Tag: comparison of irrational numbers

Questions Related to comparison of irrational numbers

Compare the following pairs of surds. $\sqrt[8]{80}, \sqrt[4]{40}$    

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $\sqrt[8]{80} < \sqrt[4]{40}$

  2. $\sqrt[8]{80} \neq \sqrt[4]{40}$

  3. $\sqrt[8]{80} = \sqrt[4]{40}$

  4. $\sqrt[8]{80} > \sqrt[4]{40}$

Show answer
Correct Option: A
Explanation:

   $\sqrt[8]{80}, \sqrt[4]{40}$
$={80}^{\frac{1}{8}}, {40}^{\frac{1}{4}}$
$={80}^{\frac{1}{8}}, {40}^{\frac{2}{8}}$
$={80}^{\frac{1}{8}}, {1600}^{\frac{1}{8}}$
Now,
   $80<1600$
$=>{80}^{\frac{1}{8}}<{1600}^{\frac{1}{8}}$
$=>\sqrt[8]{80}< \sqrt[4]{40}$

Which among the following numbers is the greatest?
$\displaystyle 0.07+\sqrt{0.16},\sqrt{1.44},1.2\times 0.83,1.02-\frac{0.6}{24}$

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $\displaystyle \sqrt {1.44}$

  2. $\displaystyle 0.07+\sqrt{0.16}$

  3. $1.2\times 0.83$

  4. $1.02-\dfrac{0.6}{24}$

Show answer
Correct Option: A
Explanation:
$\Rightarrow 0.07+\sqrt{0.16}=0.07+0.4=0.47$
$\Rightarrow \sqrt{1.44}=1.2$
$\Rightarrow 1.2 \times 0.83 = 0.996$
$\Rightarrow 1.02-\cfrac{0.6}{24}=1.02-\cfrac {6}{240}=1.02-0.025=0.995$
$ \therefore$ The greatest number is $1.2$ i.e. $\sqrt {1.44}$

State whether the following equality is true or false:

$\displaystyle \frac{2\sqrt{3}}{\sqrt{5}} = $$\displaystyle \frac{2\sqrt{15}}{\sqrt{5}}$

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

Rationalizing factor of  $\displaystyle \frac{2\sqrt{3}}{\sqrt{5}}$ is $\sqrt{5}$


$\therefore \displaystyle \frac{2\sqrt{3}}{\sqrt{5}}$  $=\displaystyle \frac{2\sqrt{3}\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$

$= \dfrac{2\sqrt{15}}{5}$

Determine the order relation between the following pairs of ratios.

$\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}}, \frac{2\sqrt{2}}{3\sqrt{3}}$

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}} > \frac{2\sqrt{2}}{3\sqrt{3}}$

  2. $\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}} < \frac{2\sqrt{2}}{3\sqrt{3}}$

  3. Cannot be determined

  4. None of These

Show answer
Correct Option: A
Explanation:

$\dfrac{3\sqrt{3}}{2\sqrt{2}}=\dfrac{3\times 1.73214}{2\times 1.41429} = \dfrac{5.19642}{2.85828}
=1.82151$
$\dfrac{2\sqrt{2}}{3\sqrt{3}}=\dfrac{2\times 1.41429}{3\times 1.73214}=\dfrac{2.85828}{5.19642}=0.55004$
$\therefore \dfrac{3\sqrt{3}}{2\sqrt{2}} >\dfrac{2\sqrt{2}}{3\sqrt{3}}$

Compare the following pairs of surds $\sqrt[8]{12}, \sqrt[4]{6}$

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $\sqrt[8]{2} < \sqrt[4]{6}$

  2. $\sqrt[8]{8} < \sqrt[4]{6}$

  3. $\sqrt[8]{12} < \sqrt[4]{6}$

  4. $\sqrt[8]{12} < \sqrt[4]{4}$

Show answer
Correct Option: C

Compare the following pair of surds:

$\sqrt[3]{6}, \sqrt[4]{8}$

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $\sqrt[3]{6} > \sqrt[4]{8}$

  2. $\sqrt[3]{6} > \sqrt[4]{4}$

  3. $\sqrt[3]{6} > \sqrt[3]{8}$

  4. $\sqrt[3]{4} > \sqrt[4]{8}$

Show answer
Correct Option: A

Arrange the following in ascending order of magnitude: 

$\displaystyle \sqrt[3]{4}, \sqrt[4]{5}, \sqrt{3}$ 

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $\displaystyle \sqrt[4]{5} < \sqrt[3]{4} < \sqrt{3}$

  2. $\displaystyle \sqrt[4]{5} > \sqrt[3]{4} > \sqrt{3}$

  3. $\displaystyle \sqrt[4]{5} > \sqrt[3]{4} < \sqrt{3}$

  4. $\displaystyle \sqrt[4]{5} < \sqrt[3]{4} > \sqrt{3}$

Show answer
Correct Option: A

What is the least value of $a$ in $ \displaystyle\frac{\sqrt 2+\sqrt 3}{\sqrt{2+3}} < a$?

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: B
Explanation:
$\dfrac { \sqrt { 2 } +\sqrt { 3 }  }{ \sqrt { 2+3 }  } =\dfrac { \sqrt { 2 } +\sqrt { 3 }  }{ \sqrt { 5 }  } =\dfrac { (\sqrt { 2 } +\sqrt { 3 } )\times \sqrt { 5 }  }{ 5 } =\dfrac { 7.02 }{ 5 } \\ =1.40$
$\Rightarrow 1.40<a$
So, least integer value of $a$ is $2$.
Hence, option B is correct.

Which of the following is the greatest?

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $\sqrt 2$

  2. $\sqrt 3$

  3. $\sqrt 4$

  4. $\sqrt 5$

Show answer
Correct Option: D
Explanation:

$\sqrt 5$ > $\sqrt 4$ > $\sqrt 3$ > $\sqrt 2$


So, option D is correct.

The greatest among $\displaystyle \sqrt[6]{3}$, $\displaystyle \sqrt{2}$, $\displaystyle \sqrt[3]{4}$, $\displaystyle \sqrt[4]{5}$ is--

comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

  1. $\displaystyle \sqrt[6]{3}$

  2. $\displaystyle \sqrt{2}$

  3. $\displaystyle \sqrt[3]{4}$

  4. $\displaystyle \sqrt[4]{5}$

Show answer
Correct Option: C
Explanation:

$\displaystyle \therefore $ $\displaystyle \sqrt[6]{3}$ = $\displaystyle \left ( 3 \right )^{\dfrac{1}{6}}$ = $\displaystyle \left ( 3^{2} \right )^{\dfrac{1}{12}}$ = $\displaystyle \left ( 9 \right )^{\dfrac{1}{12}}$
$\displaystyle \sqrt{2}$ = $\displaystyle \left ( 2 \right )^{\dfrac{1}{2}}$ = $\displaystyle \left ( 2^{6} \right )^{\dfrac{1}{12}}$ = $\displaystyle \left ( 64 \right )^{\dfrac{1}{12}}$
$\displaystyle \sqrt[3]{4}$ = $\displaystyle \left ( 4 \right )^{\dfrac{1}{3}}$ = $\displaystyle \left ( 4^{4} \right )^{\dfrac{1}{12}}$ = $\displaystyle \left ( 256 \right )^{\dfrac{1}{12}}$

$\sqrt[4]{5}=(5)^{\dfrac{1}{4}}=(5^{3})^{\dfrac{1}{12}}=(125)^{\dfrac{1}{12}}$
$\displaystyle \therefore $ The greatest number is $\displaystyle \left ( 256 \right )^{\dfrac{1}{12}}$ = $\displaystyle \sqrt[3]{4}$