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.

As we've $\sqrt{7}$ is an irrational number and $23$ is a rational number then the sum of an irrational number and a rational number is again an irrational number.

The real value of $\sqrt { 16 } =4$

Natural numbers from the given list are 87, 54,  $\sqrt { 16 } =4$ and 217

To get the sum as irrational, the numbers need to have an irrational part as well which are different from each other.

Example, the pair of numbers $ \sqrt{3} + 2 $ and $ 5 + \sqrt {2} $ have the sum $ \sqrt{3} + 2 + 5 + \sqrt {2} = 7 + \sqrt {2} + \sqrt {3} $ which is an irrational number too.

Decimal form of $\sqrt { 3 } $ is non terminating and non repeating, So, it is irrational number.

$ \sqrt{2} \times \sqrt[3] {3} \times \sqrt[4]{4}$
$=2^{ \frac { 1 }{ 2 }  } \times 3^{ \frac { 1 }{ 3 }  }\times 2^{ \frac { 2 }{ 4 }  }$
$=2^{ \frac { 1 }{ 2 }  } \times 2^{ \frac { 1 }{ 2 }  }\times 3^{ \frac { 1 }{ 3 }  }$
$=2  \times3^{ \frac { 1 }{ 3 }  }$
$=2^{ \frac { 3 }{ 3 }  }\times3^{ \frac { 1 }{ 3 }  }  $
$=\sqrt [ 3 ]{ 2^{ 3 } }\times\sqrt[3]{3}$
$=\sqrt[3]{8\times3}$
$=\sqrt[3]{24}$

We will be considering the digits only after the decimal point.

An irrational number is any real number that cannot be expressed as a ratio of integers. Irrational numbers are those real numbers that cannot be represented as terminating or repeating decimals.
Among all the options only $(D)$ $0.123223222322223$...... is non terminating and non repeating decimal.Therefore, it is a irrational number.

Rational numbers are those numbers which can be expressed in the form $ \dfrac {p}{q} $, where p and q are integers and $ q \neq 0 $
Numbers which are not rational numbers are called irrational numbers.
Since, $ \sqrt {7} $ cannot be written in $ \dfrac {p}{q} $, where $p$ and $q$ are integers and $ q \neq 0 $; it is an irrational number.

Real number consists of collection of rationals and irrationals.

Hence, every irrational number is also a real number.

Example-2 is also real.

The statement is false since real numbers consists of both rational and irrational numbers. $5,65,8/9...$ are all real numbers which are rational.

$2$ is rational

Decimal representation of an irrational number is always non terminating non repeating.

No. Not all square roots of integers are irrational. Examples are $\sqrt{4}=2$, $\sqrt{1}=1$, $\sqrt{9}=3$, etc.

$\sqrt{2} = 1.4142136...$
The decimal expansion of the number $\sqrt{2}$ is Non-terminating non recurring

$\ { (\sqrt { 2 } -2) }^{ 2 }=2+4-4\sqrt { 2 } =6-4\sqrt { 2 } \ \sqrt { 2 } =1.41421356237........\ \ \sqrt { 2 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ Multiplication\quad and\quad subtration\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\ Hence,\quad { (\sqrt { 2 } -2) }^{ 2 }\quad is\quad an\quad irrational\quad number.\ \quad $

$\ { (\sqrt { 2 } +\sqrt { 3 } ) }^{ 2 }=2+3+2\sqrt { 6 } =5+2\sqrt { 6 } \ \sqrt { 6 } =2.44948974278........\ \ \sqrt { 6 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ Multiplication\quad ans\quad addition\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\ Hence,\quad { (\sqrt { 2 } +\sqrt { 3 } ) }^{ 2 }\quad is\quad an\quad irrational\quad number.\ \quad $

$\ \sqrt { 3 } =1.73205080757......\ \sqrt { 3 } is\quad an\quad irrational\quad number,\quad since\quad it's\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ And\quad addition\quad of\quad a\quad rational\quad and\quad irrational\quad number\quad is\quad irrational.\ Hence,\quad 2+\sqrt { 3 } \quad is\quad an\quad irrational\quad number.\ \quad $

$\ \sqrt { 3 } =1.73205080757......\ Also\quad \sqrt { 2 } =1.41421356237........\ \sqrt { 3 } and\quad \sqrt { 2 } are\quad irrational\quad numbers,\quad since\quad their\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ And\quad addition\quad of\quad two\quad irrational\quad numbers\quad is\quad irrational.\ Hence,\quad \sqrt { 2 } +\sqrt { 3 } \quad is\quad an\quad irrational\quad number.\ \quad $

$\ \sqrt { 3 } =1.73205080757......\ Also\quad \sqrt { 5 } =2.2360679775........\ \sqrt { 3 } and\quad \sqrt { 5 } are\quad irrational\quad numbers,\quad since\quad their\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ And\quad addition\quad of\quad two\quad irrational\quad numbers\quad is\quad irrational.\ Hence,\quad \sqrt { 5 } +\sqrt { 3 } \quad is\quad an\quad irrational\quad number.\ \quad $

$\sqrt { 7 } =2.64575131106...\ The\quad decimal\quad representation\quad is\quad non\quad repeating\quad non\quad terminating.\ Hence,\quad \sqrt { 7 } is\quad an\quad irrational\quad number.\ \quad $

$\ { (2-\sqrt { 2 } ) }(2+\sqrt { 2 } )=4-2=2\ \  { 2 } is\quad a\quad rational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad terminating.\ Hence,\quad { (2-\sqrt { 2 } ) }(2+\sqrt { 2 } )\quad is\quad a\quad rational\quad number.\ \quad $

$\ { (\sqrt { 5 } -2) }\ \sqrt { 5 } =2.2360679775........\ \ \sqrt { 5 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ Subtraction\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\ Hence,\quad { (\sqrt { 5 } -2) }\quad is\quad an\quad irrational\quad number.\ \quad $

$\ { \frac { -2 }{ 5 } \sqrt { 8 }  }={ \frac { -4 }{ 5 } \sqrt { 2 }  }=-0.8*\sqrt { 2 } \ \sqrt { 2 } =1.41421356237........\ \ \sqrt { 2 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ Multiplication\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\ Hence,\quad { (\frac { -2 }{ 5 } \sqrt { 8 } ) }\quad is\quad an\quad irrational\quad number.\ \quad $

$\displaystyle \frac { (2+\sqrt { 2 } )(3-\sqrt { 5 } ) }{ (3+\sqrt { 5 } )(2-\sqrt { 2 } ) } =\frac { { (2+\sqrt { 2 } ) }^{ 2 }{ (3-\sqrt { 5 } ) }^{ 2 } }{ (9-5)(4-2) } =\frac { (4+2+4\sqrt { 2 } )(9+5-6\sqrt { 5 } ) }{ 8 } \$

${ (2+\sqrt { 3 } ) }({ 2-\sqrt { 3 } ) }=4-3=1\ Hnece\quad rational.\ $

${ (2+\sqrt { 3 } ) }^{ 2 }=4+3+4\sqrt { 3 } =7+4\sqrt { 3 } \ The\quad above\quad given\quad expression\quad consists\quad of\quad an\quad algebric\quad equation\quad \quad \ consisting\quad of\quad irrational\quad terms,\quad hence\quad it\quad is\quad an\quad irrational\quad expression.\ $

$\displaystyle 3.14=\frac{22}{7}=\pi(pi)$

Let $\sqrt3$ is a rational number
$\therefore \sqrt3 = \displaystyle \frac{a}{b}$ [Where a & b are co-primes]
$a^2=3b^2$ .......(i)
$\Rightarrow$ 3 divides $a^2$
$\Rightarrow$ 3 also divides a
$\Rightarrow$ a=3c
[Where c is any non-zero positive integer]
$\Rightarrow a^2 = 9c^2$
From equation (i)
$3b^2=9c^2$
$\Rightarrow b^2 = 3c^2  \Rightarrow$ 3 divides $b^2$
$\Rightarrow$ 3 also divides b
So, 3 is a common factor of a and b.
Our assumption is wrong, because a and b are not co - primes.
It means $\sqrt3$ is an irrational number.

$\cfrac{m}{n} = \sqrt{2} + \sqrt{3} $
Square both sides:
$\cfrac{m^2}{ n^2} = 5 + 2\sqrt{6} $

$1.73202002$ is the irrational number because it can not  be expressed as a fraction

Given that,$3.\overline{25}$.

Let,

$x=3.\overline{25}$

 $x=3.252525.....$

Multiply by 100 both sides,

  $ 100x=100\times 3.252525..... $

 $ 100x=325.2525..... $

 $ 100x=322+3.2525..... $

 $ 100x=322+x $

 $ 99x=322 $

 $ x=\dfrac{322}{99} $

Hence, this is the answer.

Given that,$0.\overline{05}$

Let,

  $ x=0.\overline{05} $

 $ x=0.05050505..... $

Multiply by $100$ both sides,

 $ 100x=100\times 0.05050505..... $

 $ 100x=5.050505..... $

 $ 100x=5+0.050505..... $

 $ 100x=5+x $

 $ 99x=5 $

 $ x=\dfrac{5}{99} $

Hence, this is the answer.

The square root of any prime number is irrational.
Example: $\sqrt{2}$ is a irrational number.

$\dfrac{7}{9}$ is of the form  $\dfrac {p}{q}$ form , hence it is rational no.

As per the theorem, the square root of any prime number is irrational. $\sqrt {23}$ is a prime number, so is not a rational number. It is irrational.
Therefore, $D$ is the correct answer.

Sum or difference of a rational and irrational number is irrational. Therefore, $D$ is the correct answer.

$\pi = 3.14159265358979........$ is a non-terminating and non-repeating irrational number.

$\sqrt { 3 } +\sqrt { 3 } =2\sqrt { 3 } \quad irrational\quad number\ \sqrt { 3 } -\sqrt { 3 } =0\quad rational\quad number\ \sqrt { 3 } \times \sqrt { 3 } =3\quad rational\quad number\ \frac { \sqrt { 3 }  }{ \sqrt { 3 }  } =1\quad rational\quad number$

1/22 is an irrational number hence it is a non terminating number

Real number are all numbers on number line 

False,

$\dfrac { \sqrt { 12 }  }{ \sqrt { 75 }  } =\dfrac { 2\sqrt { 3 }  }{ 5\sqrt { 3 }  } =\dfrac { 2 }{ 3 } $ which is a rational number 

$\sqrt{7}$ is an irrational number

Option (i)

A number having non-terminating and non-recurring decimal expansion is  a Irrational Number

$x$ is any irrational number 
Let $x=\sqrt [ 4 ]{ 3 } $
$\Rightarrow { x }^{ 2 }=\sqrt { 3 } $
which is irrational so option $B$ is correct.
Now let $x=\sqrt 3$
$\Rightarrow {x}^{2}=3$
which is rational so option $C$ is correct.
So correct answer is $D$

$7.4848...=7.\bar{48}\Rightarrow $Rational  number

By definition, an irrational number in decimal form goes on forever without repeating (a non-repeating, non-terminating decimal). By definition, a rational number in decimal form either terminates or repeats. 

We know that sum of two irrational number or one rational and one irrational number will be irrational number. Option A, B , C stisfies this criteria but option D have two rational number i.e. $6 + \sqrt { 9 }$ = $6+ 3=9$

${ (2+\sqrt { 5 } ) }^{ 2 }\ =4+5+4\sqrt { 5 } \ =9+4\sqrt { 5 } $

According to definition of irrational number, If written in decimal notation, an irrational number would have an infinite number of digits to the right of the decimal point, without repetition.

In the given options $\sqrt { 16 }$ and $\sqrt { 25 } $ are irrational numbers. Their real values are 4 and 5 respectively. So, option A and C are incorrect.

In the given options, only option D is non terminating non recurring decimal. 

Irrational numbers have decimal expansions that neither terminate nor repeating

In the given options, 

The value $\dfrac{22}7$ is a rational number, as it can be expressed in the form $\dfrac pq$. 

An irrational number is any real number that cannot be expressed as a ratio a/b, where a and b are integers and b is non-zero.
$\sqrt5$ is irrational as it can never be expressed in the form a/b

An irrational number is any real number that cannot be expressed as a ratio a/b, where a and b are integers and b is non-zero.
$1/(\sqrt2)$ is irrational as it can never be expressed in the form a/b

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.

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 $\dfrac pq$.

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.

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

$16^{\frac{1}{2}}=4$, which is a rational number.
The other options are irrational  numbers.
So, $16^{\frac{1}{2}}$ is different from others.

Given, for $a\ne 1$,

$A:$

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

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

$\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.

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 $\displaystyle \frac {p}{q}$.

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

$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 $

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

Let be the Number are $4\sqrt{3}  and  2\sqrt{3}$
Difference of Number  $4\sqrt{3} - 2\sqrt{3} = 2\sqrt{3}$
Which is a irrational number

Let be the Number are $\sqrt{15}  and  \sqrt{5}$
Quotient of Numbers  $\frac{\sqrt{15}}{\sqrt{5}} = \sqrt{\frac{15}{5}} = \sqrt{3} $
Which is a irrational number

Let be the Number are $2\sqrt{5}  and  3\sqrt{5}$
Sum of Number  $2\sqrt{5} + 3\sqrt{5} = 5\sqrt{5}$
Which is a irrational number

Let be the Number are $\sqrt{8}  and  \sqrt{2}$
Quotient of Numbers  $\frac{\sqrt{8}}{\sqrt{2}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2 $
Which is a rational number

Let be the Number are $\sqrt{8}  and  \sqrt{2}$
Product of Numbers  $\sqrt{8}\times \sqrt{2} = \sqrt{16} = 4$
Which is a rational number

Let be the Number are $\sqrt{2}  and  \sqrt{3}$
Product of Numbers  $\sqrt{2}\times \sqrt{3} = \sqrt{6} $
Which is a irrational number

$  Log AB = log A + log B $
Also, $ log a^b = b log a $

So, $ log _{4} 18 = \frac { log 2 \times 3^2}{log 4}  = \frac { log 2 \times 3^2}{log 2^2}  = \frac { log 2}{2log 2} + \frac {2 log 3}{2 log 2}  = \frac {1}{2} +  \frac {log 3}{log 2}   $

As both $ log 2 $ and $ log 3 $ are irrational numbers, $ log _{x} 18 $ is an irrational number too. 

For a number to be divisible by $\sqrt { 2 } $, it must be an irrational number. An integer is not an irrational,

Irrational numbers are non-repeating and non-terminating numbers.

$\sqrt {5} = \dfrac {a}{b}$

There can be infinite number of irrational numbers between any two rational numbers. Hence the answer is infinitely many.

False ,

SInce, we know that prime numbers are those which are never perfect square and not divisible by any other number except by itself.
which are $2,3,5,7,...$
Clearly, if $p$ is prime then $\sqrt p $ is irrational number.
Option $D$ is correct. 

$3\sqrt{18}=9\sqrt{2},$ which is the product of a rational and an irrational number and so an irrational number.

Let's assume that $6+\sqrt2$ is rational..... 

then 

$6+\sqrt2 = p/q $

$\sqrt2 =( p-6q)/(q) $ 

now take $p-6q$ to be P and $q$ to be Q........where P and Q are integers 

which means, $\sqrt2= P/Q$...... 

But this contradicts the fact that $\sqrt2$ is rational 

So our assumption is wrong and $6+\sqrt2$ is irrational.