Tag: rational and irrational numbers

Quizzes (6) Questions (185)

Questions Related to rational and irrational numbers

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

If $p$ is prime, then $\sqrt {p}$ is:

  1. Composite number

  2. Rational number

  3. Positive integer

  4. Irrational number
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

SInce, we know that prime numbers are those which are never perfect square and not divisible by any other number except by itself.
which are $2,3,5,7,...$
Clearly, if $p$ is prime then $\sqrt p $ is irrational number.
Option $D$ is correct. 

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

$6+\sqrt{2}$ is a rational number.

  1. True

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

Let's assume that $6+\sqrt2$ is rational..... 

then 

$6+\sqrt2 = p/q $

$\sqrt2 =( p-6q)/(q) $ 

now take $p-6q$ to be P and $q$ to be Q........where P and Q are integers 

which means, $\sqrt2= P/Q$...... 

But this contradicts the fact that $\sqrt2$ is rational 

So our assumption is wrong and $6+\sqrt2$ is irrational.

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

Compare the following pairs of surds. $\sqrt[8]{80}, \sqrt[4]{40}$    

  1. $\sqrt[8]{80} < \sqrt[4]{40}$
  2. $\sqrt[8]{80} \neq \sqrt[4]{40}$
  3. $\sqrt[8]{80} = \sqrt[4]{40}$
  4. $\sqrt[8]{80} > \sqrt[4]{40}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

   $\sqrt[8]{80}, \sqrt[4]{40}$
$={80}^{\frac{1}{8}}, {40}^{\frac{1}{4}}$
$={80}^{\frac{1}{8}}, {40}^{\frac{2}{8}}$
$={80}^{\frac{1}{8}}, {1600}^{\frac{1}{8}}$
Now,
   $80<1600$
$=>{80}^{\frac{1}{8}}<{1600}^{\frac{1}{8}}$
$=>\sqrt[8]{80}< \sqrt[4]{40}$

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?
$\displaystyle 0.07+\sqrt{0.16},\sqrt{1.44},1.2\times 0.83,1.02-\frac{0.6}{24}$

  1. $\displaystyle \sqrt {1.44}$
  2. $\displaystyle 0.07+\sqrt{0.16}$
  3. $1.2\times 0.83$
  4. $1.02-\dfrac{0.6}{24}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\Rightarrow 0.07+\sqrt{0.16}=0.07+0.4=0.47$
$\Rightarrow \sqrt{1.44}=1.2$
$\Rightarrow 1.2 \times 0.83 = 0.996$
$\Rightarrow 1.02-\cfrac{0.6}{24}=1.02-\cfrac {6}{240}=1.02-0.025=0.995$
$ \therefore$ The greatest number is $1.2$ i.e. $\sqrt {1.44}$
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

State whether the following equality is true or false:

$\displaystyle \frac{2\sqrt{3}}{\sqrt{5}} = $$\displaystyle \frac{2\sqrt{15}}{\sqrt{5}}$

  1. True

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

Rationalizing factor of  $\displaystyle \frac{2\sqrt{3}}{\sqrt{5}}$ is $\sqrt{5}$


$\therefore \displaystyle \frac{2\sqrt{3}}{\sqrt{5}}$  $=\displaystyle \frac{2\sqrt{3}\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$

$= \dfrac{2\sqrt{15}}{5}$

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

Determine the order relation between the following pairs of ratios.

$\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}}, \frac{2\sqrt{2}}{3\sqrt{3}}$

  1. $\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}} > \frac{2\sqrt{2}}{3\sqrt{3}}$
  2. $\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}} < \frac{2\sqrt{2}}{3\sqrt{3}}$

  3. Cannot be determined

  4. None of These
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{3\sqrt{3}}{2\sqrt{2}}=\dfrac{3\times 1.73214}{2\times 1.41429} = \dfrac{5.19642}{2.85828}
=1.82151$
$\dfrac{2\sqrt{2}}{3\sqrt{3}}=\dfrac{2\times 1.41429}{3\times 1.73214}=\dfrac{2.85828}{5.19642}=0.55004$
$\therefore \dfrac{3\sqrt{3}}{2\sqrt{2}} >\dfrac{2\sqrt{2}}{3\sqrt{3}}$

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

Compare the following pairs of surds $\sqrt[8]{12}, \sqrt[4]{6}$

  1. $\sqrt[8]{2} < \sqrt[4]{6}$
  2. $\sqrt[8]{8} < \sqrt[4]{6}$
  3. $\sqrt[8]{12} < \sqrt[4]{6}$
  4. $\sqrt[8]{12} < \sqrt[4]{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To compare, convert to a common root: sqrt[8]{12} is 12^(1/8) and sqrt[4]{6} is 6^(1/4) = 6^(2/8) = 36^(1/8). Since 12 < 36, sqrt[8]{12} < sqrt[4]{6}.

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

Compare the following pair of surds:

$\sqrt[3]{6}, \sqrt[4]{8}$

  1. $\sqrt[3]{6} > \sqrt[4]{8}$
  2. $\sqrt[3]{6} > \sqrt[4]{4}$
  3. $\sqrt[3]{6} > \sqrt[3]{8}$
  4. $\sqrt[3]{4} > \sqrt[4]{8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Convert to a common root (12th root): sqrt[3]{6} = 6^(4/12) = 1296^(1/12). sqrt[4]{8} = 8^(3/12) = 512^(1/12). Since 1296 > 512, the first is larger.

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

Arrange the following in ascending order of magnitude: 

$\displaystyle \sqrt[3]{4}, \sqrt[4]{5}, \sqrt{3}$ 

  1. $\displaystyle \sqrt[4]{5} < \sqrt[3]{4} < \sqrt{3}$
  2. $\displaystyle \sqrt[4]{5} > \sqrt[3]{4} > \sqrt{3}$
  3. $\displaystyle \sqrt[4]{5} > \sqrt[3]{4} < \sqrt{3}$
  4. $\displaystyle \sqrt[4]{5} < \sqrt[3]{4} > \sqrt{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Convert to 12th roots: sqrt[3]{4} = 4^(4/12) = 256^(1/12). sqrt[4]{5} = 5^(3/12) = 125^(1/12). sqrt{3} = 3^(6/12) = 729^(1/12). Ordering 125 < 256 < 729 gives the correct sequence.