Prove that the irrationals are dense in the rationals: for all rational numbers
$x < y$ there exists an irrational $\alpha$ such that $x < \alpha < y$.
I know that between any two real numbers there exists an irrational number.
Proof: Let $x < y$ be two real numbers. Then $x – \sqrt{2}$ and $y- \sqrt{ 2}$ are also real numbers. There is a rational number $r$ such that $x – \sqrt{2} < r < y- \sqrt{ 2} $ . Adding $\sqrt{2}$ to both sides of these equations, and we have $r + \sqrt{ 2 }$. We know that $r$ is rational and $\sqrt{2}$ is irrational Therefore $r + \sqrt{ 2 }$ is an irrational number between two arbitrary real numbers, and hence the claim.
Will the proof done with rationals follow the same steps, considering that rational numbers should be written in the form $p/q$ where $p, q \in\mathbb{Z}$ instead of $x$ and $y$ which can be mistaken for real numbers?
Best Answer
Write $x=\dfrac{p}{q}$ and $y=\dfrac{m}{n}$ Here ($p,q)=1$ and $(m,n)=1$
$x^2= \dfrac{p^2}{q^2}$
$y^2= \dfrac{m^2}{n^2}$
Now can you find rational numbers between them ? You can infinitely many rational numbers between $x^2$ and $y^2$ of the form $\dfrac{p_i}{q_i}$ where $(p_i,q_i)=1$
Now $ ( \dfrac{p_i}{q_i})^\frac{1}{2}$ is a irrational.(Why?), these lie between $x$ and $y$.