Tag: rational and irrational numbers

Questions Related to rational and irrational numbers

Which one of the following set of surds is correct sequence of ascending order of their values?

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

  1. $\displaystyle \sqrt[4]{10},\sqrt[3]{6},\sqrt{3}$

  2. $\displaystyle \sqrt{3},\sqrt[4]{10},\sqrt[3]{6},$

  3. $\displaystyle \sqrt{3},\sqrt{10},\sqrt[3]{6},$

  4. $\displaystyle \sqrt[4]{10},\sqrt{3},\sqrt[3]{6}$

Show answer
Correct Option: B
Explanation:

$\sqrt[4]{10},\sqrt[3]{6},\sqrt{3}$
The order of the given irrational numbers are 2,3,4.
LCM of (2,3,4)=12
Now convert each irrational number as of order 12
$\sqrt[4]{10}=\sqrt[12]{10^3}=\sqrt[12]{1000}$
$\sqrt[3]{6}=\sqrt[12]{6^4}=\sqrt[12]{1296}$
$\sqrt{3}=\sqrt[12]{3^6}=\sqrt[12]{729}$
Hence, ascending order$\sqrt{3}<\sqrt[4]{10}<\sqrt[3]{6}$

Which is the greatest out of the following ?

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

  1. $\displaystyle \sqrt[3]{1.728}$

  2. $\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1}$

  3. $\displaystyle \left ( \frac{1}{2} \right )^{-2}$

  4. $\displaystyle \frac{17}{8}$

Show answer
Correct Option: C
Explanation:
$\Rightarrow  \sqrt[3]{1.728}=1.2$

$\Rightarrow \cfrac{\sqrt{3}-1}{\sqrt{3}+1}=\cfrac{(\sqrt{3}-1)^{2}}{(\sqrt{3}+1)(\sqrt{3}-1)}=\cfrac{3+1-2\sqrt{3}}{3-1}$
$ =\cfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}$
$ =2-1.732=0.268$

$\Rightarrow \left ( \cfrac{1}{2} \right )^{-2}=2^{2}=4$

$\Rightarrow \cfrac{17}{8}=2.2125$

$ \therefore \left ( \cfrac{1}{2} \right )^{-2}$ is the greatest. 

$4\sqrt{18}$ $=$ $12\sqrt{2}$
State true or false

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Consider $4\sqrt { 18 }$ and factorize it as follows:

 
$4\sqrt { 18 } =4\sqrt { 2\times 3\times 3 } =4\sqrt { 2\times 3^{ 2 } } =4\times 3\sqrt { 2 } =12\sqrt { 2 }$
 
Hence, $4\sqrt { 18 } =12\sqrt { 2 }$    

Which is greater $\displaystyle (\sqrt{7}+\sqrt{10})$ or $\displaystyle (\sqrt{3}+\sqrt{19})$?

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

  1. $\displaystyle \sqrt{7}+\sqrt{10}$

  2. $\displaystyle \sqrt{3}+\sqrt{19}$

  3. Both are equal

  4. None of these

Show answer
Correct Option: B
Explanation:

$(\sqrt{7}+\sqrt{10})$
$2.6457+3.1622=5.8079$
$(\sqrt{3}+\sqrt{19})$
$1.732+4.358=6.090$
Hence $(\sqrt{3}+\sqrt{19})$is greater.

$\displaystyle \sqrt[4]{3},\sqrt[6]{10},\sqrt[12]{25}$, when arranged in descending order will be 

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

  1. $\displaystyle \sqrt[4]{3},\sqrt[6]{10},\sqrt[12]{25}$

  2. $\displaystyle \sqrt[6]{10},\sqrt[4]{3},\sqrt[12]{25}$

  3. $\displaystyle \sqrt[6]{10},\sqrt[12]{25},\sqrt[4]{3}$

  4. $\displaystyle \sqrt[4]{3},\sqrt[12]{25},\sqrt[6]{10}$

Show answer
Correct Option: B
Explanation:

LCM of $4, 6$ and $12 = 12$.
$ \therefore $ Raising each of the given number to power $12$, we have 
$ (3^{1/4})^{12},(10^{1/6})^{12},(25^{1/12})^{12}$
$= 3^{3},10^{2},25$
$= 27, 100, 25$
Arranging in descending order, the numbers are $ 100, 27, 25$
$\Rightarrow \sqrt[6]{10},\sqrt[4]{3},\sqrt[12]{25}$

The greatest amongst the the values $0.7 + \sqrt { 0.16 } ,  1.02 - \displaystyle\frac { 0.6 }{ 24 } ,   1.2 \times 0.83$ and $\sqrt { 1.44 } $ is

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

  1. $0.7+\sqrt { 0.16 } $

  2. $1.02-\displaystyle\frac { 0.6 }{ 24 } $

  3. $1.2\times 0.83$

  4. $\sqrt { 1.44 } $

Show answer
Correct Option: D
Explanation:

$0.7+ \sqrt { 0.16 } = 0.7 +0.4 = 1.1$ 

$1.02-\displaystyle  \frac { 0.6 }{ 24 } = 1.02 - 0.025 = 1.175$
$1.2 \times 0.83 = 0.996 $
$\sqrt { 1.44 } = 1.2$

$\therefore $ the greatest is $\sqrt { 1.44 } $.

Which of the following is smallest?

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

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

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

  3. $\sqrt {4}$

  4. $\sqrt {3}$

Show answer
Correct Option: B
Explanation:

Let us rewrite the given set of magnitudes $\sqrt [ 4 ]{ 5 } ,\sqrt [ 5 ]{ 4 } ,\sqrt { 4 },\sqrt {3}$ as follows:


$\sqrt [ 4 ]{ 5 } ={ \left( 5 \right)  }^{ \dfrac { 1 }{ 4 }  }\ \sqrt [ 5 ]{ 4 } ={ \left( 4 \right)  }^{ \dfrac { 1 }{ 5 }  }\ \sqrt { 4 } ={ \left( 4 \right)  }^{ \dfrac { 1 }{ 2 }  }\ \sqrt { 3 } ={ \left( 3 \right)  }^{ \dfrac { 1 }{ 2 }  }$

 
We now take the LCM of the denominators of the powers to make the denominators same, then the above magnitudes will be:

$\sqrt [ 4 ]{ 5 } ={ \left( 5 \right)  }^{ \dfrac { 1\times 5 }{ 4\times 5 }  }={ \left( 5 \right)  }^{ \dfrac { 5 }{ 20 }  }={ \left( { 5 }^{ 5 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 3125 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 3125 } \\ \sqrt [ 5 ]{ 4 } ={ \left( 4 \right)  }^{ \dfrac { 1\times 4 }{ 5\times 4 }  }={ \left( 4 \right)  }^{ \dfrac { 4 }{ 20 }  }={ \left( { 4 }^{ 4 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 256 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 256 } \\ \sqrt { 4 } ={ \left( 4 \right)  }^{ \dfrac { 1\times 10 }{ 2\times 10 }  }={ \left( 4 \right)  }^{ \dfrac { 10 }{ 20 }  }={ \left( { 4 }^{ 10 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 1048576 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 1048576 } \\ \sqrt { 3 } ={ \left( 3 \right)  }^{ \dfrac { 1\times 10 }{ 2\times 10 }  }={ \left( 3 \right)  }^{ \dfrac { 10 }{ 20 }  }={ \left( { 3 }^{ 10 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 2187 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 2187 }$     

Now, the descending order is as shown below:

$\sqrt [ 20 ]{ 1048576 } >\sqrt [ 20 ]{ 3125 } >\sqrt [ 20 ]{ 2187 } >\sqrt [ 20 ]{ 256 } \\ \Rightarrow \sqrt { 4 } >\sqrt [ 4 ]{ 5 } >\sqrt { 3 } >\sqrt [ 5 ]{ 4 }$ 

Hence, the smallest magnitude is $\sqrt [ 5 ]{ 4 }$.

Arrange the following surds in ascending order of their magnitudes: $\sqrt{5},\sqrt [ 3 ]{ 11 } ,2\sqrt [ 6 ]{ 3 } $

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

  1. $\sqrt [ 3 ]{ 11 } > \sqrt{5}< 2\sqrt [ 6 ]{ 3 } $

  2. $\sqrt [ 3 ]{ 11 } < \sqrt{5}< 2\sqrt [ 6 ]{ 3 } $

  3. $\sqrt [ 3 ]{ 11 } > \sqrt{5}> 2\sqrt [ 6 ]{ 3 } $

  4. $\sqrt [ 3 ]{ 11 } < \sqrt{5}> 2\sqrt [ 6 ]{ 3 } $

Show answer
Correct Option: B
Explanation:

$\sqrt{5} = 5^{1/2}$

$\sqrt[3]{11} = 11^{1/3}$

$2\sqrt[6]{3} = \sqrt[6]{12}= 12^{1/6}$

L.C.M of the denominators of the exponents is 12.

So,

$\sqrt{5} = 5^{\frac{1}{2}\times\frac{6}{6}} = \sqrt [12]{5^6}=\sqrt[12]{15625}$

$\sqrt[3]{11} = 11^{\frac{1}{3}\times\frac{4}{4}} = \sqrt [12]{11^4} =\sqrt[12]{14641}$

$2\sqrt[6]{3} = \sqrt[6]{12}= 12^{\frac{1}{6}\times\frac{2}{2}} = \sqrt[12]{12^2} =\sqrt[12]{144}$

Hence, the Ascending order is $2\sqrt[6]{3}, \sqrt[3]{11},\sqrt{5}$

Write $\displaystyle \sqrt[4]{6},\sqrt{2},\sqrt[3]{4}$ in ascending order

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

  1. $\displaystyle \sqrt{2},\sqrt[4]{6}$ and $\displaystyle \sqrt[3]{4}$

  2. $\displaystyle \sqrt[4]{6}$, $\sqrt{2}$ and $\displaystyle \sqrt[3]{4}$

  3. $\displaystyle \sqrt{2}$, $\displaystyle \sqrt[3]{4}$ and $\sqrt[4]{6}$

  4. None of these

Show answer
Correct Option: A
Explanation:

$\displaystyle 6^\cfrac 14,2^\cfrac 12,4^\cfrac 13$
The surds are of the order $4, 2$ and $3$ respectively The L.C.M. is $12$ So, we change each surd of the order $12$
The terms now are $\displaystyle \left ( 6^{3} \right )^\cfrac {1}{12},\left ( 2^{6} \right )^\cfrac{1}{12}$ and $\displaystyle \left ( 4^{4} \right )^\cfrac{1}{12}$
$\displaystyle \Rightarrow \left ( 216 \right )^\cfrac{1}{12},\left ( 64 \right )^\cfrac{1}{12}$ and $\displaystyle \left ( 256 \right )^\cfrac{1}{12}$
$\displaystyle \therefore $ Ascending order is $\displaystyle \sqrt{2},\sqrt[4]{6}$ and $\displaystyle \sqrt[3]{4}$

State true or false
$\sqrt { 17 } -\sqrt { 12 } $ is less than $\sqrt { 11 } -\sqrt { 6 } $

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

  1. True

  2. False

Show answer
Correct Option: A