present that the product of 2 irrational numbers may be irrational. You might use any kind of facts friend know around the genuine numbers.

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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