Proof that the set of irrational numbers is dense in reals I'm being asked to prove that the set of irrational number is dense in the real numbers.  While I do understand the general idea of the proof: 

Given an interval $(x,y)$, choose a positive rational number (say) $z=\sqrt{2}$.  By density of rationals, there exists a rational number $p$ in the interval $(x/z, y/z)$, which essentially means that $$\frac{x}{\sqrt{2}} &lt; p &lt; \frac{y}{\sqrt{2}}.$$  I find that $pz$ is irrational, since it is the product of a rational and irrational number. However, my instructions besides writing that proof down is to specifically verify that $y=xz$ is irrational.  What does this have to do with anything and how does it prove denseness of irrationals?  Regardless, assume that $x$ is a nonzero rational number and that $z$ is irrational. For the sake of contradiction, assume that $y=xz$ is rational. This should mean that $y/x$ rational as well, and therefore that $z$ is rational, a contradiction.
Another argument:

$\mathbb{Q}$ is dense in $\mathbb{R}$, so $\mathbb{Q} + \sqrt{2}$ is dense in $\mathbb{R} + \sqrt{2} = \mathbb{R}$. Since $\mathbb{Q} + \sqrt{2}$ is a subset of the irrationals, we conclude that the irrationals are also dense in $\mathbb{R}$.
A set $A\subset \mathbb R$ is dense in $\mathbb R$ if $Cl(A)=\mathbb R$, where $Cl(A)=A\cup der(A)$ is closure of $A$.