Tag: rational and irrational numbers

Quizzes (6) Questions (185)

Questions Related to rational and irrational numbers

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

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

  1. $A> B$
  2. $A=B$
  3. $A< B$
  4. $A\ge B$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\sqrt{7} > \sqrt{5}\Rightarrow \dfrac{1}{A} > \dfrac{1}{B}$
$\Rightarrow \boxed{B > A}$
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

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

  1. $\sqrt{17} - \sqrt{12}$
  2. $\sqrt{11} - \sqrt{6}$

  3. Both are equal

  4. Can't be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. $\sqrt [ 3 ]{ 8 } $
  2. $\sqrt [ 4 ]{ 5 } $
  3. $\sqrt [ 3 ]{ 4 } $
  4. $\sqrt [ 4 ]{ 6 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
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
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

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


State True or False.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  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. $\sqrt[6]{3} , \sqrt[9]{4} , \sqrt[3]{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

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

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

  1. $\sqrt{19}$
  2. $\sqrt{29}$
  3. $\cfrac { 4\sqrt { 3 } }{ \sqrt { 3 } } $
  4. $\sqrt{17}$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
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)

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

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

  1. $\sqrt[4]{64} > \sqrt[6]{128}$
  2. $\sqrt[4]{64} < \sqrt[6]{128}$
  3. $\sqrt[4]{64} \neq \sqrt[6]{128}$
  4. $\sqrt[4]{64} = \sqrt[6]{128}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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