Mix Examples - Number Systems Questions in English

(A) Given expression is $\left[\left(\frac{5}{6}\right)^{\frac{1}{5}}\right]^{-\frac{1}{6}}$.
Using the power rule $(a^m)^n = a^{m \times n}$,we get:
$\left(\frac{5}{6}\right)^{\frac{1}{5} \times (-\frac{1}{6})} = \left(\frac{5}{6}\right)^{-\frac{1}{30}}$.
Now,let's evaluate the options:
$A$: $\left(\frac{5}{6}\right)^{\frac{1}{5}-\frac{1}{6}} = \left(\frac{5}{6}\right)^{\frac{6-5}{30}} = \left(\frac{5}{6}\right)^{\frac{1}{30}}$. This is not equal to $\left(\frac{5}{6}\right)^{-\frac{1}{30}}$.
$B$: $\frac{1}{\left[\left(\frac{5}{6}\right)^{\frac{1}{5}}\right]^{\frac{1}{6}}} = \frac{1}{\left(\frac{5}{6}\right)^{\frac{1}{30}}} = \left(\frac{5}{6}\right)^{-\frac{1}{30}}$. This is equal.
$C$: $\left(\frac{6}{5}\right)^{\frac{1}{30}} = \left(\left(\frac{5}{6}\right)^{-1}\right)^{\frac{1}{30}} = \left(\frac{5}{6}\right)^{-\frac{1}{30}}$. This is equal.
$D$: $\left(\frac{5}{6}\right)^{-\frac{1}{30}}$. This is equal.
Therefore,option $A$ is not equal to the given expression.

(B) We know that the set of real numbers consists of all rational and irrational numbers.
Since every rational number belongs to the set of real numbers,it follows that every rational number is a real number.

(C) Between any two distinct rational numbers,there exist infinitely many rational numbers. This property is known as the density property of rational numbers.

(D) rational number is defined as a number that can be expressed in the form $p/q$,where $p$ and $q$ are integers and $q \neq 0$. The decimal expansion of a rational number is always either terminating or non-terminating repeating (recurring). $A$ non-terminating,non-repeating decimal expansion is a characteristic property of an irrational number. Therefore,the decimal representation of a rational number cannot be non-terminating,non-repeating.

(A) The product of two irrational numbers can be either rational or irrational.
For example:
$1$. If we take $\sqrt{2} \times \sqrt{2} = 2$,which is a rational number.
$2$. If we take $\sqrt{2} \times \sqrt{3} = \sqrt{6}$,which is an irrational number.
Therefore,the product is sometimes rational and sometimes irrational.

(B) The number $\sqrt{2}$ is an irrational number.
By definition,the decimal expansion of an irrational number is always non-terminating and non-recurring (non-repeating).
Therefore,the decimal expansion of $\sqrt{2}$ is $1.41421356...$,which does not end and does not repeat any pattern.

(C) To determine which number is irrational,we simplify each option:
$(a)$ $\sqrt{\frac{4}{9}} = \frac{2}{3}$,which is a rational number because it can be expressed in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$.
$(b)$ $\frac{\sqrt{12}}{\sqrt{3}} = \sqrt{\frac{12}{3}} = \sqrt{4} = 2$,which is a rational number.
$(c)$ $\sqrt{7}$ cannot be simplified into a fraction of integers. Since $7$ is not a perfect square,$\sqrt{7}$ is an irrational number.
$(d)$ $\sqrt{81} = 9$,which is a rational number.
Therefore,$\sqrt{7}$ is the irrational number. Hence,$(C)$ is the correct answer.

(D) number is irrational if and only if its decimal representation is non-terminating and non-recurring.
$(A)$ $0.14$ is a terminating decimal,so it is a rational number.
$(B)$ $0.14 \overline{16}$ is a non-terminating and recurring decimal,so it is a rational number.
$(C)$ $0. \overline{1416}$ is a non-terminating and recurring decimal,so it is a rational number.
$(D)$ $0.4014001400014 \ldots$ is a non-terminating and non-recurring decimal,which is the definition of an irrational number.
Hence,$(D)$ is the correct answer.

(A) We know that $\sqrt{2} \approx 1.414$ and $\sqrt{3} \approx 1.732$.
$A$ rational number is a number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Comparing the given options:
$1.5 = \frac{15}{10} = \frac{3}{2}$,which is a rational number.
Since $1.414 < 1.5 < 1.732$,the value $1.5$ lies between $\sqrt{2}$ and $\sqrt{3}$.
Option $B$ and $C$ involve irrational numbers,so they are not rational numbers.
Option $D$ $(1.8)$ is greater than $\sqrt{3} \approx 1.732$,so it does not lie between them.
Therefore,$1.5$ is the correct rational number.

(B) Let $x = 1.999... = 1.\overline{9} \quad \dots(1)$
Multiply both sides by $10$:
$10x = 19.999... = 19.\overline{9} \quad \dots(2)$
Subtract equation $(1)$ from equation $(2)$:
$10x - x = 19.999... - 1.999...$
$9x = 18$
Divide by $9$:
$x = \frac{18}{9} = 2$
Since $2$ can be written as $\frac{2}{1},$ where $p=2$ and $q=1$ are integers and $q \neq 0,$ the value is $2.$
Hence,option $(b)$ is the correct answer.

(C) To add the given expressions,we treat $\sqrt{3}$ as a common variable or term.
Given expression: $2 \sqrt{3} + \sqrt{3}$.
This can be written as $(2 + 1) \sqrt{3}$.
Calculating the sum inside the parentheses: $3 \sqrt{3}$.
Therefore,the correct option is $(c)$.

(D) Using the property of radicals $\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}$,we have:
$\sqrt{10} \times \sqrt{15} = \sqrt{10 \times 15}$
$= \sqrt{150}$
$= \sqrt{25 \times 6}$
$= \sqrt{25} \times \sqrt{6}$
$= 5 \sqrt{6}$
Therefore,the correct option is $D$.

(A) To rationalize the denominator of $\frac{1}{7-\sqrt{2}}$,we multiply both the numerator and the denominator by the conjugate of the denominator,which is $(7+\sqrt{2})$.
$\frac{1}{7-\sqrt{2}} = \frac{1}{7-\sqrt{2}} \times \frac{7+\sqrt{2}}{7+\sqrt{2}}$
Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator:
$= \frac{7+\sqrt{2}}{(7)^2 - (\sqrt{2})^2}$
$= \frac{7+\sqrt{2}}{49-2}$
$= \frac{7+\sqrt{2}}{47}$
Hence,option $A$ is the correct answer.

(B) Given expression: $\frac{1}{\sqrt{9}-\sqrt{8}}$
Since $\sqrt{9} = 3$ and $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$,the expression becomes:
$\frac{1}{3-2\sqrt{2}}$
To rationalize the denominator,multiply the numerator and denominator by the conjugate $(3+2\sqrt{2})$:
$\frac{1}{3-2\sqrt{2}} \times \frac{3+2\sqrt{2}}{3+2\sqrt{2}} = \frac{3+2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2}$
Simplify the denominator:
$(3)^2 - (2\sqrt{2})^2 = 9 - (4 \times 2) = 9 - 8 = 1$
Thus,the expression simplifies to:
$\frac{3+2\sqrt{2}}{1} = 3+2\sqrt{2}$
Hence,option $(B)$ is the correct answer.

(C) To rationalize the denominator of $\frac{7}{3 \sqrt{3}-2 \sqrt{2}},$ we multiply the numerator and the denominator by the conjugate of the denominator,which is $(3 \sqrt{3}+2 \sqrt{2}).$
$\frac{7}{3 \sqrt{3}-2 \sqrt{2}} \times \frac{3 \sqrt{3}+2 \sqrt{2}}{3 \sqrt{3}+2 \sqrt{2}} = \frac{7(3 \sqrt{3}+2 \sqrt{2})}{(3 \sqrt{3})^2 - (2 \sqrt{2})^2}$
$= \frac{7(3 \sqrt{3}+2 \sqrt{2})}{(9 \times 3) - (4 \times 2)} = \frac{7(3 \sqrt{3}+2 \sqrt{2})}{27 - 8}$
$= \frac{7(3 \sqrt{3}+2 \sqrt{2})}{19}$
Thus,the denominator is $19.$
Therefore,the correct option is $(C)$.

(D) Given expression: $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$
Simplify the square roots in the numerator and denominator:
$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$
$\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
$\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$
$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$
Substitute these values back into the expression:
$\frac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}$
Factor out the common terms:
$\frac{4(\sqrt{2} + \sqrt{3})}{2(\sqrt{2} + \sqrt{3})}$
Cancel the common term $(\sqrt{2} + \sqrt{3})$:
$\frac{4}{2} = 2$
Hence,the correct option is $(d)$.

(A) To simplify the expression $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$,we rationalize the denominator:
$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} = \sqrt{\frac{(\sqrt{2}-1) \times (\sqrt{2}-1)}{(\sqrt{2}+1) \times (\sqrt{2}-1)}}$
$= \sqrt{\frac{(\sqrt{2}-1)^2}{(\sqrt{2})^2 - 1^2}}$
$= \sqrt{\frac{(\sqrt{2}-1)^2}{2-1}}$
$= \sqrt{(\sqrt{2}-1)^2}$
$= \sqrt{2}-1$
Given $\sqrt{2} = 1.4142$,we have:
$1.4142 - 1 = 0.4142$
Hence,the correct option is $(A)$.

(B) To solve the expression $\sqrt[4]{\sqrt[3]{2^{2}}}$,we use the laws of exponents.
First,express the radicals as fractional exponents: $\sqrt[n]{x} = x^{\frac{1}{n}}$.
$\sqrt[4]{\sqrt[3]{2^{2}}} = \sqrt[4]{(2^{2})^{\frac{1}{3}}}$.
Using the power of a power rule $(a^{m})^{n} = a^{m \times n}$,we get $(2^{2})^{\frac{1}{3}} = 2^{\frac{2}{3}}$.
Now,apply the outer radical: $(2^{\frac{2}{3}})^{\frac{1}{4}} = 2^{\frac{2}{3} \times \frac{1}{4}}$.
Simplifying the exponent: $\frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}$.
Thus,the expression simplifies to $2^{\frac{1}{6}}$.
Hence,$(B)$ is the correct answer.

(C) We have,
$\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}} \cdot (2^5)^{\frac{1}{12}}$
$= 2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}} \cdot 2^{\frac{5}{12}}$
Using the law of exponents $a^m \cdot a^n = a^{m+n}$,we get:
$= 2^{\frac{1}{3} + \frac{1}{4} + \frac{5}{12}}$
$= 2^{\frac{4+3+5}{12}}$
$= 2^{\frac{12}{12}}$
$= 2^1 = 2$
Hence,$(c)$ is the correct answer.

(D) Given expression: $\sqrt[4]{(81)^{-2}}$
We know that $a^{-n} = \frac{1}{a^n}$,so $(81)^{-2} = \frac{1}{81^2}$.
Thus,$\sqrt[4]{(81)^{-2}} = \sqrt[4]{\frac{1}{81^2}} = \left(\frac{1}{81^2}\right)^{1/4}$.
Since $81 = 9^2$,we have $81^2 = (9^2)^2 = 9^4$.
Substituting this back,we get $\left(\frac{1}{9^4}\right)^{1/4} = \left(\frac{1}{9}\right)^{4 \times \frac{1}{4}} = \frac{1}{9}$.
Hence,$(d)$ is the correct answer.

(A) Using the law of exponents,$a^m \times a^n = a^{m+n}$.
Given expression: $(256)^{0.16} \times (256)^{0.09} = (256)^{0.16 + 0.09}$.
Adding the exponents: $(256)^{0.25}$.
Since $0.25 = \frac{1}{4}$,we have $(256)^{\frac{1}{4}}$.
Expressing $256$ as a power of $4$: $256 = 4^4$.
Therefore,$(4^4)^{\frac{1}{4}} = 4^{4 \times \frac{1}{4}} = 4^1 = 4$.
Hence,the correct option is $A$.

(B) We evaluate each option to see which simplifies to $x^1 = x$:
$(A)$ $x^{\frac{12}{7}} - x^{\frac{5}{7}}$ cannot be simplified to $x$.
$(B)$ $\left(\sqrt{x^{3}}\right)^{\frac{2}{3}} = \left((x^3)^{\frac{1}{2}}\right)^{\frac{2}{3}} = x^{3 \times \frac{1}{2} \times \frac{2}{3}} = x^1 = x$.
$(C)$ $\sqrt[12]{\left(x^{4}\right)^{\frac{1}{3}}} = \left(x^{4 \times \frac{1}{3}}\right)^{\frac{1}{12}} = x^{\frac{4}{3} \times \frac{1}{12}} = x^{\frac{1}{9}} \neq x$.
$(D)$ $x^{\frac{12}{7}} \times x^{\frac{7}{12}} = x^{\frac{12}{7} + \frac{7}{12}} = x^{\frac{144+49}{84}} = x^{\frac{193}{84}} \neq x$.
Thus,option $(B)$ is the correct answer.

(N/A) Yes,there exist such irrational numbers.
Consider two irrational numbers $a = 3 + \sqrt{2}$ and $b = 3 - \sqrt{2}$.
Sum: $(3 + \sqrt{2}) + (3 - \sqrt{2}) = 6$,which is a rational number.
Product: $(3 + \sqrt{2}) \times (3 - \sqrt{2}) = (3)^2 - (\sqrt{2})^2 = 9 - 2 = 7$,which is also a rational number.
Thus,$3 + \sqrt{2}$ and $3 - \sqrt{2}$ are two irrational numbers whose sum and product are both rational.

(TRUE) The statement is true.
Let us consider $x = \sqrt[4]{2}$.
Now,calculate $x^{2}$:
$x^{2} = (\sqrt[4]{2})^{2} = \sqrt{2}$. Since $\sqrt{2}$ cannot be expressed as a ratio of two integers,it is an irrational number.
Next,calculate $x^{4}$:
$x^{4} = (\sqrt[4]{2})^{4} = 2$. Since $2$ can be expressed as $\frac{2}{1}$,it is a rational number.
Thus,we have found a number $x = \sqrt[4]{2}$ such that $x^{2}$ is irrational and $x^{4}$ is rational.

(A) Yes,$x+y$ is necessarily an irrational number.
Proof by contradiction:
Suppose $x+y = r$,where $r$ is a rational number.
Since $x$ is a rational number,we can write $y = r - x$.
Since the difference of two rational numbers ($r$ and $x$) is always a rational number,$y$ must be a rational number.
However,this contradicts the given information that $y$ is an irrational number.
Therefore,our assumption is wrong,and $x+y$ must be an irrational number.
Example:
Let $x = 5$ (rational) and $y = \sqrt{2}$ (irrational).
Then,$x+y = 5 + \sqrt{2} = 6.4142...$,which is a non-terminating and non-repeating decimal.
Hence,$x+y$ is an irrational number.

(NO) Let $x = 0$ (a rational number) and $y = \sqrt{3}$ (an irrational number).
Then,the product $xy = 0 \times \sqrt{3} = 0$.
Since $0$ can be expressed as $\frac{0}{1}$,it is a rational number.
Therefore,$xy$ is not necessarily an irrational number.

(FALSE, FALSE) $(i)$ The given statement is false. $A$ rational number is defined as a number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$. Here,$\frac{\sqrt{2}}{3}$ has $p = \sqrt{2}$,which is an irrational number,not an integer. Therefore,$\frac{\sqrt{2}}{3}$ is an irrational number.
$(ii)$ The given statement is false. By definition,integers are whole numbers (...,$-2, -1, 0, 1, 2, ...$). Between any two consecutive integers,such as $3$ and $4$,there are no other integers.

(N/A) $(i)$ The statement is false. There are infinitely many rational numbers between any two distinct rational numbers. Therefore,the number of rational numbers between $15$ and $18$ is infinite.
$(ii)$ The statement is true. Numbers that cannot be expressed in the form $\frac{p}{q}$ (where $p, q$ are integers and $q \neq 0$) are called irrational numbers. For example,$\sqrt{2}$,$\sqrt{3}$,and $\pi$ are such numbers.

(N/A) $(i)$ The statement is false. Consider the irrational number $\sqrt[4]{2}$. Its square is $(\sqrt[4]{2})^{2} = \sqrt{2}$,which is an irrational number.
$(ii)$ The statement is false. We can simplify the expression as $\frac{\sqrt{12}}{\sqrt{3}} = \sqrt{\frac{12}{3}} = \sqrt{4} = 2$. Since $2$ can be written as $\frac{2}{1}$,it is a rational number.

(B) The given statement is false.
To determine if the number is rational,we must simplify the expression first.
$\frac{\sqrt{15}}{\sqrt{3}} = \sqrt{\frac{15}{3}} = \sqrt{5}$.
Since $\sqrt{5}$ is not a perfect square,it is an irrational number.
$A$ rational number is defined as a number that can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.
In the expression $\frac{\sqrt{5}}{1}$,although it is in the form $\frac{p}{q}$,the numerator $p = \sqrt{5}$ is not an integer.
Therefore,$\frac{\sqrt{15}}{\sqrt{3}}$ is an irrational number.

(N/A) $(i)$ $\sqrt{196} = 14$. Since $14$ can be expressed in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$ (i.e.,$\frac{14}{1}$),it is a rational number.
$(ii)$ $3\sqrt{18} = 3\sqrt{9 \times 2} = 3 \times 3\sqrt{2} = 9\sqrt{2}$. Since $\sqrt{2}$ is an irrational number and the product of a non-zero rational number and an irrational number is always irrational,$9\sqrt{2}$ is an irrational number.

(N/A) $(i)$ $\sqrt{\frac{9}{27}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$. Since $\sqrt{3}$ is an irrational number,the quotient of a rational number $(1)$ and an irrational number $(\sqrt{3})$ is an irrational number.
$(ii)$ $\frac{\sqrt{28}}{\sqrt{343}} = \sqrt{\frac{28}{343}} = \sqrt{\frac{4 \times 7}{49 \times 7}} = \sqrt{\frac{4}{49}} = \frac{2}{7}$. Since $\frac{2}{7}$ is in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$,it is a rational number.

(N/A) $(i)$ $-\sqrt{0.4} = -\sqrt{\frac{4}{10}} = -\frac{2}{\sqrt{10}}$. Since $\sqrt{10}$ is an irrational number,the quotient of a rational number $(2)$ and an irrational number $(\sqrt{10})$ is irrational. Therefore,$-\sqrt{0.4}$ is an irrational number.
$(ii)$ $\frac{\sqrt{12}}{\sqrt{75}} = \sqrt{\frac{12}{75}} = \sqrt{\frac{4}{25}} = \frac{2}{5}$. Since $\frac{2}{5}$ is in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$,it is a rational number.

(N/A) $(i)$ $0.5918$ is a terminating decimal expansion. Any number that can be expressed as a terminating decimal is a rational number because it can be written in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$. Here,$0.5918 = \frac{5918}{10000}$,which is rational.
$(ii)$ Simplify the expression: $(1+\sqrt{5})-(4+\sqrt{5}) = 1 + \sqrt{5} - 4 - \sqrt{5} = 1 - 4 = -3$. Since $-3$ can be written as $\frac{-3}{1}$,which is in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$,it is a rational number.

(N/A) $(i)$ $10.124124 \ldots$ is a decimal expansion which is non-terminating and recurring. Since a number with a non-terminating recurring decimal expansion can be expressed in the form $p/q$ where $p, q$ are integers and $q \neq 0$,it is a rational number.
$(ii)$ $1.010010001 \ldots$ is a decimal expansion which is non-terminating and non-recurring. Since it cannot be expressed in the form $p/q$ where $p, q$ are integers and $q \neq 0$,it is an irrational number.

(N/A) We write $13$ as the sum of the squares of two natural numbers:
$13 = 9 + 4 = 3^{2} + 2^{2}$
On the number line,take $OA = 3$ units.
Draw $BA = 2$ units,perpendicular to $OA$. Join $OB$.
By Pythagoras theorem,$OB = \sqrt{OA^{2} + AB^{2}} = \sqrt{3^{2} + 2^{2}} = \sqrt{9 + 4} = \sqrt{13}$.
Using a compass with centre $O$ and radius $OB$,draw an arc which intersects the number line at the point $C$. Then,$C$ corresponds to $\sqrt{13}$.

(B) Let $x = 0.12333\ldots$ (Equation $1$)
Multiply by $10$ to shift the decimal point: $10x = 1.2333\ldots$ (Equation $2$)
Multiply Equation $1$ by $100$ to shift the decimal point past the non-repeating part: $100x = 12.333\ldots$ (Equation $3$)
Subtract Equation $2$ from Equation $3$:
$100x - 10x = 12.333\ldots - 1.2333\ldots$
$90x = 11.1$
$x = \frac{11.1}{90} = \frac{111}{900}$
Simplify the fraction by dividing both numerator and denominator by $3$:
$x = \frac{111 \div 3}{900 \div 3} = \frac{37}{300}$
Thus,$0.12 \overline{3} = \frac{37}{300}$.

(A) To simplify the expression $(3 \sqrt{5}-5 \sqrt{2})(4 \sqrt{5}+3 \sqrt{2})$,we use the distributive property ($FOIL$ method):
$= (3 \sqrt{5} \times 4 \sqrt{5}) + (3 \sqrt{5} \times 3 \sqrt{2}) - (5 \sqrt{2} \times 4 \sqrt{5}) - (5 \sqrt{2} \times 3 \sqrt{2})$
$= (12 \times 5) + (9 \sqrt{10}) - (20 \sqrt{10}) - (15 \times 2)$
$= 60 + 9 \sqrt{10} - 20 \sqrt{10} - 30$
$= (60 - 30) + (9 \sqrt{10} - 20 \sqrt{10})$
$= 30 - 11 \sqrt{10}$

(B) To solve $\frac{6}{3 \sqrt{2}-2 \sqrt{3}}$,we rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(3 \sqrt{2}+2 \sqrt{3})$.
$\frac{6}{3 \sqrt{2}-2 \sqrt{3}} = \frac{6(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2}-2 \sqrt{3})(3 \sqrt{2}+2 \sqrt{3})}$
Using the identity $(x-y)(x+y) = x^2 - y^2$,the denominator becomes $(3 \sqrt{2})^2 - (2 \sqrt{3})^2 = (9 \times 2) - (4 \times 3) = 18 - 12 = 6$.
So,the expression simplifies to $\frac{6(3 \sqrt{2}+2 \sqrt{3})}{6} = 3 \sqrt{2} + 2 \sqrt{3}$.
Comparing this with the given equation $3 \sqrt{2} - a \sqrt{3} = 3 \sqrt{2} + 2 \sqrt{3}$,we get $-a \sqrt{3} = 2 \sqrt{3}$.
Dividing both sides by $-\sqrt{3}$,we find $a = -2$.

(C) Given expression: $\left[5\left(8^{\frac{1}{3}}+27^{\frac{1}{3}}\right)^{3}\right]^{\frac{1}{4}}$
First,simplify the terms inside the parentheses:
$8^{\frac{1}{3}} = (2^3)^{\frac{1}{3}} = 2$
$27^{\frac{1}{3}} = (3^3)^{\frac{1}{3}} = 3$
Substitute these values back into the expression:
$= \left[5(2+3)^{3}\right]^{\frac{1}{4}}$
$= \left[5(5)^{3}\right]^{\frac{1}{4}}$
$= \left[5^1 \cdot 5^3\right]^{\frac{1}{4}}$
$= \left[5^{1+3}\right]^{\frac{1}{4}}$
$= \left[5^4\right]^{\frac{1}{4}}$
$= 5^{4 \cdot \frac{1}{4}}$
$= 5^1 = 5$

(A) $(i)$ $x^{2}=5 \Rightarrow x=\sqrt{5},$ which is an irrational number because $\sqrt{5}$ cannot be expressed in the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$.
$(ii)$ $y^{2}=9 \Rightarrow y=\sqrt{9}=3,$ which is a rational number because it can be expressed as $\frac{3}{1}$.
$(iii)$ $z^{2}=0.04 \Rightarrow z=\sqrt{0.04}=0.2,$ which is a rational number because it is a terminating decimal and can be written as $\frac{2}{10} = \frac{1}{5}$.
$(iv)$ $u^{2}=\frac{17}{4} \Rightarrow u=\sqrt{\frac{17}{4}}=\frac{\sqrt{17}}{2}.$ Since $\sqrt{17}$ is not an integer,$u$ is an irrational number.

(N/A) To find rational numbers between $-1$ and $-2$,we can express them as fractions with a common denominator.
Let the numbers be $-1 = -10/10$ and $-2 = -20/10$.
Any three numbers between $-20/10$ and $-10/10$ are rational numbers.
For example,$-11/10, -12/10, -13/10$ are three rational numbers between $-1$ and $-2$.
Alternatively,these can be written as $-1.1, -1.2, -1.3$.

(A) To find rational numbers between $0.1$ and $0.11$,we can express them as decimals with more digits.
$0.1 = 0.100$
$0.11 = 0.110$
Now,we can easily identify numbers between $0.100$ and $0.110$.
Three such rational numbers are $0.101, 0.102,$ and $0.103$.

(A) To find three rational numbers between $\frac{5}{7}$ and $\frac{6}{7}$,we can multiply the numerator and denominator of both fractions by $(3+1) = 4$ or any larger number like $10$.
Multiplying by $10$:
$\frac{5}{7} = \frac{5 \times 10}{7 \times 10} = \frac{50}{70}$
$\frac{6}{7} = \frac{6 \times 10}{7 \times 10} = \frac{60}{70}$
Now,we can choose any three rational numbers between $\frac{50}{70}$ and $\frac{60}{70}$,such as $\frac{51}{70}, \frac{52}{70}, \text{ and } \frac{53}{70}$.
Thus,three rational numbers between $\frac{5}{7}$ and $\frac{6}{7}$ are $\frac{51}{70}, \frac{52}{70}, \text{ and } \frac{53}{70}$.

(B) To find rational numbers between $\frac{1}{4}$ and $\frac{1}{5}$,we first make their denominators equal.
Find the Least Common Multiple $(LCM)$ of $4$ and $5$,which is $20$.
$\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}$
$\frac{1}{5} = \frac{1 \times 4}{5 \times 4} = \frac{4}{20}$
To find three rational numbers,we multiply the numerator and denominator by $(3+1) = 4$.
$\frac{5}{20} = \frac{5 \times 4}{20 \times 4} = \frac{20}{80}$
$\frac{4}{20} = \frac{4 \times 4}{20 \times 4} = \frac{16}{80}$
The rational numbers between $\frac{16}{80}$ and $\frac{20}{80}$ are $\frac{17}{80}, \frac{18}{80}, \frac{19}{80}$.
Thus,three rational numbers between $\frac{1}{4}$ and $\frac{1}{5}$ are $\frac{17}{80}, \frac{18}{80}, \frac{19}{80}$ (or $\frac{9}{40}$).

(A) rational number is a number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$. The average of $2$ and $3$ is $\frac{2+3}{2} = 2.5$,which is a rational number.
An irrational number is a number that cannot be expressed in the form $\frac{p}{q}$ and has a non-terminating,non-recurring decimal expansion.
An example of an irrational number between $2$ and $3$ is $2.1010010001...$ (a non-terminating and non-recurring decimal).

(B) $0.04$ is a terminating decimal and it lies between $0$ and $0.1$. Hence,$0.04$ is a rational number which lies between $0$ and $0.1$.
Again,$0.003000300003...$ is a non-terminating and non-recurring decimal which lies between $0$ and $0.1$. Hence,$0.003000300003...$ is an irrational number between $0$ and $0.1$.

(N/A) To find a rational number between $\frac{1}{3}$ and $\frac{1}{2}$,we can express them with a common denominator:
$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$ and $\frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12}$.
Since $\frac{4}{12} < \frac{5}{12} < \frac{6}{12}$,the rational number $\frac{5}{12}$ lies between $\frac{1}{3}$ and $\frac{1}{2}$.
To find an irrational number,we convert the fractions to decimal form:
$\frac{1}{3} = 0.3333\ldots$ and $\frac{1}{2} = 0.5$.
An irrational number is a non-terminating and non-recurring decimal. We can choose a number like $0.414114111\ldots$,which is greater than $0.3333\ldots$ and less than $0.5$.

(N/A) First,convert the fractions to decimal form:
$\frac{-2}{5} = -0.4$ and $\frac{1}{2} = 0.5$
$1$. Rational number:
$A$ rational number is any number that can be expressed as $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$. Any terminating decimal is a rational number. We can choose $0$ (or $0.1$,$0.2$,etc.) as it lies between $-0.4$ and $0.5$.
Thus,$0$ is a rational number between $\frac{-2}{5}$ and $\frac{1}{2}$.
$2$. Irrational number:
An irrational number is a non-terminating and non-recurring decimal. We can construct one by following a pattern that does not repeat. For example,$0.1010010001...$ lies between $-0.4$ and $0.5$.
Thus,$0.1010010001...$ is an irrational number between $\frac{-2}{5}$ and $\frac{1}{2}$.

(N/A) rational number is a number that can be expressed in the form $p/q$ where $p$ and $q$ are integers and $q \neq 0$. Any terminating decimal is a rational number.
$1.$ To find a rational number between $0.15$ and $0.16$,we can choose any terminating decimal such as $0.151$. Since $0.15 < 0.151 < 0.16$,$0.151$ is a rational number.
$2.$ An irrational number is a number whose decimal expansion is non-terminating and non-recurring. To find an irrational number between $0.15$ and $0.16$,we can construct a pattern that does not repeat.
$0.15101101110...$ is an irrational number because it is non-terminating and non-recurring,and it lies between $0.15$ and $0.16$.