Tag: rational and irrational numbers

Questions Related to rational and irrational numbers

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

Which of the following is smallest?

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

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

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

  3. $ \displaystyle \sqrt{4} $

  4. $ \displaystyle \sqrt{3} $

Show answer
Correct Option: B
Explanation:

Since, fifth root of four is smallest.

Option $B$ is correct.

Which one of the following is the smallest surd?

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

  1. $\sqrt 4$

  2. $\sqrt {27}$

  3. $\sqrt 9$

  4. $\sqrt 5$

Show answer
Correct Option: A
Explanation:
As we know, $4 < 5 < 9 < 27$
So, $\sqrt 4$ < $\sqrt 5$ < $\sqrt 9$ < $\sqrt {27}$
Hence, the answer is $\sqrt{4}$.

If $p=\sqrt{32}-\sqrt{24}$ and $q=\sqrt{50}-\sqrt{48}$, then:

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

  1. $p< q$

  2. $p> q$

  3. $p=q$

  4. $p\le q$

Show answer
Correct Option: B
Explanation:

$p= \sqrt{32} - \sqrt{24}$

$p= 5.656-4.898 = 0.758$

$q = \sqrt{50} + \sqrt{48}$

$q= 7.071- 6.928 = 0.089$

$\therefore p> q$

If $x=\sqrt{2}+1, y=\sqrt{17}-\sqrt{2}$, then .............

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

  1. $x< y$

  2. $x>y$

  3. $x=y$

  4. $x\ge y$

Show answer
Correct Option: A
Explanation:

$\sqrt{2} =1.414$

$\sqrt{17} =4.123$

Hence,

$x = \sqrt{2}+1 =1.414+1 = 2.414$

$ y= \sqrt{17} -\sqrt{2} =4.123- 1.414 = 2.709$

Hence , $x<y$

$\sqrt{11}-\sqrt{10}$ $\Box$ $ \sqrt{12}-\sqrt{11}$

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

  1. $<$

  2. $>$

  3. $=$

  4. cannot be determined

Show answer
Correct Option: B
Explanation:

$\sqrt{11} =3.316$

$\sqrt{10} =3.162$

$\sqrt{12} =3.464$

Hence,

$\sqrt{11} -\sqrt{10} =3.316-3.162 = 0.154$

$\sqrt{12} -\sqrt{11} =3.464-3.316 = 0.148$

Hence , $\sqrt{11} -\sqrt{10} >\sqrt{12} -\sqrt{11}$

Which among the following numbers is the greatest?
$\displaystyle \sqrt[3]{4},\sqrt{2},\sqrt[6]{13},\sqrt[4]{5}$

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

  1. $\displaystyle \sqrt[3]{4}$ is the greatest

  2. $\sqrt{2}$ is the greatest

  3. $\sqrt[6]{13}$ is the greatest

  4. $\sqrt[4]{5}$ is the greatest

Show answer
Correct Option: A
Explanation:

LCM of $3, 6, 4 = 12$

So, raising each given number to power $12$.
$\Rightarrow \sqrt[3]{4}=(4)^{1/3}=(4^{1/3})^{12}=4^{4}=256$
$\Rightarrow \sqrt{2}=(2)^{1/2}=(2^{1/2})^{12}=2^{6}=64$
$\Rightarrow \sqrt[6]{13}=(13)^{1/6}=(13^{1/6})^{12}=13^{2}=169$
$\Rightarrow \sqrt[4]{5}=(5)^{1/4}=(15^{1/4})^{12}=5^{3}=125$

$\therefore \sqrt[3]{4}$ is the greatest.

If $x=\sqrt{7}-\sqrt{5}, y=\sqrt{13}-\sqrt{11}$, then:

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

  1. $x=y$

  2. $x> y$

  3. $x< y$

  4. $x\ge y$

Show answer
Correct Option: B
Explanation:

$x = \sqrt{7}-\sqrt{5}$

$= 2.64 – 2.23= 0.41$

$y=  \sqrt{13}-\sqrt{11}$

$= 3.60- 3.31 = 0.29$

$Hence, x> y$