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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?

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  • $\begingroup$ Rational numbers are real numbers. $\endgroup$ – Jonas Meyer Mar 29 '13 at 6:18
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    $\begingroup$ Rational numbers are in particular real numbers. So nothing need change. $\endgroup$ – André Nicolas Mar 29 '13 at 6:20
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    $\begingroup$ Odd phrasing; the irrationals aren't even in the rationals, so how could they be dense in the rationals? Fortunately, the question clarifies what is meant. $\endgroup$ – Hurkyl Mar 29 '13 at 7:14
  • $\begingroup$ Is there another way for proof like using fractions instead of assuming that rational numbers as real numbers. $\endgroup$ – Avinesh Mar 29 '13 at 8:13
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    $\begingroup$ This question is similar to mine. Take a look: math.stackexchange.com/questions/338328/… $\endgroup$ – Q.matin Mar 29 '13 at 8:44
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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 $y$.

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  • $\begingroup$ This what i was looking for. Thanks for your time kind help. $\endgroup$ – Avinesh Mar 29 '13 at 9:03

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