Tag: existence of irrational numbers

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

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

  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. $\left( \sqrt { 11 } -\sqrt { 7 } \right) \left( \sqrt { 11 } +\sqrt { 7 } \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

Which of the following numbers are an irrational number. 

  1. $2- \sqrt 5$
  2. $\left( {3 + \sqrt {23} } \right) - \left( {\sqrt {23} } \right)$
  3. $\frac{1}{\sqrt 2}$
  4. $2\pi $
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
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

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

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

  1. $\dfrac{p}{q}$
  2. $pq$
  3. $(p+q)^2$
  4. $\dfrac{p^2q+qp}{pq}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

$\sqrt 7 $ is irrational.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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
Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers
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
 

  1. True

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

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

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

  1. True

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

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\displaystyle \frac {p}{q}$.

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

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

  1. Rational

  2. Irrational

  3. Both

  4. Can't say
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

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 sum is a rational number

  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}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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