$\begingroup$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.
$\endgroup$36 Answers
$\begingroup$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.
$\endgroup$0$\begingroup$What about $\sqrt{2}\times \sqrt{3}$?
$\endgroup$1$\begingroup$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.
$\endgroup$1$\begingroup$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.
$\endgroup$2$\begingroup$$(\sqrt 2 + 1)^2=2+2\sqrt 2 +1=3+2\sqrt 2$ is irrational.
$\endgroup$$\begingroup$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 $x''=y''z''=\sqrt x.$
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