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

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

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