How do you simplify $sqrt(2/3)$ ?

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Answer 1 · 2017-04-21T05:19:41

$sqrt(2/3)=sqrt2/sqrt3(sqrt3/sqrt3)=(sqrt2sqrt3)/(sqrt3sqrt3)=sqrt6/3$

$sqrt(2/3)=sqrt2/sqrt3$

and now we can see that there is a square root in the denominator that doesn't belong. So let's get rid of that.

$sqrt2/sqrt3(sqrt3/sqrt3)=(sqrt2sqrt3)/(sqrt3sqrt3)=sqrt6/3$

Answer 2 · 2017-04-21T06:08:33

$sqrt(2/3) = sqrt(6)/3$

Note that if $a >= 0$ and $b > 0$ then:

$sqrt(a/b) = sqrt(a)/sqrt(b)$

Also if $a >= 0$ then:

$sqrt(a^2) = a$

Instead of breaking up the square root then rationalising the denominator, we can make the denominator square first as follows:

$sqrt(2/3) = sqrt((2*3)/(3*3)) = sqrt(6/3^2) = sqrt(6)/sqrt(3^2) = sqrt(6)/3$

How do you simplify sqrt(2/3)sqrt(2/3)?