Tag: proofs of irrationality

Questions Related to proofs of irrationality

If $ x = ( 2 + \sqrt3)^n , n \epsilon N $ and $ f = x - [x],$ then $ \dfrac {f^2}{1-f} $ is :

  1. An irrational number

  2. A non-integer rational number

  3. An odd number

  4. An even number


Correct Option: D

The product of two irrational numbers is 

  1. Always irrational

  2. Always rational

  3. Can be both rational and irrational

  4. always an integer


Correct Option: C
Explanation:

Let $p=\sqrt 3$ and $q=\sqrt 3$ be two irrational numbers 
$pq=\sqrt 3\times \sqrt 3=3$
which is rational
Now let $p=\sqrt 3$ and $q=\sqrt 2$
$pq=\sqrt 3\times \sqrt 2=\sqrt 6$
which is an irrational number
So the product can be both rational and irrational .
Option $C$ is correct.

Which of the following irrational number lies between 20 and 21

  1. $\sqrt442$

  2. $\sqrt440$

  3. $\sqrt443$

  4. $\sqrt444$


Correct Option: B
Explanation:

$20^2=400, 21^2=441\ \Rightarrow 400<440<441\ \Rightarrow 20<\sqrt{440}<21 $

State whether the given statement is True or False :

$4-5\sqrt { 2 } $ is an irrational number.
  1. True

  2. False


Correct Option: A
Explanation:

$\text{Here 4 is a rational number and }$$5\sqrt2$ is a $irrational$ number


And $\text{the difference of rational and irrational is always an irrational number}$

So that $(4-5\sqrt2)$ is an irrational number.

hence option A is correct.

State whether the given statement is True or False :

$5-2\sqrt { 3 } $ is an irrational number.
  1. True

  2. False


Correct Option: A
Explanation:

$\text{Here 5 is a rational number and }$$2\sqrt3$ is a $irrational$ number


And $\text{the difference of rational and irrational is always an irrational number}$

So that $(5 - 2\sqrt3)$ is an irrational number.

hence option A is correct.

State whether the given statement is True or False :

$3+\sqrt { 2 } $ is an irrational number.
  1. True

  2. False


Correct Option: A
Explanation:

$3 \text{ is a  rational number and }$$\sqrt2$ is a $irrational$ number


And $\text{the addition of rational and irrational is always an irrational number}$

So that $(3+\sqrt2)$ is an irrational number.

hence option A is correct

The equation $\sqrt{x+4}$- $\sqrt{x-3}$+ 1=0 has:

  1. no root

  2. one real root

  3. one real root and one imaginary root

  4. two imaginary roots

  5. two real roots


Correct Option: A
Explanation:

$\Longrightarrow \sqrt { x+4 } -\sqrt { x-3 } +1=0\ \Longrightarrow \sqrt { x+4 } +1=\sqrt { x-3 } \ \Longrightarrow x+4+1+2\sqrt { x+4 } =x-3\ \Longrightarrow 2\sqrt { x+4 } =-8\ \Longrightarrow x+4=16\ \therefore x=12$

But x = 12 will not satisfy given equation.
$\therefore$ No roots for given equation.

State whether True or False :


All the following numbers are irrationals.
(i) $\dfrac { 2 }{ \sqrt { 7 }  } $ (ii) $\dfrac { 3 }{ 2\sqrt { 5 }  }$ (iii) $4+\sqrt { 2 } $ (iv) $5\sqrt { 2 } $

  1. True

  2. False


Correct Option: A
Explanation:

In all of the above questions $\sqrt7,\sqrt5,\sqrt2$ are a $irrational$ numbers


And $\text{the addition, subtraction, division and product between  rational and irrational gives  an irrational number}$

So that all of the above are irrational numbers.

hence option A is correct.

State whether the given statement is True or False :

$2-3\sqrt { 5 }$ is an irrational number.
  1. True

  2. False


Correct Option: A
Explanation:

Here 2 is a rational number

and $3\sqrt5$ is an irrational number  and $\text{difference of rational and irrational is always an irrational number}$


So that $2 - 3\sqrt5 $  is an $irrational$ number

hence option A is correct.

State whether the given statement is True or False :

$\sqrt { 3 } +\sqrt { 4 } $ is an irrational number.
  1. True

  2. False


Correct Option: A
Explanation:

$\sqrt4= 2 \text{ is a  rational number and }$$\sqrt3$ is a $irrational$ number


And $\text{the addition of rational and irrational is always an irrational number}$

So that $(\sqrt3+\sqrt4)$ is an irrational number.

hence option A is correct.