Tag: rational and irrational numbers

Questions Related to rational and irrational numbers

The smallest of $\sqrt[3]{4},    \sqrt[4]{5},     \sqrt[4]{6},    \sqrt[3]{8}$ is:

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

  1. $\sqrt[3]{8}$

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

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

  4. $\sqrt[4]{6}$

Show answer
Correct Option: B
Explanation:

(B) $\sqrt[3]{4}, \sqrt[4]{5},  \sqrt[4]{6}, \sqrt[3]{8}$

$=4^{1/3}, 5^{1/4}, 6^{1/4}, 8^{1/3}$

L.C.M of 3 & 4 $=12$

So, the given surds can be written as,

$=4^{4/12}, 5^{3/12}, 6^{3/12}, 8^{4/12}$

$=(4^{4})^{1/12}, (5^{3})^{1/12}, (6^{3})^{1/12}, (8^{4})^{1/12}$

$=(256)^{1/12}, (125)^{1/12}, (216)^{1/12}, (4096)^{1/12}$

$\therefore $ The smallest one is $\sqrt[4]{5}$

If $a = \sqrt {15} + \sqrt {11}, b = \sqrt {14} + \sqrt {12}$ then

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

  1. $a > b$

  2. $a < b$

  3. $a = b$

  4. None

Show answer
Correct Option: B

$\sqrt{11}-\sqrt{10} .... \sqrt{12}-\sqrt{11}$,use appropriate inequality to fill the gap.

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

  1. <

  2. >

  3. $=$

  4. cannot determined

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Correct Option: B
Explanation:

We first consider $\sqrt { 11 } -\sqrt { 10 }$ as follows:


$\sqrt { 11 } -\sqrt { 10 } =3.317-3.162=0.156$

Now we find the value of $\sqrt { 12 } -\sqrt { 11 }$ as follows:

$\sqrt { 12 } -\sqrt { 11 } =3.464-3.317=0.147$


Since $0.156>0.147$

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

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

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. $p\leq q$

Show answer
Correct Option: B
Explanation:

(B) $\frac{p}{q}=\frac{\sqrt{32}-\sqrt{24}}{\sqrt{50}-\sqrt{48}}\times \frac{\sqrt{50}+\sqrt{48}}{\sqrt{50}+\sqrt{48}}$

$=\frac{(4\sqrt{2}-2\sqrt{6})(5\sqrt{2}+4\sqrt{3})}{2}$

$=(2\sqrt{2}-\sqrt{6})(5\sqrt{2}+4\sqrt{3})> 1$

$\therefore p> q$

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. $x\geq y$

Show answer
Correct Option: A
Explanation:

(A) Given, $x=\sqrt{2}+1$ and $y=\sqrt{17}-\sqrt{2}$

$\dfrac{x}{y}=\dfrac{\sqrt{2}+1}{\sqrt{17}-\sqrt{2}}\times \dfrac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}$

$=\dfrac{(\sqrt{2}+1)(\sqrt{17}+\sqrt{2})}{17-2}$

$=\dfrac{(\sqrt{2}+1)(\sqrt{17}+\sqrt{2})}{15}< 1$

$\therefore x< y$

Arrange the following in ascending order of magnitude: $\displaystyle \sqrt[4]{90}, \sqrt[3]{10}, \sqrt{6}$

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

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

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

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

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

Show answer
Correct Option: A

$if\,A\, = \sqrt 7  - \sqrt 6 \,and\,B = \,\sqrt 6  - \sqrt {5,} \,then\,$

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

  1. $A > B$

  2. $A = B$

  3. $A < B\,$

  4. $A \geqslant B$

Show answer
Correct Option: C
Explanation:
Given numbers are $A=\sqrt{7}-\sqrt{6}$ and $B=\sqrt{6}- \sqrt{5}$
Let $x =\sqrt{5}$ and $y= \sqrt{7}$ then by
$A.M$ and $G.M$
$\dfrac{x+y}{2} \le \sqrt{\dfrac{x^{2}+y^{2}}{2}}$
$\Rightarrow \dfrac{\sqrt{5}+ \sqrt{7}}{2} \le \sqrt{6}$
$\Rightarrow \sqrt{5}+ \sqrt{7} \le 2 \sqrt{6}$
$\Rightarrow \sqrt{7}- \sqrt{6} \le \sqrt{6} - \sqrt{5}$

Which of the following numbers is the least ?
$\displaystyle (0.5)^{2},\sqrt{0.49},\sqrt[3]{0.008},0.23$

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

  1. $\displaystyle (0.5)^{2}$

  2. $\displaystyle \sqrt{0.49}$

  3. $\displaystyle \sqrt[3]{0.008}$

  4. 0.23

Show answer
Correct Option: C
Explanation:

$ (0.5)^{2}=0.25$
$\sqrt{0.49}=0.7;$
$ \sqrt[3]{0.008}=\sqrt[3]{.2^3}=0.2$
$0.23$
Arranging in ascending order the numbers are $0.2< 0.23< 0.25< 0.7$
$ \therefore \sqrt[3]{0.008}=0.2$ is the least

The greatest number among $\displaystyle \sqrt[3]{2},\sqrt{3},\sqrt[3]{5}$ and $1.5$ is 

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

  1. $\displaystyle \sqrt[3]{2}$

  2. $\displaystyle \sqrt{3}$

  3. $\displaystyle \sqrt[3]{5}$

  4. $1.5$

Show answer
Correct Option: B
Explanation:

LCM of $3, 2 = 6$
Given numbers are $ \sqrt[3]{2},\sqrt{3},\sqrt[3]{5}, 1.5$ i.e,
$ 2^{1/3},3^{1/2},5^{1/3},1.5$
$ \therefore $ Raising each number to power $6$, we get
$ (2^{1/3})^{6},(3^{1/2})^{6},(5^{1/3})^{6}, (1.5)^{6}$

$= 2^{2},3^{3},5^{2}, \left(\cfrac{3}{2}\right)^{6}$
$=4,27,25,\cfrac{729}{64}$
Of all these numbers, $27$ is the greatest.
$ \Rightarrow \sqrt{3}$ is the greatest. 

The smallest of $\displaystyle \sqrt{8}+\sqrt{5},\sqrt{7}+\sqrt{6},\sqrt{10}+\sqrt{3}$ and $\displaystyle \sqrt{11}+\sqrt{2}$ is 

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

  1. $\displaystyle \sqrt{8}+\sqrt{5}$

  2. $\displaystyle \sqrt{7}+\sqrt{6}$

  3. $\displaystyle \sqrt{10}+\sqrt{3}$

  4. $\displaystyle \sqrt{11}+\sqrt{2}$

Show answer
Correct Option: D
Explanation:

$\displaystyle \sqrt{8}+\sqrt{5}=2.83+2.24=5.07$
$\displaystyle \sqrt{7}+\sqrt{6}=2.65+2.45=5.09$
$\displaystyle \sqrt{10}+\sqrt{13}=3.16+3.61=6.77$
$\displaystyle \sqrt{11}+\sqrt{12}=3.32+1.41=4.73$
$\displaystyle \therefore $ Smallest is $\displaystyle \sqrt{11}+\sqrt{2}$