You are watching: Can the product of two irrational numbers be rational

All we recognize is that $sqrt2$ is irrational and also that $sqrt2cdot sqrt2 = 2$; however this is a *rational* product that irrational numbers.

Well, if all you recognize is the $sqrt2$ is irrational, shot the pair the $sqrt2$ and $sqrt2+1$ - both of i m sorry are clearly irrational, and their product is $2+sqrt2$, i m sorry is also clearly irrational. Then us don"t have to know anything other than the $sqrt2$ is irrational and an irrational to add a reasonable is quiet irrational.

Another method to handle this is to prove that if $n$ is irrational, so is $sqrtn$. (This is straightforward from the definition of rationality.) then it"s straightforward to watch that because that irrational $n$, $$sqrtn cdot sqrtn = n$$ is an irrational product that irrational numbers.

There are uncountably plenty of points on the hyperbola $xy=sqrt2$, yet only countably countless with rational $x$-coordinate, and only countably many with reasonable $y$-coordinate.

Let $x$ be irrational with $x>0.$ let $y=sqrt x,.$ since $yin Bbb Qimplies x=y^2in Bbb Q,$ it can not be that $yin Bbb Q.$ So v $z=y$ we have actually $y,z ot in Bbb Q$ and $yz=x ot in Bbb Q.$

If you desire $y",z" ot in Bbb Q$ and also $y"z" ot in Bbb Q$ v $y" e z",$ take it $x,y,z$ as in 1. And also let $x"=2x,,y"=y=sqrt x, ,$ and $z"=2z=2sqrt x,. $

2". Also, through $0 allow $y""=x,, z""=1/sqrt x,,$ and $x""=y""z""=sqrt x.$

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just how would i go about proving the for any two genuine numbers $a,binjulianum.netbbR$ us may find a both a rational and irrational number in between them.

Please aid me spot the error in my "proof" the the amount of 2 irrational numbers need to be irrational

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