How to show that the product of two irrational numbers may be irrational? 
Show that the product of two irrational numbers may be irrational. You may use any facts you know about the real numbers.

All we know is that $\sqrt{2}$ is irrational and that $\sqrt{2}\cdot \sqrt{2} = 2$; but this  is a rational product of irrational numbers. 
 A: Well, if all you know is that $\sqrt{2}$ is irrational, try the pair of $\sqrt{2}$ and $\sqrt{2}+1$ - both of which are clearly irrational, and their product is $2+\sqrt{2}$, which is also clearly irrational. Then we don't have to know anything other than that $\sqrt{2}$ is irrational and an irrational plus a rational is still irrational.
A: What about $\sqrt{2}\times \sqrt{3}$?
A: Another way to tackle this is to prove that if $n$ is irrational, so is $\sqrt{n}$. (This is straightforward from the definition of rationality.) Then it's easy to see that for irrational $n$, $$\sqrt{n} \cdot \sqrt{n} = n$$ is an irrational product of irrational numbers. 
A: There are uncountably many points on the hyperbola $xy=\sqrt2$, but only countably many with rational $x$-coordinate, and only countably many with rational $y$-coordinate.
A: $(\sqrt 2 + 1)^2=2+2\sqrt 2 +1=3+2\sqrt 2$ is irrational.
A: *

*Let $x$ be irrational with $x>0.$ Let $y=\sqrt x\,.$ Since $y\in \Bbb Q\implies x=y^2\in \Bbb Q,$ it cannot be that $y\in \Bbb Q.$ So with $z=y$ we have $y,z\not \in \Bbb Q$ and $yz=x\not \in \Bbb Q.$

*If you want $y',z'\not \in \Bbb Q$ and $y'z' \not \in \Bbb Q$ with $y'\ne z',$ take $x,y,z$ as in 1. and let $x'=2x,\,y'=y=\sqrt x, \,$ and $z'=2z=2\sqrt x\,. $
2'. Also, with $0<x\not \in \Bbb Q,$ let $y''=x,\, z''=1/\sqrt x,\,$ and $x''=y''z''=\sqrt x.$
