Is The Product Of Two Irrational Numbers Always Irrational Numbers (Video)

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

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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.



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.



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.


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.

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

If you want $y”,z”
ot in Bbb Q$ and $y”z”
ot in Bbb Q$ with $y”
e z”,$ take $x,y,z$ as in 1. and let $x”=2x,,y”=y=sqrt x, ,$ and $z”=2z=2sqrt x,. $

2″. Also, with $0 let $y””=x,, z””=1/sqrt x,,$ and $x””=y””z””=sqrt x.$

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