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#