Tag: existence of irrational numbers

Questions Related to existence of irrational numbers

Simplify the following expressions.
Classify the following numbers as rational or irrational.

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

  1. $\left( 5+\sqrt { 7 } \right) \left( 2+\sqrt { 5 } \right)$

  2. $\left( 5+\sqrt { 5 } \right) \left( 5-\sqrt { 5 } \right)$

  3. ${ \left( \sqrt { 3 } +\sqrt { 7 } \right) }^{ 2 }$

  4. $\left( \sqrt { 11 } -\sqrt { 7 } \right) \left( \sqrt { 11 } +\sqrt { 7 } \right)$

Show answer
Correct Option: D
Explanation:

$A:$

$\left( {{\rm{5}} + \sqrt {\rm{7}} } \right)\left( {{\rm{2}} + \sqrt {\rm{5}} } \right)$  

$=10+5\sqrt5+2\sqrt7+\sqrt{35}$

Now, $10$ is rational and $\sqrt5,\sqrt7$ are non terminating , non repeating is an irrational 

and we know that $rational + irrational = irrational$ 

Therefore,  $\left( {{\rm{5}} + \sqrt {\rm{7}} } \right)\left( {{\rm{2}} + \sqrt {\rm{5}} } \right)$  is  irrational 


$B:$
$\left( {{\rm{5}} + \sqrt {\rm{5}} } \right)\left( {5 - \sqrt {\rm{5}} } \right)$

$={{\rm{5}}^2} + {\left( {\sqrt {\rm{5}} } \right)^2} = 25 - 5$

$=5$, which is rational 

So, $\left( {{\rm{5}} + \sqrt {\rm{5}} } \right)\left( {5 - \sqrt {\rm{5}} } \right)$
Is rational number.


$C:$
${\left( {\sqrt {\rm{3}}  + \sqrt {\rm{7}} } \right)^{\rm{2}}}$

$={\left( {\sqrt {\rm{3}} } \right)^2} + {\left( {\sqrt {\rm{7}} } \right)^2} + 2\sqrt {\rm{3}} \sqrt 7 $

$={\left( {\sqrt {\rm{3}} } \right)^2} + {\left( {\sqrt {\rm{7}} } \right)^2} + 2\sqrt {{\rm{21}}} =3 + 7 + 2\sqrt {{\rm{21}}} =10+2\sqrt{21}$
and $10$ and $\sqrt{21}$ are both rational.

Therefore, ${\left( {\sqrt {\rm{3}}  + \sqrt {\rm{7}} } \right)^{\rm{2}}}$ is rational.


$D:$
$\left( {{\rm{11}} - \sqrt {\rm{7}} } \right)\left( {{\rm{11 + }}\sqrt {\rm{7}} } \right)$

$={\left( {{\rm{11}}} \right)^2} - {\left( {\sqrt {\rm{7}} } \right)^2}$

$=11-7=4$, which is rational.

Therefore $\left( {{\rm{11}} - \sqrt {\rm{7}} } \right)\left( {{\rm{11 + }}\sqrt {\rm{7}} } \right)$ is rational.

Which of the following numbers are an irrational number. 

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

  1. $2- \sqrt 5$

  2. $\left( {3 + \sqrt {23} } \right) - \left( {\sqrt {23} } \right)$

  3. $\frac{1}{\sqrt 2}$

  4. $2\pi $

Show answer
Correct Option: A,C,D
Explanation:

$A$ is a irrational number as it cannot be expressed of the form $\cfrac{p}{q}$

$B$ is a rational number. As it can be expressed in the form of $\cfrac{3}{1}$
$C$ is a irrational number as it cannot be expressed of the form $\cfrac{p}{q}$
$D$ is a irrational number as it cannot be expressed of the form $\cfrac{p}{q}$ of two integers

If $p$ and $q$ are two distinct irrational numbers, then which of the following is always is an irrational number

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

  1. $\dfrac{p}{q}$

  2. $pq$

  3. $(p+q)^2$

  4. $\dfrac{p^2q+qp}{pq}$

Show answer
Correct Option: D
Explanation:

As, given $p$ and $q$ are two distinct irrational numbers.


Let $p=2+\sqrt 3$ and $q=2-\sqrt 3$

Then,

Option $A$
$\dfrac{p}{q}=\dfrac{2+\sqrt3}{2-\sqrt 3}$
$\dfrac{p}{q}=\dfrac{4+3+4\sqrt3}{4-3}=7+4\sqrt 3$

Option $B$
$pq=(2+\sqrt3)(2-\sqrt 3)=4-3=1$


Option $C$
$(p+q)^2=(2+\sqrt3+2-\sqrt 3)^2=4^2=16$

Option $D$
$\dfrac{p^2q+pq}{pq}=p+1$ is always an irrational number, because sum of rational and irrational is always irrational.

Hence, this is irrational.

Hence, this is the answer.

$\sqrt 7 $ is irrational.

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:
Lets assume that √7 is rational number. ie √7 = p/q.
suppose p/q have common factor then
we divide by the common factor to get √7 = a/b were a and b are co-prime number.
that is a and b have no common factor.
√7 =  a/b co- prime number
√7 = a/b
a = √7b
squaring
a² = 7b²                                   ....(i)
a² is divisible by 7
a = 7c
substituting values in eq (i)
(7c)² = 7b²
49c² = 7b²
7c² = b²
b² = 7c²
b² is divisible by 7
that is a and b have at least one common factor 7. 
√7 is irrational
Say true or false:
$87, 54, 0, -13, -4.7, \sqrt{5}, 2{1}{7}, \sqrt{15}, -{8}{7}, 3\sqrt{2}, 4.807, 0.002, \sqrt{16}$ and $2+\sqrt{3}.$ are rational numbers
 
maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

$\sqrt { 5 } ,\quad \sqrt { 15 } ,\quad 3\sqrt { 2 } ,\quad 2\quad +\sqrt { 3 } $ are irrational numbers as they cannot be expressed as a ratio.

Say True or False
$3+2\sqrt 5$ is an irrational number

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Let us assume, to the contrary, that $3+2\sqrt{5}$ is rational.


That is, we can find coprime integers $a$ and $b$ $(b0)$ such that $3+2\sqrt{5}=\dfrac{a}{b}$.

Therefore, $\dfrac{a}{b} - 3=2\sqrt{5}$

$\dfrac{a-3b}{b}=2\sqrt{5}$

$\dfrac{a-3b}{2b}=\sqrt{5}$

$\dfrac{a}{2b}-\frac{3}{2}=\sqrt{5}$

Since $a$ and $b$ are integers, we get $\dfrac{a}{2b}-\dfrac{3}{2}$ is rational, and so $\dfrac{a-3b}{2b}=\sqrt{5}$ is rational.

But this contradicts the fact that $\sqrt{5}$ is irrational.

This contradiction has arisen because of our incorrect assumption that $3+2\sqrt{5}$ is rational.
So, we conclude that  $3+2\sqrt{5}$ is irrational.

Say true or false:$0.120 1200 12000 120000 $....is a rational number

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

Given, $0.120 1200 12000 120000 ....$
Since, the decimal expansion is neither terminating nor non-terminating repeating, therefore, the given real number is not rational.
they are not rational, so we can't write of the form $\displaystyle \frac {p}{q}$.

State True or False.

$\sqrt{4}$ is an irrational number.

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

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

$\sqrt { 4 } =2\ The\quad decimal\quad representation\quad is\quad terminating.\ Hence,\quad \sqrt { 4 } is\quad a\quad rational\quad number.\ \quad $

The number $\displaystyle\frac{3-\sqrt{3}}{3+\sqrt{3}}$ is 

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

  1. Rational

  2. Irrational

  3. Both

  4. Can't say

Show answer
Correct Option: B
Explanation:

$Here,\quad we\quad will\quad carry\quad out\quad rationalization.\quad \ \frac { 3-\sqrt { 3 }  }{ 3+\sqrt { 3 }  } =\frac { 3-\sqrt { 3 }  }{ 3+\sqrt { 3 }  } x\frac { 3-\sqrt { 3 }  }{ 3-\sqrt { 3 }  } =\frac { { (3-\sqrt { 3 } ) }^{ 2 } }{ (3+\sqrt { 3) } (3-\sqrt { 3 } ) } =\frac { 9+3-6\sqrt { 3 }  }{ 9-3 } =\frac { 12-6\sqrt { 3 }  }{ 6 } =\frac { 2-\sqrt { 3 }  }{ 1 } \ Since\quad \sqrt { 3 } is\quad irrational\quad number\quad and\quad subtraction\quad of\quad rational\quad and\quad irrational\quad is\quad irrational.\ The\quad given\quad expression\quad is\quad irrational.\ \quad $

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

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

  1. $4 +\sqrt{5},-\sqrt{5}$

  2. $4 +\sqrt{5},\sqrt{5}$

  3. $4 -\sqrt{5},-\sqrt{5}$

  4. $ 2+\sqrt{5},2+\sqrt{5}$

Show answer
Correct Option: A
Explanation:

Let be the Number are $\sqrt{5}  and  -\sqrt{5}$
Sum of Number  $\left(\sqrt{5}\right) + \left(-\sqrt{5}\right)$
$\sqrt{5}-\sqrt{5} = 0$
Which is a rational number