Tag: rational and irrational numbers

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Questions Related to rational and irrational numbers

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

Give an example of two irrational numbers, whose product is a rational number.

  1. $\sqrt{8},\sqrt{2}$
  2. $\sqrt{5},\sqrt{2}$
  3. $2+\sqrt{8},\sqrt{2}$
  4. $\sqrt{8},2+\sqrt{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let be the Number are $\sqrt{8}  and  \sqrt{2}$
Product of Numbers  $\sqrt{8}\times \sqrt{2} = \sqrt{16} = 4$
Which is a rational number

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

Give an example of two irrational numbers, whose product is an irrational number.

  1. $\sqrt{3},\sqrt{3}$
  2. $\sqrt{2},\sqrt{2}$
  3. $\sqrt{2},-\sqrt{2}$
  4. $\sqrt{2},\sqrt{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let be the Number are $\sqrt{2}  and  \sqrt{3}$
Product of Numbers  $\sqrt{2}\times \sqrt{3} = \sqrt{6} $
Which is a irrational number

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

$\displaystyle log _{4}18$ is 

  1. an irrational number

  2. a rational number

  3. natural number

  4. whole number
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$  Log AB = log A + log B $
Also, $ log a^b = b log a $

So, $ log _{4} 18 = \frac { log 2 \times 3^2}{log 4}  = \frac { log 2 \times 3^2}{log 2^2}  = \frac { log 2}{2log 2} + \frac {2 log 3}{2 log 2}  = \frac {1}{2} +  \frac {log 3}{log 2}   $

As both $ log 2 $ and $ log 3 $ are irrational numbers, $ log _{x} 18 $ is an irrational number too. 

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

Number of integers lying between $1 $ to $102$  which are divisible by all $\displaystyle \sqrt{2},\sqrt{3},\sqrt{6}, $ is 

  1. $16$
  2. $17$
  3. $15$
  4. $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a number to be divisible by $\sqrt { 2 } $, it must be an irrational number. An integer is not an irrational,

so  there are no  numbers between  $ 1$ to  $102$ which are divisible by all  $\sqrt{2},\sqrt{3},\sqrt{6}$.

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

Simplify by combining similar terms :$\displaystyle 3\sqrt{147}-\frac{7}{3}\sqrt{\frac{1}{3}}+7\sqrt{\frac{1}{3}}$

  1. $\displaystyle \frac{189}{3\sqrt{3}}$
  2. $\displaystyle \frac{175}{3\sqrt{3}}$
  3. $\displaystyle \frac{208\sqrt{3}}{3}$
  4. $\displaystyle \frac{203}{3\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Simplify the expression: 3*sqrt(147) = 3*sqrt(49*3) = 21*sqrt(3). The other terms are -(7/3)(1/sqrt(3)) + 7(1/sqrt(3)) = (14/3)*(1/sqrt(3)) = 14/(3*sqrt(3)). Combining these requires a common denominator. The result is (21*3*sqrt(3) + 14)/(3*sqrt(3)) = (189+14)/(3*sqrt(3)) = 203/(3*sqrt(3)).

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

$\sqrt {5}$ is a\an ......... number.

  1. rational

  2. whole

  3. integer

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

$\sqrt {5} = \dfrac {a}{b}$

$b\sqrt {5} = a$ $(a$ and $b$ are co-prime i.e. they have no common factors$) ...(1)$ 
$5b^{2} = a^{2}$ (squaring both sides)
Therefore $5$ divides $a^{2}$
As per Fundamental Theorem of Arithmetic, $5$ divides $a.$
Let's take it as $a = 5c,$
$5b^{2} = 25 c^{2}$
$b^{2} = 5c^{2}$
As per Fundamental Theorem of Arithmetic, $5$ divides $a.$
So $a$ and $b$ have $5$ as a common factor but $a$ and $b$ have only $1$ common factor $1$ from equation $(1),$ so it is not rational.
So, we conclude that $\sqrt {5}$ is irrational.
Therefore, $D$ is the correct answer.

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

$\sqrt{21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}}=$...........

  1. $\sqrt{5}-2+2\sqrt{3}$
  2. $\sqrt{5}-\sqrt{4}-\sqrt{12}$
  3. $-\sqrt{5}+\sqrt{4}+\sqrt{12}$
  4. $-\sqrt{5}-\sqrt{4}+\sqrt{12}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression is sqrt(21 - 4*sqrt(5) + 8*sqrt(3) - 4*sqrt(15)). This is of the form sqrt((a+b+c)^2) = |a+b+c|. Expanding (sqrt(5) - 2 - 2*sqrt(3))^2 gives 5 + 4 + 12 - 4*sqrt(5) - 4*sqrt(15) + 8*sqrt(3) = 21 - 4*sqrt(5) + 8*sqrt(3) - 4*sqrt(15). Thus the square root is |sqrt(5) - 2 - 2*sqrt(3)|, which equals -sqrt(5) + 2 + 2*sqrt(3).

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

State whether the following statements are true or false. 
$\sqrt {n}$ is not irrational if n is a perfect square

  1. True

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

False ,

$\sqrt{4}=2$ where 2 is a rational number.Here n is perfect square the  $\sqrt{n}$ is rational number 
$\sqrt{5}=2.236..$ is not rational  number But it is irrational number . here n is not a perfect square the  $\sqrt{n}$ is  irrational  number
So $\sqrt{n}$ is not irrational number if n is perfect square