Tag: comparison of irrational numbers

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

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

  1. $\displaystyle \sqrt{7}+\sqrt{10}$
  2. $\displaystyle \sqrt{3}+\sqrt{19}$

  3. Both are equal

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  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. $\displaystyle \sqrt[4]{3},\sqrt[12]{25},\sqrt[6]{10}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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$
$\Rightarrow \sqrt[6]{10},\sqrt[4]{3},\sqrt[12]{25}$

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

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

  1. $0.7+\sqrt { 0.16 } $
  2. $1.02-\displaystyle\frac { 0.6 }{ 24 } $
  3. $1.2\times 0.83$
  4. $\sqrt { 1.44 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
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 } $.

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

Which of the following is smallest?

  1. $\sqrt [4]{5}$
  2. $\sqrt [5]{4}$
  3. $\sqrt {4}$
  4. $\sqrt {3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\sqrt [ 5 ]{ 4 }$.
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

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

  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 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $2\sqrt[6]{3}, \sqrt[3]{11},\sqrt{5}$

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\displaystyle \sqrt[3]{4}$

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

Which is greater?
${ \left( \cfrac { 1 }{ 2 }  \right)  }^{ 1/2 } $ or ${ \left( \cfrac { 2 }{ 3 }  \right)  }^{ 1/3 } $

  1. ${ \left( \cfrac { 2 }{ 3 } \right) }^{ 1/3 } $
  2. ${ \left( \cfrac { 1 }{ 2 } \right) }^{ 1/2 } $

  3. Both are equal

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ \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}$

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

The correct descending order of the following surds is 

$ \sqrt [3]{2}$, $\sqrt 3$, $\sqrt 4$, $\sqrt 5$

  1. $\sqrt 3$ > $\sqrt 4$ > $\sqrt 5$ > $\sqrt [3]{2}$
  2. $\sqrt 4$ > $\sqrt 5$ > $\sqrt [3]{2}$ > $\sqrt 3$
  3. $\sqrt 5$ > $\sqrt 4$ > $\sqrt 3$ > $\sqrt [3]{2}$
  4. $\sqrt [3]{2}$ > $\sqrt 4$ > $\sqrt 3$ > $\sqrt 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

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

$\therefore 125 > 64 > 27 > 4$

So, option $C$ is correct.