Tag: rational and irrational numbers

Questions Related to rational and irrational numbers

If $A=\sqrt{7}-\sqrt{6}$ and $B=\sqrt{6}-\sqrt{5}$, then identify the true statement.

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. $A\ge B$

Show answer
Correct Option: C
Explanation:
$A=\sqrt{7}-\sqrt{6}\Rightarrow \dfrac{1}{A}=\dfrac{\sqrt{7}+\sqrt{6}}{(\sqrt{7}-\sqrt{6})(\sqrt{7}+\sqrt{6})}$
$\Rightarrow \boxed{\dfrac{1}{A}=\sqrt{7}+\sqrt{6}}$
$\boxed{\dfrac{1}{B}=\sqrt{6}+\sqrt{5}}$
As $\sqrt{7} > \sqrt{5}\Rightarrow \dfrac{1}{A} > \dfrac{1}{B}$
$\Rightarrow \boxed{B > A}$

The smallest between $\sqrt{17} - \sqrt{12}$ and $\sqrt{11} - \sqrt{6}$ is _________.

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

  1. $\sqrt{17} - \sqrt{12}$

  2. $\sqrt{11} - \sqrt{6}$

  3. Both are equal

  4. Can't be determined

Show answer
Correct Option: A
Explanation:

$\sqrt{17} \approx 4.123$

$\sqrt{12} \approx 3.464$
$\sqrt{11} \approx 3.316$
$\sqrt{6} \approx 2.449$

$\Rightarrow \sqrt{17} - \sqrt{12} = 0.659$
$\Rightarrow \sqrt{11} - \sqrt{6} = 0.867$

Hence, $\sqrt{17}-\sqrt{12}$ is smaller.

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:
$\sqrt[3]{4}=\sqrt[12]{44}=\sqrt[12]{256}$
$\sqrt[4]{6}=\sqrt[12]{5^3}=\sqrt[12]{125}$
$\sqrt[4]{6}=\sqrt[12]{6^{3}}=\sqrt[12]{216}$
$\sqrt[3]{8}=\sqrt[12]{8^{4}}=\sqrt[12]{64^{2}}$
As $'125'$ is smallest
$\therefore \boxed{4\sqrt{5}}$ is smallest

Let x and y be rational and irrational numbers, respectively, then x + y necessarily an irrational number.


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:

Yes.

Let x $= 21, y =\sqrt{2}$ be a rational number
Now $x+y=21 +\sqrt{2}=21+1.4142....=22.4142....$ , which is non-terminating and non-recurring. Hence x+y is irrational.

If $A=\sqrt [ 3 ]{ 3 } , B=\sqrt [ 4 ]{ 5 } $, then which of the following is true?

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

  1. $A=\cfrac{3}{5}B$

  2. $A< B$

  3. $A> B$

  4. $A={B}^{4/5}$

Show answer
Correct Option: B
Explanation:
$A=3^{\dfrac{1}{3}}=\sqrt[12]{34}=\sqrt[12]{81}$
$B=5^{\dfrac{1}{4}}=\sqrt[12]{5^{3}}=\sqrt[12]{125}$
As $125 > 81$
$\Rightarrow \boxed{B > A}$

The descending order of the surds $\sqrt[3]{2} , \sqrt[6]{3} , \sqrt[9]{4}$ is _________.

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

  1. $\sqrt[9]{4} , \sqrt[6]{3} , \sqrt[3]{2}$

  2. $\sqrt[9]{4} , \sqrt[3]{2} , \sqrt[6]{3}$

  3. $\sqrt[3]{2} , \sqrt[6]{3} , \sqrt[9]{4}$

  4. $\sqrt[6]{3} , \sqrt[9]{4} , \sqrt[3]{2}$

Show answer
Correct Option: C
Explanation:

$\sqrt[3]{2} \approx 1.26$

$\sqrt[6]{3} \approx 1.201$
$\sqrt[9]{4} \approx 1.166$

$\therefore$ Ascending order is $\sqrt[9]{4} < \sqrt[6]{3} < \sqrt[3]{2}$

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:

Clearly, one of $4^{\frac{1}{5}}$ and $5^{\frac{1}{4}}$ must be the smallest.
$\frac{1}{5}$<$\frac{1}{4}$.
So, $\displaystyle 4^{\frac{4}{20}}$ < $5^{\frac{5}{20}}$

Which of the following is the greatest?
$\sqrt{12}$, $\sqrt{13}$,$\sqrt{15}$,$\sqrt{17}$.

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

  1. $\sqrt{12}$

  2. $\sqrt{13}$

  3. $\sqrt{15}$

  4. $\sqrt{17}$

Show answer
Correct Option: D
Explanation:

$12<13<15<17\ \Rightarrow \sqrt{12}<\sqrt{13}<\sqrt{15}<\sqrt{17}$

Therefore, $ \sqrt{17}$ is greatest.

Identify the irrational number(s) between $2\sqrt{3}$ and $3\sqrt{3}$

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

  1. $\sqrt{19}$

  2. $\sqrt{29}$

  3. $\cfrac { 4\sqrt { 3 } }{ \sqrt { 3 } } $

  4. $\sqrt{17}$

Show answer
Correct Option: A,D
Explanation:
$2\sqrt{3}=\sqrt{12}$
$3\sqrt{3}=\sqrt{27}$
$\therefore \sqrt{176}\sqrt{19}$ are irrational no between them $\sqrt{29}$ lie out of it.
As $\dfrac{4\sqrt{3}}{\sqrt{3}}=4$ (Rational)

Compare the following pairs of surds. $\sqrt[4]{64}, \sqrt[6]{128}$    

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

  1. $\sqrt[4]{64} > \sqrt[6]{128}$

  2. $\sqrt[4]{64} < \sqrt[6]{128}$

  3. $\sqrt[4]{64} \neq \sqrt[6]{128}$

  4. $\sqrt[4]{64} = \sqrt[6]{128}$

Show answer
Correct Option: A
Explanation:

$\sqrt[4]{64}=\sqrt[4]{2^6}=2\sqrt[4]{2^2}=2\sqrt{2}=2\sqrt[6]{2^3}=2\sqrt[6]{8}$
$\sqrt[6]{128}=\sqrt[6]{2^7}=2\sqrt[6]{2}$
$\sqrt[4]{64}>\sqrt[6]{128}$