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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