Irrational Numbers Questions in English

(A) Yes,zero is a rational number. $A$ rational number is defined as any number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \ne 0$. Since zero can be written as $\frac{0}{1}$,$\frac{0}{2}$,$\frac{0}{3}$,etc.,where the numerator $p = 0$ (an integer) and the denominator $q$ is any non-zero integer,it satisfies the definition of a rational number.

(N/A) There are infinitely many rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$.
To find $5$ rational numbers,we can multiply the numerator and denominator of both fractions by $(5 + 1) = 6$.
$\frac{3}{5} = \frac{3 \times 6}{5 \times 6} = \frac{18}{30}$
$\frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30}$
Therefore,$5$ rational numbers between $\frac{18}{30}$ and $\frac{24}{30}$ are:
$\frac{19}{30}, \frac{20}{30}, \frac{21}{30}, \frac{22}{30}, \frac{23}{30}$.

It is easy to see how the Greeks might have discovered $\sqrt{2}$. Consider a square $OABC$,with each side $1$ unit in length. Then you can see by the Pythagoras theorem that $OB = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$.
To represent $\sqrt{2}$ on the number line,place the square $OABC$ on the number line such that the vertex $O$ coincides with zero and the side $OA$ lies along the positive direction of the number line.
We have just seen that $OB = \sqrt{2}$. Using a compass with centre $O$ and radius $OB$,draw an arc intersecting the number line at the point $P$. Then the point $P$ corresponds to $\sqrt{2}$ on the number line.

(N/A) To locate $\sqrt{3}$ on the number line,follow these steps:
$1$. First,locate $\sqrt{2}$ on the number line by constructing a right-angled triangle with base $1$ unit and height $1$ unit. The hypotenuse will be $\sqrt{1^2 + 1^2} = \sqrt{2}$.
$2$. Now,construct a line segment $BD$ of unit length ($1$ unit) perpendicular to the hypotenuse $OB$ (where $OB = \sqrt{2}$).
$3$. Join $OD$. In the right-angled triangle $\triangle OBD$,by the Pythagoras theorem:
$OD^2 = OB^2 + BD^2$
$OD^2 = (\sqrt{2})^2 + 1^2 = 2 + 1 = 3$
$OD = \sqrt{3}$.
$4$. Using a compass,with centre $O$ and radius $OD$,draw an arc that intersects the number line at point $Q$. The point $Q$ represents $\sqrt{3}$ on the number line.
In the same way,you can locate $\sqrt{n}$ for any positive integer $n$,after $\sqrt{n-1}$ has been located.

(B) No,the square roots of all positive integers are not irrational.
For example,consider the square roots of perfect squares such as $\sqrt{4}$ and $\sqrt{9}$.
We know that $\sqrt{4} = 2$ and $\sqrt{9} = 3$.
Since $2$ and $3$ can be expressed in the form $\frac{p}{q}$ (where $p$ and $q$ are integers and $q \neq 0$),they are rational numbers.
Therefore,the square roots of all positive integers are not irrational.

(N/A) To represent $\sqrt{5}$ on the number line,we use the Pythagorean theorem: $\sqrt{5} = \sqrt{2^2 + 1^2}$.
$1$. Draw a number line and mark a point $O$ representing $0$ and a point $A$ representing $2$ units from $O$.
$2$. At point $A$,construct a perpendicular line segment $AB$ of length $1$ unit.
$3$. Join $O$ and $B$. In the right-angled triangle $\triangle OAB$,by the Pythagorean theorem:
$OB^2 = OA^2 + AB^2$
$OB^2 = 2^2 + 1^2 = 4 + 1 = 5$
$OB = \sqrt{5}$
$4$. Now,with $O$ as the center and $OB$ as the radius,draw an arc that intersects the number line at point $C$.
$5$. The distance $OC$ is equal to $OB$,which is $\sqrt{5}$. Thus,point $C$ represents $\sqrt{5}$ on the number line.

(N/A) The construction of the square root spiral is based on the Pythagorean theorem,which states that in a right-angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides $(h^2 = a^2 + b^2)$.
$1$. Start with point $O$ and draw $OP_1 = 1$ unit.
$2$. Draw $P_1P_2$ perpendicular to $OP_1$ such that $P_1P_2 = 1$ unit. In $\triangle OP_1P_2$,by Pythagoras theorem,$OP_2 = \sqrt{OP_1^2 + P_1P_2^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
$3$. Draw $P_2P_3$ perpendicular to $OP_2$ such that $P_2P_3 = 1$ unit. In $\triangle OP_2P_3$,$OP_3 = \sqrt{OP_2^2 + P_2P_3^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2 + 1} = \sqrt{3}$.
$4$. Similarly,draw $P_3P_4$ perpendicular to $OP_3$ such that $P_3P_4 = 1$ unit. In $\triangle OP_3P_4$,$OP_4 = \sqrt{OP_3^2 + P_3P_4^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
$5$. By continuing this process,the length of the hypotenuse $OP_n$ will be $\sqrt{n}$.

(A) We know that $\frac{1}{7} = 0.\overline{142857}$ and $\frac{2}{7} = 0.\overline{285714}$.
To find an irrational number between these two values,we need a number that is non-terminating and non-recurring.
Any number of the form $0.15015001500015...$ satisfies this condition because it lies between $0.142857...$ and $0.285714...$ and does not repeat in a fixed pattern.
Thus,one such irrational number is $0.15015001500015...$.

(N/A) number whose decimal expansion is non-terminating and non-recurring is an irrational number.
Three examples of such numbers are:
$1$. $\sqrt{2} = 1.414213562 \ldots$
$2$. $\sqrt{3} = 1.732050808 \ldots$
$3$. $\sqrt{5} = 2.236067978 \ldots$

(N/A) First,convert the rational numbers into decimal form:
$\frac{5}{7} = 0.\overline{714285}$
$\frac{9}{11} = 0.\overline{81}$
An irrational number is a non-terminating and non-repeating decimal.
We need to find three numbers between $0.714285...$ and $0.818181...$ that follow a non-terminating,non-repeating pattern.
Three such examples are:
$1) 0.73073007300073...$
$2) 0.75075007500075...$
$3) 0.79079007900079...$

(N/A) To determine if a number is irrational,we check if it can be expressed as a non-terminating,non-recurring decimal.
$1$. $7 \sqrt{5}$: Since $\sqrt{5} \approx 2.236$ is an irrational number,the product of a non-zero rational number $(7)$ and an irrational number $(\sqrt{5})$ is always irrational. Thus,$7 \sqrt{5} \approx 15.652...$ is irrational.
$2$. $\frac{7}{\sqrt{5}}$: Rationalizing the denominator,we get $\frac{7 \sqrt{5}}{5} \approx 3.1304...$. Since the quotient of a non-zero rational number and an irrational number is irrational,$\frac{7}{\sqrt{5}}$ is irrational.
$3$. $\sqrt{2}+21$: Since $\sqrt{2} \approx 1.414$ is irrational,the sum of an irrational number and a rational number $(21)$ is always irrational. Thus,$\sqrt{2}+21 \approx 22.414...$ is irrational.
$4$. $\pi-2$: Since $\pi \approx 3.1415...$ is an irrational number,the difference between an irrational number and a rational number $(2)$ is always irrational. Thus,$\pi-2 \approx 1.1415...$ is irrational.
Conclusion: All the given numbers are 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.

(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.

(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) $(i)$ $\sqrt{\frac{9}{27}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$. 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) 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 $\sqrt{13}$.

(N/A) When we measure the length of a line with a scale or any other device,we only obtain an approximate rational value. This implies that at least one of $c$ or $d$ is irrational. Therefore,the ratio $\frac{c}{d}$ is irrational,which makes $\pi$ an irrational number. Thus,there is no contradiction in stating that $\pi$ is irrational.

(N/A) To represent $\sqrt{9.3}$ on the number line,follow these steps:
$1$. Draw a line segment $AB = 9.3 \text{ units}$ on a line $l$.
$2$. From point $B$,mark a point $C$ such that $BC = 1 \text{ unit}$. Now,$AC = 9.3 + 1 = 10.3 \text{ units}$.
$3$. Find the midpoint of $AC$,let it be $O$. Draw a semicircle with center $O$ and radius $OA = OC$.
$4$. Draw a perpendicular line at point $B$ to the line $l$,which intersects the semicircle at point $D$. The length of $BD$ is $\sqrt{9.3}$.
$5$. With $B$ as the center and $BD$ as the radius,draw an arc on the number line $l$ to intersect it at point $E$. The distance $BE$ represents $\sqrt{9.3}$ on the number line.

(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.

(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) $6.3753$ (a terminating decimal) is a rational number between $6.375289$ and $6.375738$.
$6.375414114111...$ (a non-terminating and non-recurring decimal) is an irrational number lying between $6.375289$ and $6.375738$.

(N/A) $1$. Draw a line segment $AB = 4.5$ units on a number line.
$2$. From point $B$,mark a distance of $1$ unit and label the new point as $C$. Now,$AC = 4.5 + 1 = 5.5$ units.
$3$. Find the midpoint of $AC$ and label it as $O$.
$4$. Draw a semicircle with center $O$ and radius $OC$ (or $OA$).
$5$. Draw a line perpendicular to $AC$ at point $B$,which intersects the semicircle at point $D$. The length of $BD$ is $\sqrt{4.5}$.
$6$. With $B$ as the center and $BD$ as the radius,draw an arc that intersects the number line at point $E$. The point $E$ represents $\sqrt{4.5}$ on the number line.

(N/A) To represent $\sqrt{5.6}$ on the number line,follow these steps:
$1$. Draw a line segment $AB = 5.6 \text{ units}$ on the number line.
$2$. From point $B$,mark a point $C$ such that $BC = 1 \text{ unit}$. Now,$AC = 5.6 + 1 = 6.6 \text{ units}$.
$3$. Find the midpoint $O$ of $AC$. The length $AO = OC = 6.6 / 2 = 3.3 \text{ units}$.
$4$. Draw a semicircle with center $O$ and radius $OA = 3.3 \text{ units}$.
$5$. Draw a perpendicular line at point $B$ to the line $AC$,which intersects the semicircle at point $D$.
$6$. The length $BD$ is equal to $\sqrt{5.6}$.
$7$. With $B$ as the center and $BD$ as the radius,draw an arc on the number line to intersect it at point $E$. The distance $BE$ represents $\sqrt{5.6}$ on the number line.

(N/A) $1$. Draw a line segment $AB = 2.3$ units on a line.
$2$. From $B$,mark a distance of $1$ unit and label the new point as $C$. Now,$AC = 2.3 + 1 = 3.3$ units.
$3$. Find the midpoint of $AC$ and mark it as $O$.
$4$. Draw a semicircle with center $O$ and radius $OC$.
$5$. Draw a line perpendicular to $AC$ passing through $B$ that intersects the semicircle at point $D$. The length $BD$ is equal to $\sqrt{2.3}$.
$6$. With $B$ as the center and $BD$ as the radius,draw an arc that intersects the number line at point $E$. The point $E$ represents $\sqrt{2.3}$ on the number line.

(N/A) $1$. Select a point $O$ on the number line representing $0$.
$2$. Select a unit length and mark point $A$ on the number line at distance $1$ from $O$.
$3$. Draw a perpendicular segment $AB$ of unit length ($1$ unit) at point $A$.
$4$. Join $OB$. By Pythagoras's theorem,$OB = \sqrt{OA^2 + AB^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
$5$. Draw a perpendicular segment $BC$ of unit length at point $B$. Join $OC$. Then $OC = \sqrt{OB^2 + BC^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3}$.
$6$. Draw a perpendicular segment $CD$ of unit length at point $C$. Join $OD$. Then $OD = \sqrt{OC^2 + CD^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$.
$7$. Draw a perpendicular segment $DE$ of unit length at point $D$. Join $OE$. Then $OE = \sqrt{OD^2 + DE^2} = \sqrt{2^2 + 1^2} = \sqrt{5}$.
$8$. With $O$ as the center and $OE$ as the radius,draw an arc that intersects the number line at point $P$.
$9$. The point $P$ on the number line represents $\sqrt{5}$.

(A) To represent $\sqrt{6}$ on the number line,we use the Pythagorean theorem: $a^2 + b^2 = c^2$.
We can write $6 = 5 + 1$,so $\sqrt{6} = \sqrt{(\sqrt{5})^2 + 1^2}$.
$1$. First,represent $\sqrt{5}$ on the number line by constructing a right-angled triangle with base $2$ units and height $1$ unit.
$2$. Mark a point $O$ at $0$ and $A$ at $2$ on the number line.
$3$. Draw a perpendicular $AB$ of length $1$ unit at point $A$.
$4$. Join $OB$. By Pythagoras,$OB = \sqrt{2^2 + 1^2} = \sqrt{5}$.
$5$. Now,draw a perpendicular $BC$ of length $1$ unit at point $B$.
$6$. Join $OC$. By Pythagoras,$OC = \sqrt{(\sqrt{5})^2 + 1^2} = \sqrt{6}$.
$7$. With $O$ as the center and $OC$ as the radius,draw an arc that intersects the number line at point $P$. The distance $OP$ represents $\sqrt{6}$.

(N/A) To represent $\sqrt{10}$ on the number line,we use the Pythagorean theorem: $a^2 + b^2 = c^2$.
We can write $10$ as $3^2 + 1^2 = 10$,so $\sqrt{10} = \sqrt{3^2 + 1^2}$.
Step $1$: Draw a number line and mark a point $O$ at $0$ and a point $A$ at $3$ units from $O$.
Step $2$: At point $A$,construct a perpendicular line segment $AB$ of length $1$ unit.
Step $3$: Join $O$ and $B$. By the Pythagorean theorem,the length of $OB$ is $\sqrt{3^2 + 1^2} = \sqrt{10}$.
Step $4$: With $O$ as the center and $OB$ as the radius,draw an arc that intersects the number line at point $P$.
The point $P$ represents $\sqrt{10}$ on the number line.

(N/A) To represent $\sqrt{20}$ on the number line,we can use the Pythagorean theorem: $a^2 + b^2 = c^2$.
We can express $20$ as $4^2 + 2^2 = 16 + 4 = 20$.
$1$. Draw a number line and mark a point $O$ at $0$ and a point $A$ at $4$ units from $O$.
$2$. At point $A$,draw a perpendicular line segment $AB$ of length $2$ units.
$3$. Join $O$ and $B$. By the Pythagorean theorem,the length of $OB$ is $\sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20}$.
$4$. Using a compass with center $O$ and radius $OB$,draw an arc that intersects the number line at point $P$.
$5$. The point $P$ represents $\sqrt{20}$ on the number line.

(N/A) To construct a square root spiral up to $\sqrt{6}$, follow these steps:
$1$. Draw a line segment $OA$ of length $1 \text{ unit}$.
$2$. At point $A$, draw a perpendicular line $AB$ of length $1 \text{ unit}$. Join $OB$. By the Pythagorean theorem, $OB = \sqrt{1^2 + 1^2} = \sqrt{2}$.
$3$. At point $B$, draw a perpendicular line $BC$ of length $1 \text{ unit}$ to $OB$. Join $OC$. Then $OC = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3}$.
$4$. At point $C$, draw a perpendicular line $CD$ of length $1 \text{ unit}$ to $OC$. Join $OD$. Then $OD = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$.
$5$. At point $D$, draw a perpendicular line $DE$ of length $1 \text{ unit}$ to $OD$. Join $OE$. Then $OE = \sqrt{(\sqrt{4})^2 + 1^2} = \sqrt{5}$.
$6$. At point $E$, draw a perpendicular line $EF$ of length $1 \text{ unit}$ to $OE$. Join $OF$. Then $OF = \sqrt{(\sqrt{5})^2 + 1^2} = \sqrt{6}$.

(N/A) We know that $\frac{1}{4} = 0.25$ and $\frac{4}{5} = 0.8$.
To find three different irrational numbers between $\frac{1}{4}$ and $\frac{4}{5}$,we need to find three numbers between $0.25$ and $0.8$ that are non-terminating and non-recurring.
Three such numbers are:
$1. 0.3030030003\dots$
$2. 0.4040040004\dots$
$3. 0.5050050005\dots$

To find an irrational number between two positive numbers $a$ and $b$,we can use the formula $\sqrt{a \cdot b}$.
$1$. First irrational number between $\sqrt{3}$ and $\sqrt{5}$:
$= \sqrt{\sqrt{3} \cdot \sqrt{5}} = \sqrt{\sqrt{15}} = 15^{\frac{1}{4}}$.
$2$. Second irrational number between $\sqrt{3}$ and $15^{\frac{1}{4}}$:
$= \sqrt{\sqrt{3} \cdot 15^{\frac{1}{4}}} = \sqrt{3^{\frac{1}{2}} \cdot 3^{\frac{1}{4}} \cdot 5^{\frac{1}{4}}} = \sqrt{3^{\frac{3}{4}} \cdot 5^{\frac{1}{4}}} = 3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}$.
$3$. Third irrational number between $\sqrt{3}$ and $3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}$:
$= \sqrt{\sqrt{3} \cdot 3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}} = \sqrt{3^{\frac{1}{2}} \cdot 3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}} = \sqrt{3^{\frac{7}{8}} \cdot 5^{\frac{1}{8}}} = 3^{\frac{7}{16}} \cdot 5^{\frac{1}{16}}$.
Thus,the three irrational numbers are $15^{\frac{1}{4}}$,$3^{\frac{3}{8}} \cdot 5^{\frac{1}{8}}$,and $3^{\frac{7}{16}} \cdot 5^{\frac{1}{16}}$.

(N/A) First,convert the rational numbers into decimal form:
$\frac{1}{3} = 0.3333\ldots$
$\frac{7}{9} = 0.7777\ldots$
To find irrational numbers between $0.3333\ldots$ and $0.7777\ldots$,we need to write non-terminating and non-repeating decimals that fall within this range.
Three such numbers are:
$1) 0.4040040004\ldots$
$2) 0.5050050005\ldots$
$3) 0.6060060006\ldots$

(D) First,convert the rational numbers to decimal form: $\frac{3}{4} = 0.75$ and $\frac{4}{5} = 0.80$.
To find irrational numbers between $0.75$ and $0.80$,we need to write non-terminating and non-repeating decimals.
Examples of such numbers are:
$1. 0.75010010001...$
$2. 0.76010010001...$
$3. 0.77010010001...$
These numbers are clearly greater than $0.75$ and less than $0.80$,and they are irrational because their decimal expansions are non-terminating and non-recurring.

To find irrational numbers between two irrational numbers $a$ and $b$,we can use the form $\sqrt{a \cdot b}$,$\sqrt{a \cdot \sqrt{a \cdot b}}$,etc.,or simply find numbers whose decimal expansions are non-terminating and non-repeating.
$1$. First number: $\sqrt{\sqrt{2} \cdot \sqrt{5}} = \sqrt{\sqrt{10}} = 10^{\frac{1}{4}} \approx 1.778$.
$2$. Second number: $\sqrt{\sqrt{2} \cdot 10^{\frac{1}{4}}} = (2^{\frac{1}{2}} \cdot 10^{\frac{1}{4}})^{\frac{1}{2}} = 2^{\frac{1}{4}} \cdot 10^{\frac{1}{8}} = 2^{\frac{1}{4}} \cdot 2^{\frac{1}{8}} \cdot 5^{\frac{1}{8}} = 2^{\frac{3}{8}} \cdot 5^{\frac{1}{8}} \approx 1.682$.
$3$. Third number: $\sqrt{10^{\frac{1}{4}} \cdot \sqrt{5}} = (10^{\frac{1}{4}} \cdot 5^{\frac{1}{2}})^{\frac{1}{2}} = 10^{\frac{1}{8}} \cdot 5^{\frac{1}{4}} = 2^{\frac{1}{8}} \cdot 5^{\frac{1}{8}} \cdot 5^{\frac{1}{4}} = 2^{\frac{1}{8}} \cdot 5^{\frac{3}{8}} \approx 1.880$.
These values lie between $\sqrt{2} \approx 1.414$ and $\sqrt{5} \approx 2.236$.

(N/A) To represent $\sqrt{8.2}$ on the number line,follow these steps:
$1$. Draw a line segment $AB = 8.2 \text{ units}$ on the number line.
$2$. From point $B$,mark a point $C$ such that $BC = 1 \text{ unit}$. Now,$AC = 9.2 \text{ units}$.
$3$. Find the midpoint $O$ of $AC$. The distance $AO = OC = 4.6 \text{ units}$.
$4$. Draw a semicircle with center $O$ and radius $OA$ (or $OC$).
$5$. Draw a perpendicular line at point $B$ to the line $AC$,intersecting the semicircle at point $D$.
$6$. The length $BD$ is equal to $\sqrt{8.2}$.
$7$. With $B$ as the center and $BD$ as the radius,draw an arc on the number line. The point where the arc intersects the number line represents $\sqrt{8.2}$.

(N/A) To represent $\sqrt{5.6}$ on the number line,follow these steps:
$1$. Draw a line segment $AB = 5.6 \text{ units}$ on the number line.
$2$. From point $B$,mark a point $C$ such that $BC = 1 \text{ unit}$. Now,$AC = 6.6 \text{ units}$.
$3$. Find the midpoint $O$ of $AC$. The distance $AO = OC = 3.3 \text{ units}$.
$4$. Draw a semicircle with center $O$ and radius $3.3 \text{ units}$ passing through $A$ and $C$.
$5$. Draw a perpendicular line at point $B$ to $AC$,which intersects the semicircle at point $D$.
$6$. The length $BD$ is equal to $\sqrt{5.6}$.
$7$. With $B$ as the center and $BD$ as the radius,draw an arc on the number line. The point where the arc intersects the number line represents $\sqrt{5.6}$.

(A) To represent $\sqrt{7.5}$ on the number line,follow these steps:
$1$. Draw a line segment $AB$ of length $7.5 \text{ units}$.
$2$. Extend the line segment $AB$ by $1 \text{ unit}$ to point $C$,such that $BC = 1 \text{ unit}$. Now,$AC = 8.5 \text{ units}$.
$3$. Find the midpoint $O$ of $AC$. The distance $AO = OC = 4.25 \text{ units}$.
$4$. With $O$ as the center and $OA$ as the radius,draw a semicircle.
$5$. Draw a perpendicular line at point $B$ to the line $AC$,which intersects the semicircle at point $D$.
$6$. The length $BD$ is equal to $\sqrt{7.5}$.
$7$. Using $B$ as the center and $BD$ as the radius,draw an arc on the number line to mark the point representing $\sqrt{7.5}$.

(B) The statement is False.
Explanation:
We know that $\sqrt{49} = 7$.
Since $7$ can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$ (e.g.,$\frac{7}{1}$),it is a rational number.
Therefore,the statement that $\sqrt{49}$ is an irrational number is false.

(A) Given expression: $\sqrt{5} + \sqrt{5} = 2\sqrt{5}$.
Since $\sqrt{5}$ is an irrational number and the product of a non-zero rational number $(2)$ and an irrational number $(\sqrt{5})$ is always an irrational number,therefore,$2\sqrt{5}$ is an irrational number.

(B) To determine the nature of the number $(\sqrt{5}+3)^{2}$,we expand it using the algebraic identity $(a+b)^{2} = a^{2} + 2ab + b^{2}$.
Here,$a = \sqrt{5}$ and $b = 3$.
$(\sqrt{5}+3)^{2} = (\sqrt{5})^{2} + 2(\sqrt{5})(3) + (3)^{2}$
$= 5 + 6\sqrt{5} + 9$
$= 14 + 6\sqrt{5}$.
Since $\sqrt{5}$ is an irrational number,$6\sqrt{5}$ is also irrational. The sum of a rational number $(14)$ and an irrational number $(6\sqrt{5})$ is always an irrational number.
Therefore,$(\sqrt{5}+3)^{2}$ is an irrational number.

(IRRATIONAL) number is called irrational if it cannot be written in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$. Since $7$ is not a perfect square,its square root,$\sqrt{7}$,is an irrational number.

(B) The number $\pi$ is defined as the ratio of the circumference of a circle to its diameter. Its decimal expansion is non-terminating and non-recurring. Therefore,$\pi$ is an irrational number.

(A) number is called rational if it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Since $\frac{22}{7}$ is expressed as a ratio of two integers $22$ and $7$,where $7 \neq 0$,it satisfies the definition of a rational number.
Therefore,$\frac{22}{7}$ is a rational number.

(A) To determine whether $\sqrt{8^{2}+15^{2}}$ is a rational or an irrational number,we first simplify the expression inside the square root.
Step $1$: Calculate the squares of the numbers.
$8^{2} = 64$
$15^{2} = 225$
Step $2$: Add the results.
$64 + 225 = 289$
Step $3$: Find the square root of the sum.
$\sqrt{289} = 17$
Since $17$ can be expressed as $\frac{17}{1}$,which is in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$,it is a rational number.

(B) First,simplify the expression inside the square root: $8 + 15 = 23$.
Therefore,the expression becomes $\sqrt{23}$.
Since $23$ is not a perfect square,its square root cannot be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.
Thus,$\sqrt{23}$ is an irrational number.