Which is larget √2 or $\sqrt[3]{3}$?
√2
$\sqrt[3]{3}$
Both are equal
None of these
Explanation:
Given surds are of order 2 and 3, Their LCM is 6.
Changing each to a surd of order 6, we get :
√2=$2^{\dfrac{1}{2}}$
=$2^{\dfrac{1}{2}\times\dfrac{3}{3} }$
=$2^{\dfrac{3}{6}}$
=$(2^{3})^{\dfrac{1}{6}}$
=$(8)^{\dfrac{1}{6}}$
=$\sqrt[6]{8}$
$\sqrt[3]{3}=3^{\dfrac{1}{3}}$
=$3^{\dfrac{1}{3}\times\dfrac{2}{2} }$
=$3^{\dfrac{2}{6}}$
=$(3^{2})^{\dfrac{1}{6}}$
=$(9)^{\dfrac{1}{6}}$
=$\sqrt[6]{9}$
Clearly, $\sqrt[6]{9}$ > $\sqrt[6]{8}$ and hence $\sqrt[3]{3}$ > √2
Given surds are of order 2 and 3, Their LCM is 6.
Changing each to a surd of order 6, we get :
√2=$2^{\dfrac{1}{2}}$
=$2^{\dfrac{1}{2}\times\dfrac{3}{3} }$
=$2^{\dfrac{3}{6}}$
=$(2^{3})^{\dfrac{1}{6}}$
=$(8)^{\dfrac{1}{6}}$
=$\sqrt[6]{8}$
$\sqrt[3]{3}=3^{\dfrac{1}{3}}$
=$3^{\dfrac{1}{3}\times\dfrac{2}{2} }$
=$3^{\dfrac{2}{6}}$
=$(3^{2})^{\dfrac{1}{6}}$
=$(9)^{\dfrac{1}{6}}$
=$\sqrt[6]{9}$
Clearly, $\sqrt[6]{9}$ > $\sqrt[6]{8}$ and hence $\sqrt[3]{3}$ > √2