AS we know 3 is an irrational number and $\sqrt{6}$ is also a rational number 

$\sqrt{3}$ is an irrational number.

Counter-example for A: $(\sqrt{2}) + (4-\sqrt{2}) = 4$

 $ \displaystyle  \because  \frac{-5}{-3} $= $ \displaystyle  \frac{-5}{-3} $X $ \displaystyle  \frac{-5}{-5} $= $ \displaystyle  \frac{25}{15} $

An irrational number is a real number that cannot be represented as a ratio or a simple fraction.

Surds are numbers left in root form (√) to express its exact value. It has an infinite number of non-recurring decimals 

Square of (2+√2) is (4+2.2.√2+2) = 6+4√2 which is an irrational number

$3.54672$ can be written in simple fraction.$\Rightarrow $ Rational number.

$35.251522253...\Rightarrow$ It is a non-terminating decimal. This number cannot be written as a simple fraction.

$p,q$ and $r$ are all irrational 
Let $p=q=r=\sqrt2$
$p(q+r)=p.q+p.r$
$\Rightarrow \sqrt { 2 } (\sqrt { 2 } +\sqrt { 2 } )=\sqrt { 2 } .\sqrt { 2 } +\sqrt { 2 } .\sqrt { 2 } =2+2=4$
which is rational as well as integer
Let us take another case in which $p=\sqrt2$ and $q=r=\sqrt3$
$\Rightarrow \sqrt { 2 } (\sqrt { 3 } +\sqrt { 3 } )=\sqrt { 2 } .\sqrt { 3 } +\sqrt { 2 } .\sqrt { 3 } =\sqrt { 6 } +\sqrt { 6 } =2\sqrt { 6 } $
which is an irrational number.
So on applying distributive property on three irrational numbers we can get an integer,a rational as well as an irrational number .
So option $D$ is correct.

Sum of rational and an irrational number is rational (i.e., need not to be irrational)

$\text{Here 1 is a rational number and }$$2\sqrt3$ is a $irrational$ number

Assume that $\sqrt p  + \sqrt q $ is rational. So,

$\sqrt p  + \sqrt q  = \frac{a}{b}$

$p + q + 2\sqrt {pq}  = \frac{{{a^2}}}{{{b^2}}}$

$\sqrt {pq}  = \frac{1}{2}\left( {\frac{{{a^2}}}{{{b^2}}} – p - q} \right)$

Since the RHS of the above equation is rational but $\sqrt {pq} $ is an irrational number, so the assumption is wrong.

Therefore, it is true that $\sqrt p  + \sqrt q $ is irrational.

Let us assume that $\sqrt{2}$ is a rational number
$\Rightarrow \sqrt{2}=\dfrac{p}{q}\left [ \dfrac{p}{q}\      is\ in\    simplest\      form\  \right ]$

We have,

$\sqrt{14+6\sqrt 5}$

$\sqrt{(729x^{12})^{\frac{1}{3}}}-\cfrac{x^{-2}-x^{-3}}{x^{-4}-x^{-5}}$

$23.10100100010000...$ will be irrational number because it's non terminating and non repeating.

Multiplicative inverse of $\sqrt2-1$ is $\dfrac{1}{\sqrt2-1}$
Now multiply and divide with $\sqrt2+1$
We get $\dfrac{\sqrt2+1}{(\sqrt2+1)(\sqrt2-1)} $

(a) All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point. Here are some irrational numbers:

$(D)\,\, Real = Rational + Irrational$

If the sum of two  irrational numbers is 0 then the two numbers are called additive inverse.
So option $D$ is correct.

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.

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

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

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

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

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

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

Here 2 is a rational number

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

$Here \  6 \text{ is a  rational number and }$$\sqrt2$ is a $irrational$ number

$\sqrt 7 $ is a rational number

We know that difference of a rational and irrational number is also irrational, 

when we divide $100$ with $99$ we will get remainder $1$ and hence next we divide 99 with $1$ which gives us remainder $0$.