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

$3+2\sqrt{5}$ a rational number.

  1. True

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

Let's assume that $3+2\sqrt5$ is rational..... 

then 

$3+2\sqrt5 = p/q $

$\sqrt5 =( p-3q)/(2q) $ 

now take $p-3q$ to be P and $2q$ to be Q........where P and Q are integers 

which means, $\sqrt5= P/Q$...... 

But this contradicts the fact that $\sqrt5$ is rational 

So our assumption is wrong and $3+2\sqrt5$ is irrational.

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

$\sqrt { 2 } ,\sqrt { 3 }$ are

  1. Whole numbers

  2. Rational numbers

  3. Irrational numbers

  4. Integers
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A rational number is any number that can be expressed as a fraction $\dfrac pq$ of two integers with $q$ not equal to zero.
As in the case of $\sqrt2$ and $\sqrt3$, it cannot be expressed as a fraction $\dfrac pq$.

Hence, option $A$ is the correct answer.

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

If $p$ is prime, then $\sqrt{p}$ is irrational and if $a, b$ are two odd prime numbers, then $a^2 -b^2$ is composite. As per the above passage mark the correct answer to the following question.
$\sqrt{7}$ is:

  1. a rational number

  2. an irrational number

  3. not a real number

  4. terminating decimal
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The basic definition for a rational number is that it can be represented in the form of $p/q$, where p and q are integers and q is a non-zero integer. Here, $\sqrt7$ is not a perfect square and thus cannot be expressed in the form of $p/q$, thus it is an irrational number.

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

Consider the given statements:
I. All surds are irrational numbers.
II. All irrationals numbers are surds.
Which of the following is true.

  1. Only I

  2. Only II

  3. Both I and II

  4. Neither I nor II
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A surd, by its very definition is an irrational number.

However, not every irrational number can be expressed as a surd.
Hence, only statement 1 is true.

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

If $a\neq 1$ and $ln{ a }^{ 2 }+{ \left( ln{ a }^{ 2 } \right)  }^{ 2 }+{ \left( ln{ a }^{ 2 } \right)  }^{ 3 }+........=3\left( lna+{ \left( ln{ a } \right)  }^{ 2 }+{ \left( ln{ a } \right)  }^{ 3 }+{ \left( ln{ a } \right)  }^{ 4 }+...... \right)$ then $a$ is

  1. $an\ irrational\ number$
  2. $a\ transcendental\ number$
  3. $an\ algeberaic\ number$
  4. $a\ surd$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, for $a\ne 1$,

$ln{ a }^{ 2 }+{ \left( ln{ a }^{ 2 } \right)  }^{ 2 }+{ \left( ln{ a }^{ 2 } \right)  }^{ 3 }+........=3\left( lna+{ \left( ln{ a } \right)  }^{ 2 }+{ \left( ln{ a } \right)  }^{ 3 }+{ \left( ln{ a } \right)  }^{ 4 }+...... \right)$
or, $\dfrac{\ln a^2}{1-\ln a^2}=3\times \dfrac{\ln a}{1-\ln a}$
or, $\dfrac{2\ln a}{1-2\ln a}=3\times \dfrac{\ln a}{1-\ln a}$
or, $2(1-\ln a)=3(1-2\ln a)$ [Since $a\ne 1\Rightarrow \ln a \ne 0$ ]
or, $4\ln a =1$
or, $a=\sqrt[4]{e}$.
So clearly $a$ is an irrational number.

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