Number Systems

Yes 0 (zero) is a rational number. We can write it in the form of $\dfrac{p}{q}$ like $\dfrac{0}{1}$, $\dfrac{0}{2}$, $\dfrac{0}{3}$, $\dfrac{0}{4}$, ........

where p = 0 , q ≠ 0

Since we want six rational numbers, we write 3 and 4 as rational numbers with denominator 3 + 4 = 7,

i.e.,

$3 = \dfrac{3 \times 7}{7} = \dfrac{21}{7}$

and

$4 = \dfrac{4 \times 7} {7} = \dfrac{28}{7}$

∴ Rational numbers between 3 and 4 are, $\dfrac{22}{7}$, $\dfrac{23}{7}$, $\dfrac{24}{7}$, $\dfrac{25}{7}$, $\dfrac{26}{7}$, $\dfrac{27}{7}$

Since we want five rational numbers, we write $\dfrac{3}{5}$ and $\dfrac{4}{5}$ as rational numbers with denominator 5 + 1 = 6, i.e.,

$\dfrac{3}{5} = \dfrac{3 \times 6} {5 \times 6} = \dfrac{18}{30}$

and

$\dfrac{4}{5} = \dfrac{4 \times 6} {5 \times 6} = \dfrac{24}{30}$

∴ Rational numbers between $\dfrac{3}{5}$ and $\dfrac{4}{5}$ are, $\dfrac{19}{30}$, $\dfrac{20}{30}$, $\dfrac{21}{30}$, $\dfrac{22}{30}$, $\dfrac{23}{30}$

(i) True
Reason — Whole numbers are = 0, 1, 2, 3, 4, 5,.......∞
Natural numbers are = 1, 2, 3, 4, 5,......∞
From above we see all natural numbers are whole numbers.

(ii) False
Reason — Integers can be positive as well as negative i.e., integers are = 0, +(-1), +(-2), +(-3), +(-4)......∞ whereas whole number are 0 or positive only i.e., whole numbers are = 0, 1, 2, 3, 4,.......∞

(iii) False,
Reason — Rational numbers are $\dfrac{p}{q}$ = $\dfrac{1}{2}$, $\dfrac{2}{3}$, $\dfrac{1}{4}$, $\dfrac{4}{7}$,.....
Whole numbers are = 0, 1, 2, 3, 4,.......∞

(i) True
Reason — Every irrational number is a real number, because rational number, natural number, whole number all the numbers comes into real number.

(ii) False
Reason — Every point on the number line represents a unique real number.

(iii) False
Reason — For example $\sqrt{2}$ is real but not irrational.

No, square root of all positive integers are not always irrational.

Example = $\sqrt{4} = (\sqrt{2 \times 2})$ = 2, and 2 is a rational number

Representing $\sqrt{5}$ as the sum of squares of two natural numbers:

5 = 4 + 1 = (2)² + (1)²

Let l be the number line. If point O represents number 0 and point A represents number 2 then draw a line segment OA = 2 units.

At A, draw AC ⊥ OA. From AC, cut off AB = 1 unit.

We observe that OAB is a right angled triangle at A. By Pythagoras theorem, we get:

(OB)² = (OA)² + (AB)²

(OB)² = (2)² + (1)²

(OB)² = 4 + 1

(OB)² = 5

OB = $\sqrt{5}$ units

With O as centre and radius = OB, we draw an arc of a circle to meet the number line l at point P.

As, OP = OB = $\sqrt{5}$ units, the point P will represent the number $\sqrt{5}$ on the number line as shown in the figure below:

(i) $\dfrac{36}{100}$ = 0.36, terminating decimal expansion

(ii) $\dfrac{1}{11}$

$\begin{array}{l} \phantom{11)}{0.0909} \\ 11\overline{\smash{\big)}\quad100\quad} \\ \phantom{11)}\phantom{22}\underline{-99} \\ \phantom{{11}x^3-2}100 \\ \phantom{{11)}x^322}\underline{-99} \\ \phantom{{11)}{x^32222}}1 \\ \phantom{{11)}{x^322}}\overline{\phantom{1111}} \end{array}$

The remainder 1 keeps repeating.
Hence, $\dfrac{1}{11}$ = $0.\overline{09}$. This is a non-terminating repeating decimal expansion

(iii) $4\dfrac{1}{8} = \dfrac{33}{8}$

$\begin{array}{l} \phantom{8)}{\enspace 4.125} \\ 8\overline{\smash{\big)}\enspace 33\quad} \\ \phantom{8)}\underline{-32\space} \\ \phantom{8)-}10 \\ \phantom{8-}\underline{-8} \\ \phantom{8)-8}20 \\ \phantom{8-}\underline{-16} \\ \phantom{8)-82}40 \\ \phantom{8-8}\underline{-40\enspace} \\ \phantom{8-822}0 \\ \phantom{8-8}\overline{\phantom{11111}} \end{array}$

Hence, $4\dfrac{1}{8}$ = 4.125. This is a terminating decimal expansion.

(iv) $\dfrac{3}{13}$

$\begin{array}{l} \phantom{8)}{\enspace 0.230769230769} \\ 13\overline{\smash{\big)}\enspace 30\qquad\qquad\quad} \\ \phantom{13)}\underline{-26\space} \\ \phantom{13)-}40 \\ \phantom{8-}\underline{-39} \\ \phantom{8)-8}100 \\ \phantom{13303}\underline{-91\enspace} \\ \phantom{13303333}90 \\ \phantom{1330331}\underline{-78\enspace} \\ \phantom{133033333}120 \\ \phantom{13303333}\underline{-117\enspace} \\ \phantom{13303333333}30 \\ \phantom{133033333)}\underline{-26\enspace} \\ \phantom{13303333333}40 \\ \phantom{1330333333}\overline{\phantom{11111}} \end{array}$

Hence, $\dfrac{3}{13}$ = $0.\overline{230769}$. This is a non-terminating repeating decimal expansion.

(v) $\dfrac{2}{11}$

$\begin{array}{l} \phantom{8)}{\enspace 0.1818} \\ 11\overline{\smash{\big)}\enspace 20\qquad} \\ \phantom{11)}\underline{-11\space} \\ \phantom{11)-}90 \\ \phantom{8-}\underline{-88} \\ \phantom{8)-8)}20 \\ \phantom{1330)}\underline{-11\enspace} \\ \phantom{1330333)}90 \\ \phantom{133033}\underline{-88\enspace} \\ \phantom{13303333}2 \\ \phantom{133033}\overline{\phantom{11111}} \end{array}$

Hence, $\dfrac{2}{11} = 0.\overline{18}$. This is a non-terminating repeating decimal expansion.

(vi) $\dfrac{329}{400}$ = 0.8225, terminating decimal expansion

$\begin{array}{l} \phantom{400)}{0.8225} \\ 400\overline{\smash{\big)}\enspace 3290\quad} \\ \phantom{400)}\underline{-3200\space} \\ \phantom{400)-}900 \\ \phantom{400)2}\underline{-800} \\ \phantom{400)218}1000 \\ \phantom{400)21}\underline{-800} \\ \phantom{400)212)}2000 \\ \phantom{400)21}\underline{-2000\enspace} \\ \phantom{400)21111}0 \\ \phantom{400)21}\overline{\phantom{1111111}} \end{array}$

Hence, $\dfrac{329}{400}$ = 0.8225. This is a terminating decimal expansion.

Given,

$\dfrac{1}{7}$ = $0.\overline{142857}$

(i) $\dfrac{2}{7}$ = $2 \times \dfrac{1}{7}$

= $2 \times 0.\overline{142857}$

= $0.\overline{285714}$

(ii) $\dfrac{3}{7}$ = $3 \times \dfrac{1}{7}$

= $3 \times 0.\overline{142857}$

= $0.\overline{428571}$

(iii) $\dfrac{4}{7}$ = $4 \times \dfrac{1}{7}$

= $4 \times 0.\overline{142857}$

= $0.\overline{571428}$

(iv) $\dfrac{5}{7}$ = $5 \times \dfrac{1}{7}$

= $5 \times 0.\overline{142857}$

= $0.\overline{714285}$

(v)$\dfrac{6}{7}$ = $6 \times \dfrac{1}{7}$

= $6 \times 0.\overline{142857}$

= $0.\overline{857142}$

(i) $0.\overline{6}$

Let x = $0.\overline{6}$

x = 0.666666........... (1)

Here one digit 6 is repeated so we multiply both side by 10 in equation (1)

10x = 6.6666.......

we can write

10x = 6.66666.......

10x = 6 + 0.666666...........

From equation (1)

10x = 6 + x

10x - x = 6

9x = 6

x = $\dfrac{6}{9}$ = $\dfrac{2}{3}$

Hence, $0.\overline{6}$ = $\dfrac{2}{3}$

(ii) $0.4 \overline{7}$

Let x = 0.47777777........ (1)

Here one digit 7 is repeated so we multiply both side by 10 in equation (1)

We can write

10x = 4.77777..... (2)

On subtracting equation (2) from equation (1)

10x - x = 4.777777........ - 0.477777.........

9x = 4.300000......

x = $\dfrac{4.3}{9}$

Multiplying numerator and denominator by 10 we get,

x = $\dfrac{4.3 \times 10}{9 \times 10}$ = $\dfrac{43}{90}$

Hence, $0.4\overline{7}$ = $\dfrac{43}{90}$

(iii) $0.\overline{001}$

Let x = 0.001001001...... (1)

Here three digit 001 are repeating so we multiply both side by 1000 in equation (1)

1000x = 0001.001001........ (2)

On subtracting equation (2) from equation (1)

1000x - x = 0001.001001...... - 0.001001......

999x = 1.0000000.....

x = $\dfrac{1}{999}$

Hence, $0.\overline{001}$ = $\dfrac{1}{999}$

Let x = 0.99999 ........ (1)

Here one digit 9 is repeating so we multiply both side by 10 in equation (1)

10 x = 9.99999.......... (2)

By subtracting equation (2) - equation (1)

10 x -x = 9.99999........ - 0.99999........

9x = 9

x = $\dfrac{9}{9}$

x = 1

Therefore, (0.99999...) is equivalent to (1), which might seem surprising at first glance, but it makes sense mathematically. This result can be demonstrated by the fact that any number infinitesimally close to 1 but less than 1, when infinitely added to itself, equals 1.
For example, if we take (0.9) and add (0.09), we get (0.99), and if we add (0.009) to that, we get (0.999), and so on. As we keep adding these infinitesimal increments, we approach (1). So, in a way, (0.99999...) is just another way to represent the number (1).

$\dfrac{1}{17}$ = $0.\overline{0588235294117647}$

sixteen digit are of repeating blocks.

$\begin{array}{l} \phantom{17)}{0.\overline{0588235294117647}} \\ 17\overline{\smash{\big)}\enspace 100\qquad\qquad\qquad} \\ \phantom{17))}\underline{-85\space} \\ \phantom{17)-}150 \\ \phantom{17))}\underline{-136} \\ \phantom{17))-}140 \\ \phantom{17)))}\underline{-136\enspace} \\ \phantom{13303333}40 \\ \phantom{1330331}\underline{-34\enspace} \\ \phantom{133033333}60 \\ \phantom{13303333}\underline{-51\enspace} \\ \phantom{13303333333}90 \\ \phantom{133033333)}\underline{-85\enspace} \\ \phantom{133033333333)}50 \\ \phantom{13303333333}\underline{-34\enspace} \\ \phantom{1330333333333)}160 \\ \phantom{133033333333}\underline{-153\enspace} \\ \phantom{1330333333333333}70 \\ \phantom{13303333333333)}\underline{-68\enspace} \\ \phantom{13303333333333333)}20 \\ \phantom{1330333333333333)}\underline{-17\enspace} \\ \phantom{1330333333333333333)}30 \\ \phantom{13303333333333333))}\underline{-17\enspace} \\ \phantom{1330333333333333333)))}130 \\ \phantom{1330333333333333333)}\underline{-119\enspace} \\ \phantom{13303333333333333333)))}110 \\ \phantom{13303333333333333333)}\underline{-102\enspace} \\ \phantom{133033333333333333333333)}80 \\ \phantom{13303333333333333333333}\underline{-68\enspace} \\ \phantom{133033333333333333333333)}120 \\ \phantom{13303333333333333333333}\underline{-119\enspace} \\ \phantom{1330333333333333333333333)}1 \\ \phantom{13303333333333333333333}\overline{\phantom{111111}} \end{array}$

Suppose, $\dfrac{p}{q}$ = $\dfrac{2}{5}$ = 0.4

or

$\dfrac{9}{10}$ = 0.9

or

$\dfrac{3}{2}$ = 1.5

q must contain factors of 2 or 5 or both.

For example,

In $\dfrac{7}{8}$, 8 = 2 x 2 x 2

and

in $\dfrac{7}{10}$, 10 = 2 x 5

π = 3.1415......

$\sqrt{2}$ = 1.414213......

$\sqrt{3}$ = 1.732050......

$\dfrac{5}{7}$ = $0.\overline{714285}$

$\begin{array}{l} \phantom{7)}{\enspace 0.714285} \\ 7\overline{\smash{\big)}\enspace 50\qquad\quad} \\ \phantom{7)}\underline{-49\space} \\ \phantom{7)-}10 \\ \phantom{8-}\underline{-7} \\ \phantom{8)-8}30 \\ \phantom{1330}\underline{-28\enspace} \\ \phantom{1330333}20 \\ \phantom{13303)}\underline{-14\enspace} \\ \phantom{13303333)}60 \\ \phantom{1330333)}\underline{-56\enspace} \\ \phantom{1330333333}40 \\ \phantom{13303333)}\underline{-35\enspace} \\ \phantom{13303333333}5 \\ \phantom{133033333}\overline{\phantom{11111}} \end{array}$

$\dfrac{9}{11}$ = $0.\overline{81}$

$\begin{array}{l} \phantom{11)}{\enspace 0.8181} \\ 11\overline{\smash{\big)}\enspace 90\qquad} \\ \phantom{11)}\underline{-88\space} \\ \phantom{11)-}20 \\ \phantom{11)1}\underline{-11} \\ \phantom{8)-81}90 \\ \phantom{13301}\underline{-88\enspace} \\ \phantom{1330333}20 \\ \phantom{133033}\underline{-11\enspace} \\ \phantom{13303333}9 \\ \phantom{133033}\overline{\phantom{11111}} \end{array}$

Irrational numbers between the rational numbers $\dfrac{5}{7}$ and $\dfrac{9}{11}$

0.75075007500075000075. . .

0.767076700767000767. . .

0.808008000800008. . .

(i) $\sqrt{23}$ is a prime number. It is irrational because $\sqrt{23}$ does not have a complete root.

(ii) $\sqrt{225}$ is not a prime number. 15 is root of $\sqrt{225}$ so it is rational.

(iii) 0.3796 is terminating decimal expansion. Hence, it is rational.

(iv) 7.478478... = $7.\overline{478}$, it is non-terminating but repeating decimal expansion. Hence, it is rational.

(v) 1.101001000100001... is non-terminating and non-repeating decimal expansion. Hence, it is irrational.

(i) 2 − $\sqrt{5}$

If in any rational number we add, subtract, multiply or divide any irrational number it becomes irrational number.

Hence, 2 − ($\sqrt{5})$ is an irrational number.

(ii) 3 + $\sqrt{23}$ − $\sqrt{23}$

= 3 + $\sqrt{23}$ − $\sqrt{23}$

= 3

Hence, (3 + $\sqrt{23}$) − $\sqrt{23}$ is a rational number

(iii) $\dfrac{2\sqrt{7}}{7\sqrt{7}}$

$\dfrac{2\sqrt{7}}{7\sqrt{7}}$ = $\dfrac{2}{7}$

It is in $\dfrac{p}{q}$ where q ≠ 0.

Hence, is a rational number

(iv) $\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}} = \dfrac{\text{rational}}{\sqrt{\text{irrational}}}$ = irrational number

If in any rational number we add, subtract, multiply or divide any irrational number it becomes irrational

Hence, $\dfrac{1}{\sqrt{2}}$ is an irrational number

(v) 2π

2 x π = irrational number [∵ rational x irrational number = irrational number]

Hence, 2π is an irrational number

$(3 + \sqrt{3}) (2 + \sqrt{2})$

= 3 x 2 + 3 x $\sqrt{2}$ + 2 x $\sqrt{3}$ + $\sqrt{3}$ x $\sqrt{2}$

= 6 + 3 $\sqrt{2}$ + 2 $\sqrt{3}$ + $\sqrt{6}$

Hence, $(3 + \sqrt{3}) (2 + \sqrt{2}) = 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}$

$(3 + \sqrt{3}) (3 - \sqrt{3})$

= (3)² - ($\sqrt{3}$)² [∵ (a + b)(a - b) = (a)² - (b)²]

= 9 - 3

= 6

Hence, $(3 + \sqrt{3}) (3 - \sqrt{3})$ = 6

$(\sqrt{5} + \sqrt{2})$ ²

= (a + b)²

= a² + 2ab + b²

= $(\sqrt{5})$ ² + 2 x $(\sqrt{5})$ x $(\sqrt{2})$ + $(\sqrt{2})$ ² [∵ (a + b)² = a² + 2ab + b²]

= 5 + 2 $(\sqrt{10})$ + 2

= 7 + 2 $(\sqrt{10})$

Hence,$(\sqrt{5} + \sqrt{2})$ ² = 7 + 2 $(\sqrt{10})$

$(\sqrt{5}$ - $\sqrt{2})$ $(\sqrt{5}$ + $\sqrt{2})$

= a² - b²

= $(\sqrt{5})^2 - (\sqrt{2})^2$ [∵ (a + b)(a - b) = (a)² - (b)²]

= 5 - 2

= 3

Hence, $(\sqrt{5}$ - $\sqrt{2})$ $(\sqrt{5}$ + $\sqrt{2})$ = 3

In the question given π is irrational, π = $\dfrac{c}{d}$ There is no contradiction. When we measure a length with a scale or any other device we only get an approximate rational value. So, we may not realize that either c or d is irrational.

Steps:

1. Draw a line and take AB = 9.3 units on it.
2. From B, measure a distance of 1 unit and mark C on the number line. Mark the midpoint of AC as O.
3. With 'O' as center and OC as radius, draw a semicircle.
4. At B, draw a perpendicular to cut the semicircle at D.
5. With B as center and BD as radius draw an arc to cut the number line at E. Thus, taking B as origin the distance BE = $\sqrt{9.3}$

Hence, point E represents $\sqrt{9.3}$ on the number line.

(i)$\dfrac{1}{\sqrt{7}}$

Multiply by $(\sqrt{7})$ in numerator and denominator

= $\dfrac{1}{\sqrt{7}}$ x $\dfrac{\sqrt{7}}{\sqrt{7}}$

= $\dfrac{\sqrt{7}}{7}$

Hence, $\dfrac{1}{\sqrt{7}}$ = $\dfrac{\sqrt{7}}{7}$

(ii) $\dfrac{1}{\sqrt{7} - \sqrt{6}}$

Multiply by $\dfrac{1}{\sqrt{7} + \sqrt{6}}$ in numerator and denominator

= $\dfrac{1}{\sqrt{7} - \sqrt{6} }$ x $\dfrac{\sqrt{7}+ \sqrt{6}}{\sqrt{7} + \sqrt{6}}$

= $\dfrac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (6)^2 }$

= $\dfrac{\sqrt{7} + \sqrt{6} }{7 - 6}$

= $\dfrac{\sqrt{7} + \sqrt{6} }{1}$

= $(\sqrt{7} + \sqrt{6})$

=Hence, $\dfrac{1}{\sqrt{7} - \sqrt{6}}$ = $(\sqrt{7} + \sqrt{6})$

(iii)$\dfrac{1}{\sqrt{5} + \sqrt{2}}$

Multiply by ${\sqrt{5} + \sqrt{2}}$ in numerator and denominator

= $\dfrac{1}{\sqrt{5} + \sqrt{2}}$ x $\dfrac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}}$

= $\dfrac{\sqrt{5} - \sqrt{2}} {(\sqrt{5})^2 - (\sqrt{2})^2 }$

= $\dfrac{\sqrt{5} - \sqrt{2}} {5 - 2}$

= $\dfrac{\sqrt{5} - \sqrt{2}} {3}$

=Hence, $\dfrac{1}{\sqrt{5} + \sqrt{2}}$ = $\dfrac{\sqrt{5} - \sqrt{2}} {3}$

(iv)$\dfrac{1}{\sqrt{7} - 2}$

Multiply by ${\sqrt{7} + {2}}$ in numerator and denominator

= $\dfrac{1}{\sqrt{7} - 2}$ X $\dfrac{\sqrt{7} + 2 }{\sqrt{7} + 2 }$

= $\dfrac{\sqrt{7} + 2} {(\sqrt{7})^2 - (2)^2 }$

= $\dfrac{\sqrt{7} + 2}{7 - 4}$

= $\dfrac{\sqrt{7} + 2}{3}$

= Hence, $\dfrac{1}{\sqrt{7} - 2 }$ = $\dfrac{\sqrt{7} + 2}{3}$

(i) $\left(64^{\smash{\frac{1}{2}}} \right)$

= (8²) ^1/2

= 8

Hence, $\left(64^{\smash{\frac{1}{2}}} \right)$ = 8

(ii) $\left(32^{\smash{\frac{1}{5}}} \right)$

= (2⁵) ^1/5

= 2

Hence, $\left(32^{\smash{\frac{1}{5}}} \right)$ = 2

(iii) $\left(125^{\smash{\frac{1}{3}}} \right)$

= (5³) ^1/3

= 5

Hence, $\left(125^{\smash{\frac{1}{3}}} \right)$ = 5

(i) $\left(9^{\smash{\frac{3}{2}}} \right)$

= (3²) ^3/2

= (3)³

= 3 x 3 x 3

= 27

Hence, $\left(9^{\smash{\frac{3}{2}}} \right)$ = 27

(ii) $\left(32^{\smash{\frac{2}{5}}} \right)$

= (2⁵) ^2/5

= (2)²

= 2 x 2

= 4

Hence, $\left(32^{\smash{\frac{2}{5}}} \right)$ = 4

(iii) $\left(16^{\smash{\frac{3}{4}}} \right)$

= (2⁴) ^3/4

= (2)³

= 2 x 2 x 2

= 8

Hence, $\left(16^{\smash{\frac{3}{4}}} \right)$ = 8

(iv) $\left(125^{\smash{\frac{-1}{3}}} \right)$

= (5³) ^-1/3

= (5)^-1

= $\dfrac{1}{5}$

Hence, $\left(125^{\smash{\frac{-1}{3}}} \right)$ = $\dfrac{1}{5}$

(i) $\left(2^{\smash{\frac{2}{3}}} \right)$. $\left(2^{\smash{\frac{1}{5}}} \right)$

= $\left(2^{\smash{\frac{2}{3} + \frac{1}{5}}} \right)$

= $\left((2)^{\smash{\frac{10+3}{15}}} \right)$

= $\left((2)^{\smash{\frac{13}{15}}} \right)$

Hence, $\left(2^{\smash{\frac{2}{3}}} \right)$. $\left(2^{\smash{\frac{1}{5}}} \right)$ = $\left((2)^{\smash{\frac{13}{15}}} \right)$

(ii) $\Big(\dfrac{1}{3^3}\Big)^7$

$= \Big(\dfrac{1}{3^3}\Big)^7 \\[1em] = \Big(\dfrac{1^7}{3^{3 \times 7}}\Big) \\[1em] = \Big(\dfrac{1}{3^{21}}\Big)$

Hence, $\Big(\dfrac{1}{3^3}\Big)^7 = \Big(\dfrac{1}{3^{21}}\Big)$

(iii) $\dfrac{11^\frac{1}{2}}{11^\frac{1}{4}}$

= $\left(11^{\smash{\frac{1}{2} - \frac{1}{4}}} \right)$

= $\left(11^{\smash{\frac{2-1}{4}}} \right)$

= $\left(11^{\smash{\frac{1}{4}}} \right)$

Hence, $\dfrac{11^\frac{1}{2}}{11^\frac{1}{4}}$ = $\left(11^{\smash{\frac{1}{4}}} \right)$

(iv) $\left(7^{\smash{\frac{1}{2}}} \right)$. $\left(8^{\smash{\frac{1}{2}}} \right)$

$= (7 \times 8)^{\dfrac{1}{2}}\quad[\because \text{a}^m \times \text{b}^m = (\text{ab})^m] \\[1em] = 56^{\frac{1}{2}}$

Hence, $\left(7^{\smash{\frac{1}{2}}} \right).\left(8^{\smash{\frac{1}{2}}} \right) = 56^{\frac{1}{2}}$