Tag: existence of irrational numbers

Questions Related to existence of irrational numbers

Use ______________ to represent an irrational number on number line.

  1. Isosceles-angle theorem

  2. Scalene angle theorem

  3. Right-angled theorem

  4. None of the above


Correct Option: C
Explanation:
Using the Pythagoras Theorem, we can represent some irrational numbers, which are surds, on a number line.
Since it involves Pythagoras Theorem, we get to use Right Angle Theorem.

Hence, to represent an irrational number, we generally use right angled theorem.

$D$ is a real number with non terminating digits $a _1$ and $a _2$ after the decimal point. Let $D = 0, a _1 a _2 a _1 a _2 ........ $  with $a _1 & a _2$ both not zero which of the following when multiplied by $D$ will necessarily give an integer ?

  1. $99$

  2. $18$

  3. $125$

  4. $75$


Correct Option: A
Explanation:

its straight question
give $D=0.abababab$ $(say - 1)$ 
Multiply both sides by $100$ $i.e.$ 
$100D = ab.abababab$ $(say - 2)$
now subtract $1$ from $2 .$ That gives
$99D = ab => D = ab/99$ hence it should be multiplied by $99k$ to get an integer ab$.$

The number $5\sqrt{34}$ lies between

  1. $29$ and $30$

  2. $30$ and $31$

  3. $31$ and $32$

  4. $32$ and $33$


Correct Option: A
Explanation:
$5\sqrt {34}\ \le \ 5\sqrt {36} \ = \ 5\times 6=30$
less than $30$
i.e., $=29$ to $30$


Can $\sqrt { 3 } -3$ be represented on the number line. 

  1. True

  2. False


Correct Option: A
Explanation:

Yes, 


Number line consists of all real numbers i.e., Both rational and Irrational Numbers

Since $\sqrt 3-3$ is a real number It can be represented on number line 

Give an example of two irrational numbers whose difference is an irrational number.

  1. $\sqrt{3},-\sqrt{3}$

  2. $\sqrt{5,}-\sqrt{5}$

  3. $4\sqrt{3},-2\sqrt{3}$

  4. None of the above


Correct Option: C
Explanation:

$4\sqrt{3},2\sqrt{3}$ are the irrational numbers and thier difference,


$4\sqrt{3}-2\sqrt{3}=2\sqrt 3$ is also an irrational number.

Which is the wrong step that shows $\displaystyle 5-\sqrt{3}$ is irrational?
(I) Contradiction : Assume that $\displaystyle 5-\sqrt{3}$ is rational
(II) Find coprime a & b $\displaystyle \left ( b\neq 0 \right )$ such that $\displaystyle 5-\sqrt{3}=\frac{a}{b},\therefore 5-\frac{a}{b}=\sqrt{3}$
Rearranging above equation $\displaystyle \sqrt{3}=5-\frac{a}{b}=\frac{5b-a}{b}$
(III) Since a & b are integers we get $\displaystyle 5-\frac{a}{b}$ is irrational and so $\displaystyle \sqrt{3}$ is irrational
(IV) But this contradicts the fact that $\displaystyle \sqrt{3}$ is irrational Hence $\displaystyle 5-\sqrt{3}$ is irrational

  1. Both I and II

  2. Only III

  3. Only II

  4. Both II and III


Correct Option: B

Which of the following irrational numbers lie between $4$ and $7$?

  1. $\sqrt{25}$

  2. $\sqrt{19}$

  3. $\sqrt{47}$

  4. $\sqrt{50}$


Correct Option: B,C
Explanation:

$4^{2} = 16$

$5^{2} = 25$
$6^{2} = 36$
$7^{2} = 49$


$\Rightarrow \sqrt19$ and $\sqrt47$ are irrational numbers which lie between $4$ and $7$


$\sqrt25 = 5$ which is a rational number

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


Correct Option: A
Explanation:

Lets keep in mind the following common log values:

$log(2)=0.30$
$log(3)=0.47$
Lets compare these three surds by their log values.
Reason: Since $log(x) $ is an increasing function  when $x>1$ so we can compare these values by comparing their log values.

$log(\sqrt[3]{2})=log(2^{\frac{1}3})=\left(\dfrac{1}3\right)log(2)=\left(\dfrac{1}3\right) \times 0.3=0.100$

$log(\sqrt[6]{3})=log(3^{\frac{1}6})=\left(\dfrac{1}6\right)log(3)=\left(\dfrac{1}6\right) \times 0.47=0.078$

$log(\sqrt[9]{4})=log(\sqrt[9]{2^2})=log(2^{\frac{2}9})=\left(\dfrac{2}9\right)log(2)=\left(\dfrac{2}9\right) \times 0.3=0.060$

By looking at the log values,it is clear that,

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

$A,B,C$ and $D$ are all different digits between $0$ and $9$. If $AB+DC=7B\ (AB,DC$ and $7B$ are two digit numbers), then the value of $C$ is

  1. $0$

  2. $1$

  3. $2$

  4. $3$

  5. $5$


Correct Option: A

If $\sqrt{a}$ is an irrational number, what is a? 

  1. Rational

  2. Irrational

  3. $0$

  4. Real


Correct Option: A
Explanation:

Consider the given irrational number$\sqrt{a}$ ,

Definition  of rational number- which number can be write in the form of $\dfrac{p}{q}$ but $q\ne 0$ is called rational number.

Hence, $a=\dfrac{a}{1}$

That why  $a$ is rational number

 

Hence, this is the answer.