Tag: rational and irrational numbers

Compare (sqrt(7)-sqrt(3)) and (sqrt(5)-1). (sqrt(7)-sqrt(3)) = 4/(sqrt(7)+sqrt(3)) approx 4/(2.64+1.73) approx 0.91. (sqrt(5)-1) approx 2.23-1 = 1.23. Thus, the statement is false.

$ \left(\dfrac{1}{2}\right)^{1/2} \; or \; \left(\dfrac{2}{3}\right)^{1/3}$

$= \left(\left(\dfrac{1}{2}\right)^{1/2}\right)^6 \; or \; \left(\left(\dfrac{2}{3}\right)^{1/3}\right)^6$

$= \left(\dfrac{1}{2}\right)^3 \; or \; \left(\dfrac{2}{3}\right)^2$

$= \left(\dfrac{1}{8}\right) \; or \; \left(\dfrac{4}{9}\right)$

= 0.125 or 0.44

Since, 0.44 is greater and so is $ \left(\dfrac{2}{3}\right)^{1/3}$

LCM of $3$ and $2$ is $6$.
Therefore, multiplying the index of all numbers by $6$.
${\sqrt [3]{2}}^6 = 2^\dfrac 63 = 2^2 = 4$

${\sqrt 3}^6 = 3^\dfrac 62 = 3^3 = 27$

${\sqrt 4}^6 = 4^ \dfrac 62 = 4^3 = 64$

${\sqrt 5}^6 = 5^{\dfrac 62} =  5^3 = 125$

$\therefore 125 > 64 > 27 > 4$

Option A: $\sqrt [ 3 ]{ -27 } ={ (-3) }^{ 3\times \frac { 1 }{ 3 }  }=-3$
Option B: $\sqrt { 2 } (3\sqrt { 2 } +2\sqrt { 8 } )=\sqrt { 2 } (3\sqrt { 2 } +4\sqrt { 2 } )=\sqrt { 2 } (7\sqrt { 2 } )=14$
Option C: $\dfrac { 3\sqrt { 18 }  }{ 2\sqrt { 6 }  } =\dfrac { 3\sqrt { 3 }  }{ 2 } $
Option D: $\sqrt { \dfrac { 1 }{ 2 }  } \sqrt { \dfrac { 25 }{ 2 }  } =\dfrac { 5 }{ 2 } $
Option E: $\dfrac { 2\sqrt { 5 }  }{ \sqrt { 45 }  } =\dfrac { 2\sqrt { 5 }  }{ 3\sqrt { 5 }  } =\dfrac { 2 }{ 3 } $
Therefore, all are rational except option $C$.