You are watching: Is the product of two irrational numbers always irrational
All we recognize is that $sqrt2$ is irrational and that $sqrt2cdot sqrt2 = 2$; but this is a rational product of irrational numbers.
Well, if all you understand is that $sqrt2$ is irrational, attempt the pair of $sqrt2$ and $sqrt2+1$ - both of which are plainly irrational, and also their product is $2+sqrt2$, which is also clearly irrational. Then we do not need to recognize anypoint various other than that $sqrt2$ is irrational and also an irrational plus a rational is still irrational.
Another means to tackle this is to prove that if $n$ is irrational, so is $sqrtn$. (This is straightforward from the interpretation of rationality.) Then it"s basic to see that for irrational $n$, $$sqrtn cdot sqrtn = n$$ is an irrational product of irrational numbers.
Tright here are uncountably many kind of points on the hyperbola $xy=sqrt2$, however only countably many type of through rational $x$-coordinate, and also just countably many kind of via rational $y$-coordinate.
Let $x$ be irrational via $x>0.$ Let $y=sqrt x,.$ Due to the fact that $yin Bbb Qmeans x=y^2in Bbb Q,$ it cannot be that $yin Bbb Q.$ So through $z=y$ we have $y,z ot in Bbb Q$ and $yz=x ot in Bbb Q.$
If you desire $y",z" ot in Bbb Q$ and $y"z" ot in Bbb Q$ via $y" e z",$ take $x,y,z$ as in 1. and let $x"=2x,,y"=y=sqrt x, ,$ and also $z"=2z=2sqrt x,. $
2". Also, through $0 let $y""=x,, z""=1/sqrt x,,$ and also $x""=y""z""=sqrt x.$
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How would certainly I go about proving that for any type of two genuine numbers $a,binrianowens.tvbbR$ we might find a both a rational and also irrational number between them.
Please aid me spot the error in my "proof" that the amount of 2 irrational numbers need to be irrational
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