# 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.

• math.stackexchange.com/questions/977493/… Commented Oct 24, 2014 at 5:52
• Commented Jan 11, 2015 at 17:20
• Just try $\sqrt{2}(1+\sqrt{2})$ Commented Feb 22, 2018 at 9:38

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.

What about $\sqrt{2}\times \sqrt{3}$?

• This is a good example. Proving it irrational is similar to proving $\sqrt 2$ irrational, which OP understands. Commented Oct 24, 2014 at 2:24

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.

• I think this is the best answer - it shows that every positive irrational number is the product of two irrational numbers with a very, very simple proof. Commented Oct 25, 2014 at 23:33

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.

• @user2357112 take any positive real number $x$. Then $(x, y) = (x, \frac{\sqrt{2}}{x})$ is a point on the hyperbola as $x\frac{\sqrt{2}}{x} = \sqrt{2}$. Since there are uncountably many possible values of $x$, and each leads to a unique point, there must be uncountably many points. Commented Oct 25, 2014 at 9:51
• -1 because you obviously can't assume that. If we know the reals are uncountable, we may as well just use the fact that multiplication by nonzero numbers (like root 2) is injective and therefore can't land in the rationals all the time, since I'm sure proving they are countable was easier than proving that the reals are uncountable. Voila! Commented Oct 26, 2014 at 5:24

$(\sqrt 2 + 1)^2=2+2\sqrt 2 +1=3+2\sqrt 2$ is irrational.

1. 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 $$yz=x\not \in \Bbb Q.$$

2. 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 $$0 let $$y''=x,\, z''=1/\sqrt x,\,$$ and $$x''=y''z''=\sqrt x.$$