where, $x=\frac{x'}{\sqrt{2}}$ and so on.

So, we have an irrational number between $x,y$ such that $x<z<y$ .

Now, we took $\sqrt{2}$ as an example, but we can replace by any $n\in\R$ . Note that some $n=4,16$ are rational, but we have infinitely many $n\in\R$ (as $\R$ is an infinite set), such that $\frac{q}{\sqrt{2}}$ $\in\mathbb{Q^c}$ .