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Page No 9:

Yes, 0 is a rational number.

0 can be expressed in the form of the fraction $\frac{p}{q}$, where $p=0$ and q can be any integer except 0.

Page No 9:

(i) $\frac{5}{7}$

(ii) $\frac{8}{3}$
$\frac{8}{3}=2\frac{2}{3}$

(iii) $-\frac{23}{6}=-3\frac{5}{6}$

(iv) 1.3
$1.3=\frac{13}{10}=1\frac{3}{10}$

(v) – 2.4
$-2.4=\frac{-24}{10}=\frac{-12}{5}=-2\frac{2}{5}$

Page No 9:

(i)
Let:
x = $\frac{3}{8}$ and y = $\frac{2}{5}$
Rational number lying between x and y:

=

(ii) 1.3 and 1.4
Let:
x = 1.3 and y = 1.4
Rational number lying between x and y:

=

(iii)
Let:
x = $-$1 and y = $\frac{1}{2}$
Rational number lying between x and y:

= $-\frac{1}{4}$

(iv)
Let:
x$-\frac{3}{4}$ and y = $-\frac{2}{5}$
Rational number lying between x and y:

=

(v)
A rational number lying between  will be
$\frac{1}{2}\left(\frac{1}{9}+\frac{2}{9}\right)=\frac{1}{2}×\frac{1}{3}=\frac{1}{6}$

Page No 9:

n = 3
$d=\frac{\left(y-x\right)}{n+1}=\frac{\frac{7}{8}-\frac{3}{5}}{3+1}=\frac{11}{40}×\frac{1}{4}=\frac{11}{160}$
Rational numbers between  will be
$\left(x+d\right),\left(x+2d\right),...,\left(x+nd\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{3}{5}+\frac{11}{160}\right),\left(\frac{3}{5}+2×\frac{11}{160}\right),\left(\frac{3}{5}+3×\frac{11}{160}\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{107}{160}\right),\left(\frac{118}{160}\right),\left(\frac{129}{160}\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{107}{160}\right),\left(\frac{59}{80}\right),\left(\frac{129}{160}\right)$
There are infinitely many rational numbers between two given rational numbers.

Page No 9:

n = 4
n + 1 = 4 + 1 = 5

Thus, rational numbers between  are $\frac{16}{35},\frac{17}{35},\frac{18}{35},\frac{19}{35}$.

Page No 9:

x = 2, y = 3 and n = 6
$d=\frac{y-x}{n+1}=\frac{3-2}{6+1}=\frac{1}{7}$
Thus, the required numbers are
$\left(x+d\right),\left(x+2d\right),\left(x+3d\right),...,\left(x+nd\right)\phantom{\rule{0ex}{0ex}}=\left(2+\frac{1}{7}\right),\left(2+2×\frac{1}{7}\right),\left(2+3×\frac{1}{7}\right),\left(2+4×\frac{1}{7}\right),\left(2+5×\frac{1}{7}\right),\left(2+6×\frac{1}{7}\right)\phantom{\rule{0ex}{0ex}}=\frac{15}{7},\frac{16}{7},\frac{17}{7},\frac{18}{7},\frac{19}{7},\frac{20}{7}$

Page No 9:

n = 5
n + 1 = 6
$x=\frac{3}{5},y=\frac{2}{3}$
$d=\frac{y-x}{n+1}=\frac{\frac{2}{3}-\frac{3}{5}}{6}=\frac{10-9}{90}=\frac{1}{90}$

Thus, rational numbers between  will be
$\left(x+d\right),\left(x+2d\right),\left(x+3d\right),\left(x+4d\right),\left(x+5d\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{3}{5}+\frac{1}{90}\right),\left(\frac{3}{5}+\frac{2}{90}\right),\left(\frac{3}{5}+\frac{3}{90}\right),\left(\frac{3}{5}+\frac{4}{90}\right),\left(\frac{3}{5}+\frac{5}{90}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{55}{90}\right),\left(\frac{56}{90}\right),\left(\frac{57}{90}\right),\left(\frac{58}{90}\right),\left(\frac{59}{90}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{11}{18}\right),\left(\frac{28}{45}\right),\left(\frac{19}{30}\right),\left(\frac{29}{45}\right),\left(\frac{59}{90}\right)\phantom{\rule{0ex}{0ex}}$

Page No 10:

Let:
x = 2.1, y = 2.2 and n = 16

We know:
d = $\frac{y-x}{n+1}=\frac{2.2-2.1}{16+1}=\frac{0.1}{17}=\frac{1}{170}$= 0.005 (approx.)
So, 16 rational numbers between 2.1 and 2.2 are:
(x + d), (x + 2d), ...(x + 16d)
= [2.1 + 0.005], [2.1 + 2(0.005)],...[2.1 + 16(0.005)]
= 2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17, 2.175 and 2.18

Page No 10:

(i) Every natural number is a whole number.
True, since natural numbers are counting numbers i.e N = 1, 2,...
Whole numbers are natural numbers together with 0. i.e W = 0, 1, 2,...
So, every natural number is a whole number

(ii) Every whole number is a natural number.
False, as whole numbers contain natural numbers and 0 whereas natural numbers only contain the counting numbers except 0.

(iii) Every integer is a whole number.
False, whole numbers are natural numbers together with a zero whereas integers include negative numbers also.

(iv) Every integer is a rational number.
True, as rational numbers are of the form . All integers can be represented in the form .
(v) Every rational number is an integer.
False, as rational numbers are of the form . Integers are negative and positive numbers which are not in $\frac{p}{q}$ form.
For example, $\frac{1}{2}$ is a rational number but not an integer.

(vi) Every rational number is a whole number.
False, as rational numbers are of the form . Whole numbers are natural numbers together with a zero.
For example, $\frac{5}{7}$ is a rational number but not a whole number.

Page No 18:

(i) $\frac{13}{80}$
Denominator of $\frac{13}{80}$ is 80.
And,
80 = 24$×$5
Therefore, 80 has no other factors than 2 and 5.
Thus, $\frac{13}{80}$ is a terminating decimal.

(ii) $\frac{7}{24}$
Denominator of $\frac{7}{24}$ is 24.
And,
24 = 23$×$3
So, 24 has a prime factor 3, which is other than 2 and 5.
Thus, $\frac{7}{24}$ is not a terminating decimal.

(iii) $\frac{5}{12}$
Denominator of $\frac{5}{12}$ is 12.
And,
12 = 22$×$3
So, 12 has a prime factor 3, which is other than 2 and 5.
Thus, $\frac{5}{12}$ is not a terminating decimal.

(iv) $\frac{31}{375}$
Denominator of $\frac{31}{375}$ is 375.
$375={5}^{3}×3$
So, the prime factors of 375 are 5 and 3.
Thus, $\frac{31}{375}$ is not a terminating decimal.

(v) $\frac{16}{125}$
Denominator of $\frac{16}{125}$ is 125.
And,
125 = 53
Therefore, 125 has no other factors than 2 and 5.
Thus, $\frac{16}{125}$ is a terminating decimal.

Page No 19:

(i) $\frac{5}{8}$ = 0.625
By actual division, we have:

It is a terminating decimal expansion.

(ii) $\frac{7}{25}$
$\frac{7}{25}$ = 0.28
By actual division, we have:

It is a terminating decimal expansion.

(iii) $\frac{3}{11}$ = $0.\overline{27}$

It is a non-terminating recurring decimal.
(iv) $\frac{5}{13}$ = $0.\overline{384615}$

It is a non-terminating recurring decimal.

(v) $\frac{11}{24}$
$\frac{11}{24}$ =
By actual division, we have:

It is nonterminating recurring decimal expansion.

(vi) $\frac{261}{400}$$=0.6525$

It is a terminating decimal expansion.

(vii) $\frac{231}{625}$$=0.3696$

It is a terminating decimal expansion.

(viii) $2\frac{5}{12}$
2$\frac{5}{12}$ = $\frac{29}{12}$ =
By actual division, we have:

It is non-terminating decimal expansion.

Page No 19:

(i) $0.\overline{2}$
Let x = 0.222...                               .....(i)
Only one digit is repeated so, we multiply x by 10.
10x = 2.222...                                 .....(ii)
Subtracting (i) from (ii) we get
$9x=2\phantom{\rule{0ex}{0ex}}⇒x=\frac{2}{9}$

(ii) $0.\overline{53}$
Let x = 0.5353...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 53.5353...                                 .....(ii)
Subtracting (i) from (ii) we get
$99x=53\phantom{\rule{0ex}{0ex}}⇒x=\frac{53}{99}$

(iii) $2.\overline{93}$
Let x = 2.9393...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 293.9393...                                 .....(ii)
Subtracting (i) from (ii) we get
$99x=291\phantom{\rule{0ex}{0ex}}⇒x=\frac{291}{99}=\frac{97}{33}$

(iv) $18.\overline{48}$
Let x = 18.4848...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 1848.4848...                                 .....(ii)
Subtracting (i) from (ii) we get
$99x=1830\phantom{\rule{0ex}{0ex}}⇒x=\frac{1830}{99}=\frac{610}{33}$

(v) $0.\overline{235}$
Let x = 0.235235...                               .....(i)
Three digits are repeated so, we multiply x by 1000.
1000x = 235.235235...                                 .....(ii)
Subtracting (i) from (ii) we get
$999x=235\phantom{\rule{0ex}{0ex}}⇒x=\frac{235}{999}$

(vi) $0.00\overline{32}$
Let x = 0.003232...                               .....(i)
we multiply x by 100.
100x = 0.3232...                                 .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 32.3232...                           .....(iii)
Subtracting (ii) from (iii) we get
$9900x=32\phantom{\rule{0ex}{0ex}}⇒x=\frac{32}{9900}=\frac{8}{2475}$

(vii) $1.3\overline{23}$
Let x = 1.32323...                               .....(i)
we multiply x by 10.
10x = 13.2323...                                 .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
1000x = 1323.2323...                           .....(iii)
Subtracting (ii) from (iii) we get
$990x=1310\phantom{\rule{0ex}{0ex}}⇒x=\frac{131}{99}$

(viii) $0.3\overline{178}$
Let x = 0.3178178...                               .....(i)
we multiply x by 10.
10x = 3.178178...                                 .....(ii)
Again multiplying by 1000 as there are 3 repeating numbers after decimals we get
10000x = 3178.178178...                           .....(iii)
Subtracting (ii) from (iii) we get
$9990x=3175\phantom{\rule{0ex}{0ex}}⇒x=\frac{3175}{9990}=\frac{635}{1998}$

(ix) $32.12\overline{35}$
Let x = 32.123535...                               .....(i)
we multiply x by 100.
100x = 3212.3535...                                 .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 321235.35...                           .....(iii)
Subtracting (ii) from (iii) we get
$9900x=318023\phantom{\rule{0ex}{0ex}}⇒x=\frac{318023}{9900}$

(x) $0.40\overline{7}$
Let x = 0.40777...                               .....(i)
we multiply x by 100.
100x = 40.7777...                                 .....(ii)
Again multiplying by 10 as there is 1 repeating number after decimals we get
1000x = 407.777...                           .....(iii)
Subtracting (ii) from (iii) we get
$900x=367\phantom{\rule{0ex}{0ex}}⇒x=\frac{367}{900}$

Page No 19:

Given: $2.\overline{36}+0.\overline{23}$
Let

First we take x and convert it into $\frac{p}{q}$
100x = 236.3636...        ...(iii)
Subtracting (i) from (iii) we get
$99x=234\phantom{\rule{0ex}{0ex}}⇒x=\frac{234}{99}$
Similarly, multiply y with 100 as there are 2 decimal places which are repeating themselves.
$100y=23.2323...$               ...(iv)
Subtracting (ii) from (iv) we get
$99y=23\phantom{\rule{0ex}{0ex}}⇒y=\frac{23}{99}$
Adding x and y we get
$2.\overline{36}+0.\overline{23}$$x+y=\frac{234}{99}+\frac{23}{99}=\frac{257}{99}$

Page No 19:

x = 0.3838...                                  ...(i)
Multiply with 100 as there are 2 repeating digits after decimals
100x = 38.3838...                          ...(ii)
Subtracting (i) from (ii) we get
99x = 38
$⇒x=\frac{38}{99}$
Similarly, we take
y = 1.2727...                                  ...(iii)
Multiply y with 100 as there are 2 repeating digits after decimal.
100y = 127.2727...                       ...(iv)
Subtract (iii) from (iv) we get
99y = 126

Page No 23:

A number that can neither be expressed as a terminating decimal nor be expressed as a repeating decimal is called an irrational number. A rational number, on the other hand, is always a terminating decimal, and if not, it is a repeating decimal.
Examples of irrational numbers:
0.101001000...
0.232332333...

Page No 23:

(i) $\sqrt{\frac{3}{81}}$
$\sqrt{\frac{3}{81}}=\sqrt{\frac{1}{27}}=\frac{1}{3}\sqrt{\frac{1}{3}}$
It is an irrational number.

(ii) $\sqrt{361}$ = 19
So, it is rational.

(iii) $\sqrt{21}$

(iv) $\sqrt{1.44}$ = 1.2
So, it is rational.

(v) $\frac{2}{3}\sqrt{6}$
It is an irrational number

(vi) 4.1276
It is a terminating decimal. Hence, it is rational.

(vii) $\frac{22}{7}$

(viii) 1.232332333...

(xi) 6.834834... is a rational number because it is repeating.

Page No 23:

x be a rational number and y be an irrational number then x + y necessarily will be an irrational number.
Example: 5 is a rational number but $\sqrt{2}$ is irrational.
So, 5 + $\sqrt{2}$ will be an irrational number.

Page No 23:

a be a rational number and b be an irrational number then ab necessarily will be an irrational number.
Example: 6 is a rational number but $\sqrt{5}$ is irrational. And 6$\sqrt{5}$ is also an irrational number.

Page No 23:

Product of two irrational numbers is not always an irrational number.
Example: $\sqrt{5}$ is irrational number. And $\sqrt{5}×\sqrt{5}=5$ is a rational number. But the product of another two irrational numbers is $\sqrt{6}$ which is also an irrational numbers.

Page No 23:

(i) 2 irrational numbers with difference an irrational number will be .
(ii) 2 irrational numbers with difference is a rational number will be
(iii) 2 irrational numbers with sum an irrational number
(iv) 2 irrational numbers with sum a rational number is
(v) 2 irrational numbers with product an irrational number will be
(vi) 2 irrational numbers with product a rational number will be
(vii) 2 irrational numbers with quotient an irrational number will be
(viii) 2 irrational numbers with quotient a rational number will be .

Page No 23:

(i) Let us assume, to the contrary, that $3+\sqrt{3}$ is rational.
Then, $3+\sqrt{3}=\frac{p}{q}$, where p and q are coprime and $q\ne 0$.
$⇒\sqrt{3}=\frac{p}{q}-3\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}=\frac{p-3q}{q}$
Since, p and q are are integers.
$⇒\frac{p-3q}{q}$ is rational.
So, $\sqrt{3}$ is also rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $3+\sqrt{3}$ is rational.
Hence, $3+\sqrt{3}$ is irrational.

(ii) Let us assume, to the contrary, that $\sqrt{7}-2$ is rational.
Then, $\sqrt{7}-2=\frac{p}{q}$, where p and q are coprime and $q\ne 0$.
$⇒\sqrt{7}=\frac{p}{q}+2\phantom{\rule{0ex}{0ex}}⇒\sqrt{7}=\frac{p+2q}{q}$
Since, p and q are are integers.
$⇒\frac{p+2q}{q}$ is rational.
So, $\sqrt{7}$ is also rational.
But this contradicts the fact that $\sqrt{7}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $\sqrt{7}-2$ is rational.
Hence, $\sqrt{7}-2$ is irrational.

(iii) As, $\sqrt[3]{5}×\sqrt[3]{25}$

Hence, $\sqrt[3]{5}×\sqrt[3]{25}$ is rational.

(iv) As, $\sqrt{7}×\sqrt{343}$

Hence, $\sqrt{7}×\sqrt{343}$ is rational.

(v) As,
Hence, $\sqrt{\frac{13}{117}}$ is rational.

(vi) As, $\sqrt{8}×\sqrt{2}$

Hence, $\sqrt{8}×\sqrt{2}$ is rational.

Page No 23:

As, few rational numbers between 2 and 2.5 are: 2.1, 2.2, 2.3, 2.4, ...
And,
Since,
So, irrational number between 2 ans 2.5 are:

Hence, a rational and an irrational number can be 2.1 and $\sqrt{5}$, respectively.

Disclaimer: There are infinite rational and irrational numbers between any two rational numbers.

Page No 23:

There are infinite number of irrational numbers lying between .

As,
So, the three irrational numbers lying between  are:
1.420420042000..., 1.505005000... and 1.616116111...

Page No 23:

The two rational numbers between 0.5 and 0.55 are: 0.51 and 0.52

The two irrational numbers between 0.5 and 0.55 are: 0.505005000... and 0.5101100111000...

Disclaimer: There are infinite number of rational and irrational numbers between 0.5 and 0.55.

Page No 23:

So, the three different irrational numbers are: 0.72020020002..., 0.7515511555111... and 0.808008000...

Disclaimer: There are an infinite number of irrational numbers between two rational numbers.

Page No 24:

The rational numbers between the numbers 0.2121121112... and 0.2020020002... are:

Disclaimer: There are an infinite number of rational numbers between two irrational numbers.

Page No 24:

The two irrational numbers between 0.16 and 0.17 are 0.161161116... and 0.1606006000...

Disclaimer: There are an infinite number of irrational numbers between two rational numbers.

Page No 24:

(i) True

(ii) False
Example:

(iii) True

(iv) False

(v) True

(vi) False
Example:

(vii) False
Real numbers can be divided into rational and irrational numbers.

(viii) True

(ix) True

Page No 28:

(i)

(ii)

(iii) ${\left(3-\sqrt{3}\right)}^{2}$

(iv) ${\left(\sqrt{5}-\sqrt{3}\right)}^{2}$

(v)

$\left(5+\sqrt{7}\right)\left(2+\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=5×2+5×\sqrt{5}+\sqrt{7}×2+\sqrt{7}×\sqrt{5}\phantom{\rule{0ex}{0ex}}=10+5\sqrt{5}+2\sqrt{7}+\sqrt{35}$

(vi)

Page No 28:

$=3×6+3×4\sqrt{2}+\sqrt{3}×6+\sqrt{3}×4\sqrt{2}\phantom{\rule{0ex}{0ex}}=18+12\sqrt{2}+6\sqrt{3}+4\sqrt{6}$

Page No 28:

(i)

Hence,  is rational.

(ii) ${\left(\sqrt{3}+2\right)}^{2}$

Since, the sum and product of rational numbers and an irrational number is always an irrational.

$⇒$$7+4\sqrt{3}$ is irrational.

Hence, ${\left(\sqrt{3}+2\right)}^{2}$​ is irrational.

(iii) $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$

$=\frac{2\sqrt{13}}{3\sqrt{13×4}-4\sqrt{13×9}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{13}}{\sqrt{13}\left(3\sqrt{4}-4\sqrt{9}\right)}\phantom{\rule{0ex}{0ex}}=\frac{2}{\left(3×2-4×2\right)}$

Hence, $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$ is rational.

(iv) $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$

$=2\sqrt{2}+4×4\sqrt{2}-6\sqrt{2}\phantom{\rule{0ex}{0ex}}=2\sqrt{2}+16\sqrt{2}-6\sqrt{2}\phantom{\rule{0ex}{0ex}}=12\sqrt{2}$

Since, the product of a rational number and an irrational number is always an irrational.

Hence, $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$ is rational.

Page No 28:

(i) As,

Hence, the number of chocolates distributed by Reema is 14.

(ii) The moral values depicted here by Reema is helpfulness and caring.

Disclaimer: The moral values may vary from person to person.

Page No 28:

(i) $3\sqrt{45}-\sqrt{125}+\sqrt{200}-\sqrt{50}$

$=3\sqrt{9×5}-\sqrt{25×5}+\sqrt{100×2}-\sqrt{25×2}\phantom{\rule{0ex}{0ex}}=3×3\sqrt{5}-5\sqrt{5}+10\sqrt{2}-5\sqrt{2}\phantom{\rule{0ex}{0ex}}=9\sqrt{5}-5\sqrt{5}+5\sqrt{2}\phantom{\rule{0ex}{0ex}}=4\sqrt{5}+5\sqrt{2}$

(ii) $\frac{2\sqrt{30}}{\sqrt{6}}-\frac{3\sqrt{140}}{\sqrt{28}}+\frac{\sqrt{55}}{\sqrt{99}}$

$=\frac{2\sqrt{6×5}}{\sqrt{6}}-\frac{3\sqrt{28×5}}{\sqrt{28}}+\frac{\sqrt{5×11}}{\sqrt{9×11}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}×\sqrt{5}}{\sqrt{6}}-\frac{3\sqrt{28}×\sqrt{5}}{\sqrt{28}}+\frac{\sqrt{5}×\sqrt{11}}{\sqrt{9}×\sqrt{11}}\phantom{\rule{0ex}{0ex}}=2\sqrt{5}-3\sqrt{5}+\frac{\sqrt{5}}{3}$

$=-\sqrt{5}+\frac{\sqrt{5}}{3}\phantom{\rule{0ex}{0ex}}=\frac{-3\sqrt{5}+\sqrt{5}}{3}\phantom{\rule{0ex}{0ex}}=\frac{-2\sqrt{5}}{3}$

(iii) $\sqrt{72}+\sqrt{800}-\sqrt{18}$

$=\sqrt{36×2}+\sqrt{400×2}-\sqrt{9×2}\phantom{\rule{0ex}{0ex}}=6\sqrt{2}+20\sqrt{2}-3\sqrt{2}\phantom{\rule{0ex}{0ex}}=23\sqrt{2}$

Page No 35:

To represent $\sqrt{5}$ on the number line, follow the following steps of construction:

(i) Mark points 0 and 2 as O and P, respectively.
(ii) At point A, draw AB $\perp$ OA such that AB = 1 units.
(iii) Join OB.
(iv) With O as centre and radius OB, draw an arc intersecting the number line at point P.
Thus, point represents $\sqrt{5}$ on the number line.

Justification:

In right $∆$OAB,

Using Pythagoras theorem,

$\mathrm{OB}=\sqrt{{\mathrm{OA}}^{2}+{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{2}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+1}\phantom{\rule{0ex}{0ex}}=\sqrt{5}$

Page No 35:

To represent $\sqrt{3}$
on the number line, follow the following steps of construction:

(i) Mark points 0 and 1 as O and A, respectively.
(ii) At point A, draw AB $\perp$ OA such that AB = 1 units.
(iii) Join OB.
(iv) At point B, draw DB $\perp$ OA such that DB = 1 units.
(v) Join OD.
(vi)
With O as centre and radius OD, draw an arc intersecting the number line at point Q.
Thus, point Q represents $\sqrt{3}$ on the number line.

Justification:

In right $∆$OAB,

Using Pythagoras theorem,

$\mathrm{OB}=\sqrt{{\mathrm{OA}}^{2}+{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{1}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+1}\phantom{\rule{0ex}{0ex}}=\sqrt{2}$

Again, in right
$∆$ODB,

Using Pythagoras theorem,

$\mathrm{OD}=\sqrt{{\mathrm{OB}}^{2}+{\mathrm{DB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\sqrt{2}\right)}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{2+1}\phantom{\rule{0ex}{0ex}}=\sqrt{3}$

Page No 35:

To represent $\sqrt{10}$
on the number line, follow the following steps of construction:

(i) Mark points 0 and 3 as O and B, respectively.
(ii) At point A, draw AB $\perp$ OA such that AB = 1 units.
(iii) Join OA.
(iv) With O as centre and radius OA, draw an arc intersecting the number line at point P.
Thus, point P represents $\sqrt{10}$ on the number line.

Justification:

In right $∆$OAB,

Using Pythagoras theorem,

$\mathrm{OA}=\sqrt{{\mathrm{OB}}^{2}+{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{3}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+1}\phantom{\rule{0ex}{0ex}}=\sqrt{10}$

Page No 35:

To represent $\sqrt{8}$
on the number line, follow the following steps of construction:

(i) Mark points 0 and 2 as O and B, respectively.
(ii) At point B, draw AB $\perp$ OA such that AB = 2 units.
(iii) Join OA.
(iv) With O as centre and radius OA, draw an arc intersecting the number line at point P.
Thus, point P represents $\sqrt{8}$ on the number line.

Justification:

In right $∆$OAB,

Using Pythagoras theorem,

$\mathrm{OA}=\sqrt{{\mathrm{OB}}^{2}+{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{2}^{2}+{2}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+4}\phantom{\rule{0ex}{0ex}}=\sqrt{8}$

Page No 35:

To represent $\sqrt{4.7}$ on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 4.7 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi)  With B as centre and radius BD, draw an arc intersecting the given line at point E.

Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents $\sqrt{4.7}$.

Justification:

Here, in semi-circle, radii OA = OC = OD =

And, OB = AB $-$ AO = 4.7 $-$ 2.85 = 1.85 units

In a right angled triangle OBD,

Page No 35:

To represent $\sqrt{10.5}$ on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 10.5 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi)  With B as centre and radius BD, draw an arc intersecting the given line at point E.

Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents $\sqrt{10.5}$.

Justification:

Here, in semi-circle, radii OA = OC = OD =

And, OB = AB $-$ AO = 10.5 $-$ 5.75 = 4.75 units

In a right angled triangle OBD,

Page No 35:

To represent $\sqrt{7.28}$ on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 7.28 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi)  With B as centre and radius BD, draw an arc intersecting the given line at point E.

Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents $\sqrt{7.28}$.

Justification:

Here, in semi-circle, radii OA = OC = OD =

And, OB = AB $-$ AO = 7.28 $-$ 4.14 = 3.14 units

In a right angled triangle OBD,

Page No 35:

To represent $\left(1+\sqrt{9.5}\right)$ on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 9.5 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi)  With B as centre and radius BD, draw an arc intersecting the given line at point E.

(vii) From E, mark a point F on the same given line such that EF = 1 unit.

Thus, let us treat the given line as the number line, with B as 0, C as 1, E as $\sqrt{9.5}$ and so on, then point F represents $\left(1+\sqrt{9.5}\right)$.

Justification:

Here, in semi-circle, radii OA = OC = OD =

And, OB = AB $-$ AO = 9.5 $-$ 5.25 = 4.25 units

In a right angled triangle OBD,

Page No 35:

3 < 3.765 < 4
Divide the gap between 3 and 4 on the number line into 10 equal parts.
Now, 3.7 < 3.765 < 3.8
In order to locate the point 3.765 on the number line, divide the gap between 3.7 and 3.8 into 10 equal parts.
Further, 3.76 < 3.765 < 3.77
So, to locate the point 3.765 on the number line, again divide the gap between 3.76 and 3.77 into 10 equal parts.
Now, the number 3.765 can be located on the number line. This can be shown as follows:

Here, the marked point represents the point 3.765 on the number line.

Page No 35:

$4.\overline{67}=4.6767$  (Upto 4 decimal places)
4 < 4.6767 < 5
Divide the gap between 4 and 5 on the number line into 10 equal parts.
Now, 4.6 < 4.6767 < 4.7
In order to locate the point 4.6767 on the number line, divide the gap between 4.6 and 4.7 into 10 equal parts.
Further, 4.67 < 4.6767 < 4.68
To locate the point 4.6767 on the number line, again divide the gap between 4.67 and 4.68 into 10 equal parts.
Again, 4.676 < 4.6767 < 4.677
To locate the point 4.6767 on the number line, again divide the gap between 4.676 and 4.677 into 10 equal parts.
Now, the number 4.6767 can be located on the number line. This can be shown as follows:

Here, the marked point represents the point $4.\overline{67}$ on the number line up to 4 decimal places.

Page No 43:

$\frac{1}{\sqrt{2}+\sqrt{3}}$
$=\frac{1}{\sqrt{3}+\sqrt{2}}×\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}-\sqrt{2}}{1}$
Here, the denominator i.e. 1 is a rational number. Thus, the rationalising factor of the denominator in $\frac{1}{\sqrt{2}+\sqrt{3}}$ is $\sqrt{3}-\sqrt{2}$.

Page No 43:

(i) $\frac{1}{\sqrt{7}}$
On multiplying the numerator and denominator of the given number by $\sqrt{7}$, we get:

(ii) $\frac{\sqrt{5}}{2\sqrt{3}}$
On multiplying the numerator and denominator of the given number by $\sqrt{3}$, we get:

(iii) $\frac{1}{2+\sqrt{3}}$
On multiplying the numerator and denominator of the given number by $2-\sqrt{3}$, we get:

(iv) $\frac{1}{\sqrt{5}-2}$
On multiplying the numerator and denominator of the given number by $\sqrt{5}+2$, we get:

(v) $\frac{1}{5+3\sqrt{2}}$
On multiplying the numerator and denominator of the given number by $5-3\sqrt{2}$, we get:

(vi) $\frac{1}{\sqrt{7}-\sqrt{6}}$
Multiplying the numerator and denominator by $\sqrt{7}+\sqrt{6}$, we get
$\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}×\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}}{{\left(\sqrt{7}\right)}^{2}-{\left(\sqrt{6}\right)}^{2}}$
$=\frac{\sqrt{7}+\sqrt{6}}{7-6}\phantom{\rule{0ex}{0ex}}=\sqrt{7}+\sqrt{6}$

(vii) $\frac{4}{\sqrt{11}-\sqrt{7}}$
Multiplying the numerator and denominator by $\sqrt{11}+\sqrt{7}$, we get
$\frac{4}{\sqrt{11}-\sqrt{7}}=\frac{4}{\sqrt{11}-\sqrt{7}}×\frac{\sqrt{11}+\sqrt{7}}{\sqrt{11}+\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{{\left(\sqrt{11}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}$
$=\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{11-7}\phantom{\rule{0ex}{0ex}}=\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{4}\phantom{\rule{0ex}{0ex}}=\sqrt{11}+\sqrt{7}$

(viii) $\frac{1+\sqrt{2}}{2-\sqrt{2}}$
Multiplying the numerator and denominator by $2+\sqrt{2}$, we get
$\frac{1+\sqrt{2}}{2-\sqrt{2}}=\frac{1+\sqrt{2}}{2-\sqrt{2}}×\frac{2+\sqrt{2}}{2+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{2+\sqrt{2}+2\sqrt{2}+2}{{\left(2\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{4+3\sqrt{2}}{4-2}\phantom{\rule{0ex}{0ex}}=\frac{4+3\sqrt{2}}{2}$
(ix) $\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$
Multiplying the numerator and denominator by $3-2\sqrt{2}$, we get
$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}=\frac{3-2\sqrt{2}}{3+2\sqrt{2}}×\frac{3-2\sqrt{2}}{3-2\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(3-2\sqrt{2}\right)}^{2}}{{\left(3\right)}^{2}-{\left(2\sqrt{2}\right)}^{2}}$

Page No 43:

(i)
$\frac{2}{\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{2}{\sqrt{5}}×\frac{\sqrt{5}}{\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{5}}{5}$
$=\frac{2×2.236}{5}\phantom{\rule{0ex}{0ex}}=0.894$
(ii)
$\frac{2-\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2-\sqrt{3}}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}-3}{3}$
$=\frac{2×1.732-3}{3}\phantom{\rule{0ex}{0ex}}=0.155$
(iii)
$\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}×\frac{\sqrt{2}}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2}\left(\sqrt{10}-\sqrt{5}\right)}{2}$
$=\frac{1.414×\left(3.162-2.236\right)}{2}\phantom{\rule{0ex}{0ex}}=0.655$

Page No 43:

(i)
$\frac{\sqrt{2}-1}{\sqrt{2}+1}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2}-1}{\sqrt{2}+1}×\frac{\sqrt{2}-1}{\sqrt{2}-1}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{2}-1\right)}^{2}}{{\left(\sqrt{2}\right)}^{2}-{1}^{2}}$
$=\frac{2+1-2\sqrt{2}}{2-1}\phantom{\rule{0ex}{0ex}}=3-2\sqrt{2}$

(ii)
$\frac{2-\sqrt{5}}{2+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{2-\sqrt{5}}{2+\sqrt{5}}×\frac{2-\sqrt{5}}{2-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(2-\sqrt{5}\right)}^{2}}{{\left(2\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{4+5-4\sqrt{5}}{4-5}\phantom{\rule{0ex}{0ex}}=\frac{9-4\sqrt{5}}{-1}\phantom{\rule{0ex}{0ex}}=-9+4\sqrt{5}$

(iii)
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}×\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{3+2+2×\sqrt{3}×\sqrt{2}}{3-2}\phantom{\rule{0ex}{0ex}}=5+2\sqrt{6}$

(iv)
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{5+2\sqrt{3}}{7+4\sqrt{3}}×\frac{7-4\sqrt{3}}{7-4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{35-20\sqrt{3}+14\sqrt{3}-24}{{\left(7\right)}^{2}-{\left(4\sqrt{3}\right)}^{2}}$
$=\frac{11-6\sqrt{3}}{49-48}\phantom{\rule{0ex}{0ex}}=11-6\sqrt{3}$
$\therefore \frac{5+2\sqrt{3}}{7+4\sqrt{3}}=11+\left(-6\right)\sqrt{3}=a+b\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒a=11,b=-6$

Page No 43:

(i)
$\frac{1}{\sqrt{6}+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{6}+\sqrt{5}}×\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{6}-\sqrt{5}}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{\sqrt{6}-\sqrt{5}}{6-5}\phantom{\rule{0ex}{0ex}}=\sqrt{6}-\sqrt{5}\phantom{\rule{0ex}{0ex}}=2.449-2.236$
$=0.213$
(ii)
$\frac{6}{\sqrt{5}+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{6}{\sqrt{5}+\sqrt{3}}×\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}$
$=\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{5-3}\phantom{\rule{0ex}{0ex}}=\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{2}\phantom{\rule{0ex}{0ex}}=3\left(\sqrt{5}-\sqrt{3}\right)$
$=3×\left(2.236-1.732\right)\phantom{\rule{0ex}{0ex}}=1.512$
(iii)
$\frac{1}{4\sqrt{3}-3\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{1}{4\sqrt{3}-3\sqrt{5}}×\frac{4\sqrt{3}+3\sqrt{5}}{4\sqrt{3}+3\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{3}+3\sqrt{5}}{{\left(4\sqrt{3}\right)}^{2}-{\left(3\sqrt{5}\right)}^{2}}$
$=\frac{4\sqrt{3}+3\sqrt{5}}{48-45}\phantom{\rule{0ex}{0ex}}=\frac{4×1.732+3×2.236}{3}\phantom{\rule{0ex}{0ex}}=4.545$
(iv)
$\frac{3+\sqrt{5}}{3-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{3+\sqrt{5}}{3-\sqrt{5}}×\frac{3+\sqrt{5}}{3+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(3+\sqrt{5}\right)}^{2}}{{\left(3\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{9+5+6\sqrt{5}}{9-5}\phantom{\rule{0ex}{0ex}}=\frac{14+6\sqrt{5}}{4}\phantom{\rule{0ex}{0ex}}=\frac{7+3\sqrt{5}}{2}$
$=\frac{7+3×2.236}{2}\phantom{\rule{0ex}{0ex}}=6.854$
(v)
$\frac{1+2\sqrt{3}}{2-\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{1+2\sqrt{3}}{2-\sqrt{3}}×\frac{2+\sqrt{3}}{2+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2+\sqrt{3}+4\sqrt{3}+6}{{\left(2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}$
$=\frac{8+5\sqrt{3}}{4-3}\phantom{\rule{0ex}{0ex}}=8+5\sqrt{3}\phantom{\rule{0ex}{0ex}}=8+5×1.732$
$=16.660$
(vi)
$\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}×\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{5}+\sqrt{2}\right)}^{2}}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{5+2+2×\sqrt{5}×\sqrt{2}}{5-2}\phantom{\rule{0ex}{0ex}}=\frac{7+2\sqrt{10}}{3}\phantom{\rule{0ex}{0ex}}=\frac{7+2×3.162}{3}$
$=4.441$

Page No 44:

(i)
$\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{16×3}+\sqrt{9×2}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}$
$=\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}×\frac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{3}×4\sqrt{3}-7\sqrt{3}×3\sqrt{2}-5\sqrt{2}×4\sqrt{3}+5\sqrt{2}×3\sqrt{2}}{{\left(4\sqrt{3}\right)}^{2}-{\left(3\sqrt{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{84-21\sqrt{6}-20\sqrt{6}+30}{48-18}$
$=\frac{114-41\sqrt{6}}{30}$
(ii)
$\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}×\frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}×3\sqrt{5}+2\sqrt{6}×2\sqrt{6}-\sqrt{5}×3\sqrt{5}-\sqrt{5}×2\sqrt{6}}{{\left(3\sqrt{5}\right)}^{2}-{\left(2\sqrt{6}\right)}^{2}}$
$=\frac{6\sqrt{30}+24-15-2\sqrt{30}}{45-24}\phantom{\rule{0ex}{0ex}}=\frac{9+4\sqrt{30}}{21}$

Page No 44:

(i)
$\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{4+\sqrt{5}}{4-\sqrt{5}}×\frac{4+\sqrt{5}}{4+\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}×\frac{4-\sqrt{5}}{4-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(4+\sqrt{5}\right)}^{2}}{{\left(4\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}+\frac{{\left(4-\sqrt{5}\right)}^{2}}{{\left(4\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{16+5+8\sqrt{5}+16+5-8\sqrt{5}}{16-5}\phantom{\rule{0ex}{0ex}}=\frac{42}{11}$
(ii)
$\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{3}+\sqrt{2}}×\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}×\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}×\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{{\left(\sqrt{2}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{2-5}\phantom{\rule{0ex}{0ex}}=\sqrt{3}-\sqrt{2}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{\left(-3\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{3}-\sqrt{2}-\sqrt{5}-\sqrt{3}+\sqrt{2}+\sqrt{5}$
$=0$
(iii)
$\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}\phantom{\rule{0ex}{0ex}}=\frac{2+\sqrt{3}}{2-\sqrt{3}}×\frac{2+\sqrt{3}}{2+\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}×\frac{2-\sqrt{3}}{2-\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}×\frac{\sqrt{3}-1}{\sqrt{3}-1}\phantom{\rule{0ex}{0ex}}=\frac{{\left(2+\sqrt{3}\right)}^{2}}{{\left(2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}+\frac{{\left(2-\sqrt{3}\right)}^{2}}{{\left(2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}+\frac{{\left(\sqrt{3}-1\right)}^{2}}{{\left(\sqrt{3}\right)}^{2}-{1}^{2}}$
$=\frac{4+3+4\sqrt{3}}{4-3}+\frac{4+3-4\sqrt{3}}{4-3}+\frac{3+1-2\sqrt{3}}{3-1}\phantom{\rule{0ex}{0ex}}=7+4\sqrt{3}+7-4\sqrt{3}+\frac{4-2\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}=14+2-\sqrt{3}$
$=16-\sqrt{3}$
(iv)
$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}×\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}×\frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}×\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}×\sqrt{3}-2\sqrt{6}×\sqrt{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}+\frac{6\sqrt{2}×\sqrt{6}-6\sqrt{2}×\sqrt{3}}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}-\frac{8\sqrt{3}×\sqrt{6}-8\sqrt{3}×\sqrt{2}}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{2\sqrt{18}-2\sqrt{12}}{3-2}+\frac{6\sqrt{12}-6\sqrt{6}}{6-3}-\frac{8\sqrt{18}-8\sqrt{6}}{6-2}\phantom{\rule{0ex}{0ex}}=2\sqrt{18}-2\sqrt{12}+\frac{6\sqrt{12}-6\sqrt{6}}{3}-\frac{8\sqrt{18}-8\sqrt{6}}{4}\phantom{\rule{0ex}{0ex}}=2\sqrt{18}-2\sqrt{12}+2\sqrt{12}-2\sqrt{6}-2\sqrt{18}+2\sqrt{6}$
$=0$

Page No 44:

(i)
$\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1}\phantom{\rule{0ex}{0ex}}=\frac{1}{3+\sqrt{7}}×\frac{3-\sqrt{7}}{3-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}×\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}×\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{3}+1}×\frac{\sqrt{3}-1}{\sqrt{3}-1}\phantom{\rule{0ex}{0ex}}=\frac{3-\sqrt{7}}{{\left(3\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}+\frac{\sqrt{7}-\sqrt{5}}{{\left(\sqrt{7}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}+\frac{\sqrt{5}-\sqrt{3}}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}+\frac{\sqrt{3}-1}{{\left(\sqrt{3}\right)}^{2}-{1}^{2}}$
$=\frac{3-\sqrt{7}}{9-7}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{3}-1}{3-1}\phantom{\rule{0ex}{0ex}}=\frac{3-\sqrt{7}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2}\phantom{\rule{0ex}{0ex}}=\frac{3-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-1}{2}$
$=\frac{2}{2}\phantom{\rule{0ex}{0ex}}=1$
(ii)
$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}\phantom{\rule{0ex}{0ex}}=\frac{1}{1+\sqrt{2}}×\frac{1-\sqrt{2}}{1-\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}×\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}×\frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}×\frac{\sqrt{4}-\sqrt{5}}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}×\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}×\frac{\sqrt{6}-\sqrt{7}}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}×\frac{\sqrt{7}-\sqrt{8}}{\sqrt{7}-\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}×\frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}}$
$=\frac{1-\sqrt{2}}{{1}^{2}-{\left(\sqrt{2}\right)}^{2}}+\frac{\sqrt{2}-\sqrt{3}}{{\left(\sqrt{2}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}+\frac{\sqrt{3}-\sqrt{4}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{4}\right)}^{2}}+\frac{\sqrt{4}-\sqrt{5}}{{\left(\sqrt{4}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}+\frac{\sqrt{5}-\sqrt{6}}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{6}\right)}^{2}}+\frac{\sqrt{6}-\sqrt{7}}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}+\frac{\sqrt{7}-\sqrt{8}}{{\left(\sqrt{7}\right)}^{2}-{\left(\sqrt{8}\right)}^{2}}+\frac{\sqrt{8}-\sqrt{9}}{{\left(\sqrt{8}\right)}^{2}-{\left(\sqrt{9}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+\frac{\sqrt{4}-\sqrt{5}}{4-5}+\frac{\sqrt{5}-\sqrt{6}}{5-6}+\frac{\sqrt{6}-\sqrt{7}}{6-7}+\frac{\sqrt{7}-\sqrt{8}}{7-8}+\frac{\sqrt{8}-\sqrt{9}}{8-9}\phantom{\rule{0ex}{0ex}}=\frac{1-\sqrt{2}}{\left(-1\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(-1\right)}+\frac{\sqrt{3}-\sqrt{4}}{\left(-1\right)}+\frac{\sqrt{4}-\sqrt{5}}{\left(-1\right)}+\frac{\sqrt{5}-\sqrt{6}}{\left(-1\right)}+\frac{\sqrt{6}-\sqrt{7}}{\left(-1\right)}+\frac{\sqrt{7}-\sqrt{8}}{\left(-1\right)}+\frac{\sqrt{8}-\sqrt{9}}{\left(-1\right)}$
$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}+\sqrt{7}-\sqrt{6}+\sqrt{8}-\sqrt{7}+\sqrt{9}-\sqrt{8}\phantom{\rule{0ex}{0ex}}=3-1\phantom{\rule{0ex}{0ex}}=2$

Page No 44:

$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{7+3\sqrt{5}}{3+\sqrt{5}}×\frac{3-\sqrt{5}}{3-\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}×\frac{3+\sqrt{5}}{3+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{7\left(3-\sqrt{5}\right)+3\sqrt{5}\left(3-\sqrt{5}\right)}{{3}^{2}-{\left(\sqrt{5}\right)}^{2}}-\frac{7\left(3+\sqrt{5}\right)-3\sqrt{5}\left(3+\sqrt{5}\right)}{{3}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{21-7\sqrt{5}+9\sqrt{5}-15}{9-5}-\frac{21+7\sqrt{5}-9\sqrt{5}-15}{9-5}\phantom{\rule{0ex}{0ex}}=\frac{6+2\sqrt{5}}{4}-\frac{6-2\sqrt{5}}{4}$
$=\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{5}}{4}\phantom{\rule{0ex}{0ex}}=\sqrt{5}$
$\therefore \frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=0+1×\sqrt{5}$
Comparing with the given expression, we get

a = 0 and b = 1

Thus, the values of a and b are 0 and 1, respectively.

Page No 44:

$\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}+\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}×\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}-\sqrt{11}}+\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}×\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}+\sqrt{11}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{13}-\sqrt{11}\right)}^{2}}{{\left(\sqrt{13}\right)}^{2}-{\left(\sqrt{11}\right)}^{2}}+\frac{{\left(\sqrt{13}+\sqrt{11}\right)}^{2}}{{\left(\sqrt{13}\right)}^{2}-{\left(\sqrt{11}\right)}^{2}}$
$=\frac{13+11-2×\sqrt{13}×\sqrt{11}}{13-11}+\frac{13+11+2×\sqrt{13}×\sqrt{11}}{13-11}\phantom{\rule{0ex}{0ex}}=\frac{24-2\sqrt{143}}{2}+\frac{24+2\sqrt{143}}{2}\phantom{\rule{0ex}{0ex}}=\frac{24-2\sqrt{143}+24+2\sqrt{143}}{2}$
$=\frac{48}{2}\phantom{\rule{0ex}{0ex}}=24$

Page No 44:

$⇒\frac{1}{x}=\frac{1}{3+2\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=\frac{1}{3+2\sqrt{2}}×\frac{3-2\sqrt{2}}{3-2\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=\frac{3-2\sqrt{2}}{{3}^{2}-{\left(2\sqrt{2}\right)}^{2}}$

Adding (1) and (2), we get
$x+\frac{1}{x}=3+2\sqrt{2}+3-2\sqrt{2}=6$, which is a rational number
Thus, $x+\frac{1}{x}$ is rational.

Page No 44:

Subtracting (2) from (1), we get
$x-\frac{1}{x}=\left(2-\sqrt{3}\right)-\left(2+\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}⇒x-\frac{1}{x}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒{\left(x-\frac{1}{x}\right)}^{3}={\left(-2\sqrt{3}\right)}^{3}=-24\sqrt{3}$
Thus, the value of ${\left(x-\frac{1}{x}\right)}^{3}$ is $-24\sqrt{3}$.

Page No 44:

Adding (1) and (2), we get
$x+\frac{1}{x}=9-4\sqrt{5}+9+4\sqrt{5}\phantom{\rule{0ex}{0ex}}⇒x+\frac{1}{x}=18$
Squaring on both sides, we get
${\left(x+\frac{1}{x}\right)}^{2}={18}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\frac{1}{{x}^{2}}+2×x×\frac{1}{x}=324\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\frac{1}{{x}^{2}}=324-2=322$
Thus, the value of ${x}^{2}+\frac{1}{{x}^{2}}$ is 322.

Page No 44:

$⇒\frac{1}{x}=\frac{2}{5-\sqrt{21}}×\frac{5+\sqrt{21}}{5+\sqrt{21}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=\frac{2\left(5+\sqrt{21}\right)}{{5}^{2}-{\left(\sqrt{21}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=\frac{2\left(5+\sqrt{21}\right)}{25-21}$

Adding (1) and (2), we get
$x+\frac{1}{x}=\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2}\phantom{\rule{0ex}{0ex}}⇒x+\frac{1}{x}=\frac{5-\sqrt{21}+5+\sqrt{21}}{2}\phantom{\rule{0ex}{0ex}}⇒x+\frac{1}{x}=\frac{10}{2}=5$
Thus, the value of $x+\frac{1}{x}$ is 5.

Page No 44:

$\therefore \frac{1}{{a}^{2}}=\frac{1}{17-12\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{a}^{2}}=\frac{1}{17-12\sqrt{2}}×\frac{17+12\sqrt{2}}{17+12\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{a}^{2}}=\frac{17+12\sqrt{2}}{{17}^{2}-{\left(12\sqrt{2}\right)}^{2}}$

Subtracting (2) from (1), we get
${a}^{2}-\frac{1}{{a}^{2}}=\left(17-12\sqrt{2}\right)-\left(17+12\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}⇒{a}^{2}-\frac{1}{{a}^{2}}=17-12\sqrt{2}-17-12\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}-\frac{1}{{a}^{2}}=-24\sqrt{2}$
Thus, the value of ${a}^{2}-\frac{1}{{a}^{2}}$ is $-24\sqrt{2}$.

Page No 45:

Subtracting (2) from (1), we get

Thus, the value of $x-\frac{1}{x}$ is $4\sqrt{3}$.

Page No 45:

Adding (1) and (2), we get

Cubing both sides, we get
${\left(x+\frac{1}{x}\right)}^{3}={4}^{3}\phantom{\rule{0ex}{0ex}}⇒{x}^{3}+\frac{1}{{x}^{3}}+3×x×\frac{1}{x}\left(x+\frac{1}{x}\right)=64$
$⇒{x}^{3}+\frac{1}{{x}^{3}}+3×4=64$             [Using (3)]
$⇒{x}^{3}+\frac{1}{{x}^{3}}=64-12=52$
Thus, the value of ${x}^{3}+\frac{1}{{x}^{3}}$ is 52.

Page No 45:

Disclaimer: The question is incorrect.

$x=\frac{5-\sqrt{3}}{5+\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒x=\frac{5-\sqrt{3}}{5+\sqrt{3}}×\frac{5-\sqrt{3}}{5-\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒x=\frac{{\left(5-\sqrt{3}\right)}^{2}}{{5}^{2}-{\left(\sqrt{3}\right)}^{2}}$
$⇒x=\frac{25+3-10\sqrt{3}}{25-3}\phantom{\rule{0ex}{0ex}}⇒x=\frac{28-10\sqrt{3}}{22}\phantom{\rule{0ex}{0ex}}⇒x=\frac{14-5\sqrt{3}}{11}$

$y=\frac{5+\sqrt{3}}{5-\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒y=\frac{5+\sqrt{3}}{5-\sqrt{3}}×\frac{5+\sqrt{3}}{5+\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒y=\frac{{\left(5+\sqrt{3}\right)}^{2}}{{5}^{2}-{\left(\sqrt{3}\right)}^{2}}$
$⇒y=\frac{25+3+10\sqrt{3}}{25-3}\phantom{\rule{0ex}{0ex}}⇒y=\frac{28+10\sqrt{3}}{22}\phantom{\rule{0ex}{0ex}}⇒y=\frac{14+5\sqrt{3}}{11}$

$\therefore {x}^{2}-{y}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{14-5\sqrt{3}}{11}\right)}^{2}-{\left(\frac{14+5\sqrt{3}}{11}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{196+75-140\sqrt{3}}{121}-\frac{196+75+140\sqrt{3}}{121}\phantom{\rule{0ex}{0ex}}$
$=\frac{271-140\sqrt{3}}{121}-\frac{271+140\sqrt{3}}{121}\phantom{\rule{0ex}{0ex}}=\frac{271-140\sqrt{3}-271-140\sqrt{3}}{121}\phantom{\rule{0ex}{0ex}}=\frac{-280\sqrt{3}}{121}\phantom{\rule{0ex}{0ex}}$
The question is incorrect. Kindly check the question.
The question should have been to show that $x-y=-\frac{10\sqrt{3}}{11}$.
$\therefore x-y\phantom{\rule{0ex}{0ex}}=\frac{14-5\sqrt{3}}{11}-\frac{14+5\sqrt{3}}{11}\phantom{\rule{0ex}{0ex}}=\frac{14-5\sqrt{3}-14-5\sqrt{3}}{11}\phantom{\rule{0ex}{0ex}}=\frac{-10\sqrt{3}}{11}\phantom{\rule{0ex}{0ex}}$

Page No 45:

According to question,

Now,

Hence, $3{a}^{2}+4ab-3{b}^{2}=4+\frac{56\sqrt{10}}{3}$.

Page No 45:

According to question,

Now,

Hence, the value of a2 + b2 – 5ab is 93.

Page No 45:

According to question,

Now,

Hence, the value of p2 + q2 is 47.

Page No 45:

$\left(\mathrm{i}\right)\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\sqrt{7}+\sqrt{6}\right)-\sqrt{13}}×\frac{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}\phantom{\rule{0ex}{0ex}}=\frac{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}{{\left(\sqrt{7}+\sqrt{6}\right)}^{2}-{\left(\sqrt{13}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{{\left(\sqrt{7}\right)}^{2}+{\left(\sqrt{6}\right)}^{2}+2\left(\sqrt{7}\right)\left(\sqrt{6}\right)-{\left(\sqrt{13}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{7+6+2\sqrt{42}-13}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}×\frac{\sqrt{42}}{\sqrt{42}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{6}+6\sqrt{7}+\left(\sqrt{13}\right)\left(\sqrt{42}\right)}{84}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}$
Hence, the rationalised form is $\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}$.
$\left(\mathrm{ii}\right)\frac{3}{\sqrt{3}+\sqrt{5}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{3}{\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{5}}×\frac{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}}{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{3\left\{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right\}}{{\left(\sqrt{3}-\sqrt{2}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{{\left(\sqrt{3}\right)}^{2}+{\left(\sqrt{2}\right)}^{2}-2\left(\sqrt{3}\right)\left(\sqrt{2}\right)-{\left(\sqrt{5}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{3+2-2\sqrt{6}-5}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{-2\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{-2\sqrt{6}}×\frac{\sqrt{6}}{\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[3\sqrt{2}-2\sqrt{3}-\sqrt{30}\right]}{-12}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{30}+2\sqrt{3}-3\sqrt{2}}{4}$
Hence, the rationalised form is $\frac{\sqrt{30}+2\sqrt{3}-3\sqrt{2}}{4}$.
$\left(\mathrm{iii}\right)\frac{4}{2+\sqrt{3}+\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{4}{\left(2+\sqrt{3}\right)+\sqrt{7}}×\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\left(2+\sqrt{3}\right)-\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{{\left(2+\sqrt{3}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{{\left(2\right)}^{2}+{\left(\sqrt{3}\right)}^{2}+2\left(2\right)\left(\sqrt{3}\right)-{\left(\sqrt{7}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{4+3+4\sqrt{3}-7}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}+3-\sqrt{21}}{3}$
Hence, the rationalised form is $\frac{2\sqrt{3}+3-\sqrt{21}}{3}$.

Page No 45:

Hence, the value of $\frac{1}{\sqrt{3}-\sqrt{2}-1}$ correct to 3 places of decimal is −1.465.

Page No 45:

Now,

Hence, the value of x3 – 2x2 – 7x + 5 is 3.

Page No 45:

Hence, $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$ = 5.398 .

Page No 53:

(i) ${2}^{\frac{2}{3}}×{2}^{\frac{1}{3}}$

(ii) ${2}^{\frac{2}{3}}×{2}^{\frac{1}{5}}$

(iii) ${7}^{\frac{5}{6}}×{7}^{\frac{2}{3}}$

(iv) ${\left(1296\right)}^{\frac{1}{4}}×{\left(1296\right)}^{\frac{1}{2}}$

(i) (ab + ba)–1

(ii) (aa + bb)–1

Page No 53:

(i) ${\left(\frac{81}{49}\right)}^{-\frac{3}{2}}$

(ii) (14641)0.25

(iii) ${\left(\frac{32}{243}\right)}^{-\frac{4}{5}}$

(iv) ${\left(\frac{7776}{243}\right)}^{-\frac{3}{5}}$

Page No 54:

(i) $\frac{4}{{\left(216\right)}^{-\frac{2}{3}}}+\frac{1}{{\left(256\right)}^{-\frac{3}{4}}}+\frac{2}{{\left(243\right)}^{-\frac{1}{5}}}$
$\frac{4}{{\left(216\right)}^{-\frac{2}{3}}}+\frac{1}{{\left(256\right)}^{-\frac{3}{4}}}+\frac{2}{{\left(243\right)}^{-\frac{1}{5}}}\phantom{\rule{0ex}{0ex}}=\frac{4}{{\left[{\left(6\right)}^{3}\right]}^{-\frac{2}{3}}}+\frac{1}{{\left[{\left(4\right)}^{4}\right]}^{-\frac{3}{4}}}+\frac{2}{{\left[{\left(3\right)}^{5}\right]}^{-\frac{1}{5}}}\phantom{\rule{0ex}{0ex}}=\frac{4}{{\left(6\right)}^{-2}}+\frac{1}{{\left(4\right)}^{-3}}+\frac{2}{{\left(3\right)}^{-1}}\phantom{\rule{0ex}{0ex}}=4{\left(6\right)}^{2}+{\left(4\right)}^{3}+2\left(3\right)\phantom{\rule{0ex}{0ex}}=144+64+6\phantom{\rule{0ex}{0ex}}=214$

(ii) ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+{\left(\frac{256}{625}\right)}^{-\frac{1}{4}}+{\left(\frac{3}{7}\right)}^{0}$
${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+{\left(\frac{256}{625}\right)}^{-\frac{1}{4}}+{\left(\frac{3}{7}\right)}^{0}\phantom{\rule{0ex}{0ex}}={\left[{\left(\frac{4}{5}\right)}^{3}\right]}^{-\frac{2}{3}}+{\left[{\left(\frac{4}{5}\right)}^{4}\right]}^{-\frac{1}{4}}+1\phantom{\rule{0ex}{0ex}}={\left(\frac{4}{5}\right)}^{-2}+{\left(\frac{4}{5}\right)}^{-1}+1\phantom{\rule{0ex}{0ex}}={\left(\frac{5}{4}\right)}^{2}+\left(\frac{5}{4}\right)+1\phantom{\rule{0ex}{0ex}}=\frac{25}{16}+\frac{5}{4}+1\phantom{\rule{0ex}{0ex}}=\frac{25+20+16}{16}\phantom{\rule{0ex}{0ex}}=\frac{61}{16}$

(iii)

(iv) $\frac{{\left(25\right)}^{\frac{5}{2}}×{\left(729\right)}^{\frac{1}{3}}}{{\left(125\right)}^{\frac{2}{3}}×{\left(27\right)}^{\frac{2}{3}}×{8}^{\frac{4}{3}}}$
$\frac{{\left(25\right)}^{\frac{5}{2}}×{\left(729\right)}^{\frac{1}{3}}}{{\left(125\right)}^{\frac{2}{3}}×{\left(27\right)}^{\frac{2}{3}}×{8}^{\frac{4}{3}}}\phantom{\rule{0ex}{0ex}}=\frac{{\left[{\left(5\right)}^{2}\right]}^{\frac{5}{2}}×{\left[{\left(9\right)}^{3}\right]}^{\frac{1}{3}}}{{\left[{\left(5\right)}^{3}\right]}^{\frac{2}{3}}×{\left[{\left(3\right)}^{3}\right]}^{\frac{2}{3}}×{\left[{\left(2\right)}^{3}\right]}^{\frac{4}{3}}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(5\right)}^{5}×{\left(9\right)}^{1}}{{\left(5\right)}^{2}×{\left(3\right)}^{2}×{\left(2\right)}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{5×5×5×5×5×9}{5×5×3×3×2×2×2×2}\phantom{\rule{0ex}{0ex}}=\frac{125}{16}$

Page No 54:

(i) ${\left({1}^{3}+{2}^{3}+{3}^{3}\right)}^{\frac{1}{2}}$

(ii) ${\left[5{\left({8}^{\frac{1}{3}}+{27}^{\frac{1}{3}}\right)}^{3}\right]}^{\frac{1}{4}}$

(iii) $\frac{{2}^{0}+{7}^{0}}{{5}^{0}}$

(iv) ${\left[{\left(16\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}$

Page No 54:

(i) $\left[{8}^{-\frac{2}{3}}×{2}^{\frac{1}{2}}×{25}^{-\frac{5}{4}}\right]÷\left[{32}^{-\frac{2}{5}}×{125}^{-\frac{5}{6}}\right]=\sqrt{2}$

(ii) ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left(\frac{256}{625}\right)}^{\frac{1}{4}}}+\frac{\sqrt{25}}{\sqrt[3]{64}}=\frac{65}{16}$

(iii) ${\left[7{\left\{{\left(81\right)}^{\frac{1}{4}}+{\left(256\right)}^{\frac{1}{4}}\right\}}^{\frac{1}{4}}\right]}^{4}=16807$

Page No 54:

Hence, the result in the exponential form is ${x}^{\frac{1}{6}}$.

Page No 54:

(i) ${\left(\frac{{15}^{\frac{1}{3}}}{{9}^{\frac{1}{4}}}\right)}^{-6}$

(ii) ${\left(\frac{{12}^{\frac{1}{5}}}{{27}^{\frac{1}{5}}}\right)}^{\frac{5}{2}}$

(iii) ${\left(\frac{{15}^{\frac{1}{4}}}{{3}^{\frac{1}{2}}}\right)}^{-2}$

Page No 54:

Hence, the value of x is 6.

Hence, the value of x is 22.

Hence, the value of x is 5.

Hence, the value of x is 5.

Hence, the value of x is $\frac{5}{4}$.

Page No 55:

(i) $\sqrt{{x}^{-1}y}·\sqrt{{y}^{-1}z}·\sqrt{{z}^{-1}x}=1$

Hence, $\sqrt{{x}^{-1}y}·\sqrt{{y}^{-1}z}·\sqrt{{z}^{-1}x}=1$.

(ii) ${\left({x}^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left({x}^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}·{\left({x}^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}}=1$

Hence, ${\left({x}^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left({x}^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}·{\left({x}^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}}=1$.

(iii) $\frac{{x}^{a\left(b-c\right)}}{{x}^{b\left(a-c\right)}}÷{\left(\frac{{x}^{b}}{{x}^{a}}\right)}^{c}=1$

Hence, $\frac{{x}^{a\left(b-c\right)}}{{x}^{b\left(a-c\right)}}÷{\left(\frac{{x}^{b}}{{x}^{a}}\right)}^{c}=1$.

(iv)

Hence, .

Page No 55:

${\left(\frac{{x}^{b}}{{x}^{c}}\right)}^{b+c-a}·{\left(\frac{{x}^{c}}{{x}^{a}}\right)}^{c+a-b}·{\left(\frac{{x}^{a}}{{x}^{b}}\right)}^{a+b-c}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{b+c-a}·{\left({x}^{c-a}\right)}^{c+a-b}·{\left({x}^{a-b}\right)}^{a+b-c}\phantom{\rule{0ex}{0ex}}=\left[{\left({x}^{b-c}\right)}^{b}.{\left({x}^{b-c}\right)}^{c-a}\right]·{\left({x}^{c-a}\right)}^{c+a-b}·\left[{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\right]\phantom{\rule{0ex}{0ex}}=\left[{\left({x}^{b-c}\right)}^{b}.{\left({x}^{b-c}\right)}^{c-a}\right]·{\left({x}^{c+a-b}\right)}^{c-a}·\left[{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\right]\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{b}.\left[{\left({x}^{b-c}\right)}^{c-a}·{\left({x}^{c+a-b}\right)}^{c-a}\right]·{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{b}.\left[{\left({x}^{b-c+c+a-b}\right)}^{c-a}\right]·{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{b}.{\left({x}^{a}\right)}^{c-a}·{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b}\right)}^{b-c}.{\left({x}^{a}\right)}^{c-a}·{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}=\left[{\left({x}^{b}\right)}^{b-c}·{\left({x}^{a-b}\right)}^{b-c}\right].{\left({x}^{a}\right)}^{c-a}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}=\left[{\left({x}^{b+a-b}\right)}^{b-c}\right].{\left({x}^{a}\right)}^{c-a}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{a}\right)}^{b-c}.{\left({x}^{a}\right)}^{c-a}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{a}.{\left({x}^{c-a}\right)}^{a}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c+c-a+a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={x}^{0}\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}$

Page No 55:

$\frac{{9}^{n}×{3}^{2}×{\left({3}^{\frac{-n}{2}}\right)}^{-2}-{\left(27\right)}^{n}}{{3}^{3m}×{2}^{3}}=\frac{1}{27}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left({3}^{2}\right)}^{n}×{3}^{2}×{\left({3}^{-n}\right)}^{-1}-{\left({3}^{3}\right)}^{n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{2n}×{3}^{2}×{3}^{n}-{3}^{3n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{2n+2+n}-{3}^{3n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n+2}-{3}^{3n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n}×{3}^{2}-{3}^{3n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n}\left(9-1\right)}{{3}^{3m}×8}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n}\left(8\right)}{{3}^{3m}×8}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n}}{{3}^{3m}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒{3}^{3n-3m}={3}^{-3}\phantom{\rule{0ex}{0ex}}⇒3n-3m=-3\phantom{\rule{0ex}{0ex}}⇒3\left(n-m\right)=-3\phantom{\rule{0ex}{0ex}}⇒n-m=-1\phantom{\rule{0ex}{0ex}}⇒m-n=1$
Hence, m – n = 1.

Page No 57:

Since, the sum and product of a rational and an irrational is always irrational.

So, $1+\sqrt{3}$ and $2\sqrt{3}$ are irrational numbers.

Also, π is an irrational number.

And, 0 is an integer.

So, 0 is a rational number.

Hence, the correct option is (d).

Page No 57:

Since, –4.3 < –3.4 < –3 < 0 < 1.101100110001... < 3

But 1.101100110001... is an irrational number

So, the rational number between –3 and 3 is 0.

Hence, the correct option is (a).

Page No 57:

We have,

And,

Also,

Since, $\frac{1}{6}<\frac{2}{6}<\frac{3}{6}\left(\frac{1}{2}\right)<\frac{4}{6}\left(=\frac{2}{3}\right)<\frac{5}{6}<\frac{7}{6}<\frac{8}{6}\left(=\frac{4}{3}\right)<\frac{10}{6}\left(=\frac{5}{3}\right)<\frac{12}{6}\left(=\frac{2}{1}\right)$

So, the two rational numbers between  are .

Hence, the correct opion is (c).

Page No 57:

As, all rational numbers, all natural numbers and all irrational numbers can be represented on a nuumber line in an unique way.

So, every point on a number line represents a unique number.

Hence, the correct option is (d).

Page No 57:

(c) $\sqrt{225}$

Because 225 is a square of 15, i.e., $\sqrt{225}$ = 15, and it can be expressed in the $\frac{p}{q}$ form, it is a rational number.

Page No 57:

(d) a real number

Every rational number is a real number, as every rational number can be easily expressed on the real number line.

Page No 57:

(c) are infinitely many rational numbers

Because the range between any two rational numbers can be easily divided into any number of divisions, there can be an infinite number of rational numbers between any two rational numbers.

Page No 57:

(b) either terminating or repeating

As per the definition of rational numbers, they are either repeating or terminating decimals.

Page No 58:

(d) neither terminating nor repeating

As per the definition of irrational numbers, these are neither terminating nor repeating decimals.

Page No 58:

As, any number which have a terminating or non-terminating recurring decimal expansion is a rational number.

So, 0.5030030003... which is non-termintaing non-recurring decimal expansion is not a rational number.

Hence, the correct option is (d).

Page No 58:

(d) 3.141141114...

Because 3.141141114... is neither a repeating decimal nor a terminating decimal, it is an irrational number.

Page No 58:

Since, $\frac{7}{19}=\frac{7×3}{19×3}=\frac{21}{57}$

Hence, the correct option is (d).

Page No 58:

We have,

And,

Since, $-\frac{40}{60}\left(=-\frac{2}{3}\right)<-\frac{21}{60}\left(=-\frac{7}{20}\right)<-\frac{18}{60}\left(=-\frac{3}{10}\right)<-\frac{15}{60}\left(=-\frac{1}{4}\right)<-\frac{12}{60}\left(=-\frac{1}{5}\right)<\frac{18}{60}\left(=\frac{3}{10}\right)$

So, the rational number which does not lie between  is $\frac{3}{10}$.

Hence, the correct option is (b).

Page No 58:

Since, π has a non-terminating non-recurring decimal expansion.

So, π is an irrational number.

Hence, the correct option is (c).

Page No 58:

(c) a non-terminating and non-repeating decimal

Because $\sqrt{2}$ is an irrational number, its decimal expansion is non-terminating and non-repeating.

Page No 58:

Since, $\sqrt{225}$ = 15, which is an integer,

0.3799 is a number with terminating decimal expansion, and

$7.\overline{478}$ is a number with non-terminating recurring decimal expansion

Also, 23 is a prime number.

So, $\sqrt{23}$ is an irrational number.

Hence, the correct option is (a).

Page No 58:

So, there are 6 digits in the repeating block of digits in the decimal expansion of $\frac{17}{7}$.

Hence, the correct option is (b).

Page No 58:

Since,

$\sqrt{\frac{4}{9}}=\frac{2}{3}$, which is a rational number,

$\frac{\sqrt{1250}}{\sqrt{8}}=\sqrt{\frac{1250}{8}}=\sqrt{\frac{625}{4}}=\frac{25}{2}$, which is a rational number,

$\sqrt{8}=2\sqrt{2}$, which is an irrational number, and

$\frac{\sqrt{24}}{\sqrt{6}}=\sqrt{\frac{24}{6}}=\sqrt{4}=2$, which is a rational number

Hence, the correct option is (c).

Page No 59:

(d) sometimes rational and sometimes irrational

For example:
$\sqrt{2}$ is an irrational number, when it is multiplied with itself  it results into 2, which is a rational number.
$\sqrt{2}$  when multiplied with $\sqrt{3}$, which is also an irrational number, results into $\sqrt{6}$, which is an irrational number.

Page No 59:

(d) Every real number is either rational or irrational.

Because a real number can be further categorised into either a rational number or an irrational number, every real number is either rational or irrational.

Page No 59:

(d) $\mathrm{\pi }$ is irrational and $\frac{22}{7}$ is rational.
Because the value of $\mathrm{\pi }$ is neither repeating nor terminating, it is an irrational number. $\frac{22}{7}$, on the other hand, is of the form $\frac{p}{q}$, so it is a rational number.

Page No 59:

Since, $\frac{\left(\sqrt{2}+\sqrt{3}\right)}{2}$ and $\sqrt{6}$ are irrational numbers,

And,

So, the rational number lying between  is 1.6 .

Hence, the correct option is (c).

Page No 59:

Since, a number whose decimal expansion is terminating or non-terminating recurring is rational number.

So, 0.853853853... is a rational number.

Hence, the correct option is (d).

Page No 59:

Since, the product of a non-zero rational number with an irrational number is always an irrational number.

Hence, the correct option is (a).

Page No 59:

Let
Multiplying both sides by 10, we get

Subtracting (1) from (2), we get
$10x-x=2.\overline{)2}-0.\overline{)2}\phantom{\rule{0ex}{0ex}}⇒9x=2\phantom{\rule{0ex}{0ex}}⇒x=\frac{2}{9}\phantom{\rule{0ex}{0ex}}\therefore 0.\overline{)2}=\frac{2}{9}$
Hence, the correct answer is option (b).

Page No 59:

(c) $\frac{5}{3}$
Let x = 1.6666666...       ...(i)
Multiplying by 10 on both sides, we get:
10x = 16.6666666...       ...(ii)
Subtracting (i) from (ii), we get:
9x = 15
$⇒$x =

Page No 59:

(b) $\frac{6}{11}$
Let x = 0.545454...               ...(i)
Multiplying both sides by 100, we get:
100x = 54.5454545...           ...(ii)
Subtracting (i) from (ii), we get:
99x = 540
$⇒$x$\frac{54}{99}$ = $\frac{6}{11}$

Page No 60:

(c) $\frac{29}{90}$
Let x = 0.3222222222...          ...(i)
Multiplying by 10 on both sides, we get:
10x = 3.222222222...              ...(ii)
Again, multiplying by 10 on both sides, we get:
100x = 32.222222222...          ...(iii)
On subtracting (ii) from (iii), we get:
90x = 29
x = $\frac{29}{90}$

Page No 60:

(d) none of these
Let x = 0.12333333333...         ...(i)
Multiplying by 100 on both sides, we get:
100x = 12.33333333...             ...(ii)
Multiplying by 10 on both sides, we get:
1000x = 123.33333333...         ...(iii)
Subtracting (ii) from (iii), we get:
900x = 111

$⇒$x = $\frac{111}{900}$

Page No 60:

(c) $\sqrt{5×6}$

An irrational number between a and b is given as $\sqrt{ab}$.

Page No 60:

(d) 61/4
An irrational number between

Page No 60:

(c) $\sqrt{\frac{1}{7}×\frac{2}{7}}$

An irrational number between a and b is given as $\sqrt{ab}$.

Page No 60:

Let
Multiplying both sides by 10, we get

Subtracting (1) from (2), we get
$10x-x=3.\overline{)3}-0.\overline{)3}\phantom{\rule{0ex}{0ex}}⇒9x=3\phantom{\rule{0ex}{0ex}}⇒x=\frac{3}{9}\phantom{\rule{0ex}{0ex}}\therefore 0.\overline{)3}=\frac{3}{9}$
Let
Multiplying both sides by 10, we get

Subtracting (3) from (4), we get
$10y-y=4.\overline{)4}-0.\overline{)4}\phantom{\rule{0ex}{0ex}}⇒9y=4\phantom{\rule{0ex}{0ex}}⇒y=\frac{4}{9}\phantom{\rule{0ex}{0ex}}\therefore 0.\overline{)4}=\frac{4}{9}$
Sum of $0.\overline{3}$ and $0.\overline{4}$ = $0.\overline{)3}+0.\overline{)4}=\frac{3}{9}+\frac{4}{9}=\frac{7}{9}$
Hence, the correct answer is option (b).

Page No 60:

Let
Multiplying both sides by 100, we get

Subtracting (1) from (2), we get
$100x-x=245.\overline{)45}-2.\overline{)45}\phantom{\rule{0ex}{0ex}}⇒99x=245-2=243\phantom{\rule{0ex}{0ex}}⇒x=\frac{243}{99}\phantom{\rule{0ex}{0ex}}\therefore 2.\overline{)45}=\frac{243}{99}$
Let
Multiplying both sides by 100, we get

Subtracting (3) from (4), we get
$100y-y=36.\overline{)36}-0.\overline{)36}\phantom{\rule{0ex}{0ex}}⇒99y=36\phantom{\rule{0ex}{0ex}}⇒y=\frac{36}{99}\phantom{\rule{0ex}{0ex}}\therefore 0.\overline{)36}=\frac{36}{99}$
So, $2.\overline{)45}+0.\overline{)36}=\frac{243}{99}+\frac{36}{99}=\frac{243+36}{99}=\frac{279}{99}=\frac{31}{11}$
Hence, the correct answer is option (c).

Page No 60:

Hence, the correct answer is option (b).

Page No 60:

$\left(-2-\sqrt{3}\right)\left(-2+\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=-\left(2+\sqrt{3}\right)×\left[-\left(2-\sqrt{3}\right)\right]\phantom{\rule{0ex}{0ex}}=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)$

Thus, the given expression when simplified is positive and rational.

Hence, the correct answer is option (b).

Page No 60:

$\left(6+\sqrt{27}\right)-\left(3+\sqrt{3}\right)+\left(1-2\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=6+\sqrt{3×3×3}-\left(3+\sqrt{3}\right)+\left(1-2\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=6+3\sqrt{3}-3-\sqrt{3}+1-2\sqrt{3}\phantom{\rule{0ex}{0ex}}=4$
Thus, the given expression when simplified is positive and rational.

Hence, the correct answer is option (b).

Page No 60:

Hence, the correct answer is option (c).

Page No 60:

$\sqrt{20}×\sqrt{5}\phantom{\rule{0ex}{0ex}}=\sqrt{2×2×5}×\sqrt{5}\phantom{\rule{0ex}{0ex}}=2\sqrt{5}×\sqrt{5}$
$=2×5\phantom{\rule{0ex}{0ex}}=10$
Hence, the correct answer is option (a).

Page No 60:

$\frac{4\sqrt{12}}{12\sqrt{27}}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{2×2×3}}{12\sqrt{3×3×3}}\phantom{\rule{0ex}{0ex}}=\frac{4×2\sqrt{3}}{12×3\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2}{9}$
Hence, the correct answer is option (b).

Page No 61:

$=5×\sqrt{2×3}\phantom{\rule{0ex}{0ex}}=5\sqrt{6}$
Hence, the correct answer is option (ii).

Page No 61:

$=\frac{4\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}\phantom{\rule{0ex}{0ex}}=\frac{4}{2}\phantom{\rule{0ex}{0ex}}=2$
Hence, the correct answer is option (b).

Page No 61:

Hence, the correct answer is option (c).

Page No 61:

Hence, the correct answer is option (b).

Page No 61:

Hence, the correct answer is option (d).

Page No 61:

$\therefore$ The value of $\sqrt[4]{{\left(64\right)}^{-2}}$ is $\frac{1}{8}$.

Hence, the correct option is (a).

Page No 61:

$\therefore$ The value of $\frac{{2}^{0}+{7}^{0}}{{5}^{0}}$ is 2.

Hence, the correct option is (b).

Page No 61:

$\therefore$ The value of ${\left(243\right)}^{\frac{1}{5}}$ is 3.

Hence, the correct option is (a).

Page No 61:

Hence, the correct option is (c).

Page No 61:

$\therefore$Simplified value of ${\left(16\right)}^{-\frac{1}{4}}×\sqrt[4]{16}$ is 1.

Hence, the correct option is (b).

Page No 61:

$\therefore$The value of  $\sqrt[4]{\sqrt[3]{{2}^{2}}}$ is ${2}^{\frac{1}{6}}$.

Hence, the correct option is (c).

Page No 61:

$\therefore$Simplified value of ${\left(25\right)}^{\frac{1}{3}}×{5}^{\frac{1}{3}}$ is 5.

Hence, the correct option is (d).

Page No 61:

$\therefore$The value of ${\left[{\left(81\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}$ is 3.

Hence, the correct option is (a).

Page No 61:

$\therefore$ x can be $\sqrt[4]{2}$.

Hence, the correct option is (d).

Page No 62:

Hence, the correct option is (b).

Page No 62:

Hence, the correct option is (b).

Page No 62:

Hence, the correct option is (b).

Page No 62:

Hence, the correct option is (c).

Page No 62:

Hence, the correct option is (a).

Page No 62:

Hence, the correct answer is option (d).

Page No 62:

Hence, the correct answer is option (d).

Page No 62:

$\frac{{5}^{n+2}-6×{5}^{n+1}}{13×{5}^{n}-2×{5}^{n+1}}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{n}×{5}^{2}-6×{5}^{n}×5}{13×{5}^{n}-2×{5}^{n}×5}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{n}×5\left[5-6\right]}{{5}^{n}\left[13-2×5\right]}\phantom{\rule{0ex}{0ex}}=\frac{-5}{3}$
Hence, the correct answer is option (b).

Page No 62:

$\sqrt[3]{500}=\sqrt[3]{5×5×2×5×2}=\sqrt[3]{{5}^{3}×{2}^{2}}=5\sqrt[3]{4}$
So, the simplest rationalisation factor of $\sqrt[3]{500}$ is $\sqrt[3]{2}$.
Hence, the correct answer is option (d).

Page No 62:

Simplest rationalisation ractor of $\left(2\sqrt{2}-\sqrt{3}\right)$ is $2\sqrt{2}+\sqrt{3}$.
Hence, the correct answer is option (b).

Page No 62:

Rationalisation factor of $\frac{1}{2\sqrt{3}-\sqrt{5}}$ will be $2\sqrt{3}+\sqrt{5}=\sqrt{4×3}+\sqrt{5}=\sqrt{12}+\sqrt{5}$.
Hence, the correct answer is option (d).

Page No 62:

$\frac{1}{\sqrt{5}+\sqrt{2}}$
Rationalisation of denominator gives
$\frac{1}{\sqrt{5}+\sqrt{2}}×\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}$
Hence, the correct answer is option (d).

Page No 62:

$x=2+\sqrt{3}\phantom{\rule{0ex}{0ex}}\frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}×\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$
$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4$
Hence, the correct answer is option (c).

Page No 63:

$\frac{1}{\left(3+2\sqrt{2}\right)}=\frac{1}{\left(3+2\sqrt{2}\right)}×\frac{\left(3-2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)}=\frac{\left(3-2\sqrt{2}\right)}{9-8}=\left(3-2\sqrt{2}\right)$
Hence, the correct answer is option (c).

Page No 63:

Given: $x=\left(7+4\sqrt{3}\right)$
$\frac{1}{x}=\frac{1}{7+4\sqrt{3}}=\frac{1}{7+4\sqrt{3}}×\frac{7-4\sqrt{3}}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}}{49-48}=7-4\sqrt{3}$
$\left(x+\frac{1}{x}\right)=7+4\sqrt{3}+\left(7-4\sqrt{3}\right)=14$
Hence, the correct answer is option (b).

Page No 63:

Hence, the correct answer is option (c).

Page No 63:

$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}×\frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$
Given that

Hence, the correct answer is option (b).

Page No 63:

$3-2\sqrt{2}=2+1-2×\sqrt{2}×1\phantom{\rule{0ex}{0ex}}={\left(\sqrt{2}\right)}^{2}+{1}^{2}-2×\sqrt{2}×1$
This is of the form
${a}^{2}+{b}^{2}-2ab={\left(a-b\right)}^{2}$

Hence, the correct answer is option (d).

Page No 63:

$5+2\sqrt{6}=2+3+2×\sqrt{3}×\sqrt{2}\phantom{\rule{0ex}{0ex}}={\left(\sqrt{2}\right)}^{2}+{\left(\sqrt{3}\right)}^{2}+2×\sqrt{3}×\sqrt{2}$
This is in the form
${a}^{2}+{b}^{2}+2ab={\left(a+b\right)}^{2}$
So, we have
${\left(\sqrt{2}\right)}^{2}+{\left(\sqrt{3}\right)}^{2}+2×\sqrt{3}×\sqrt{2}={\left(\sqrt{2}+\sqrt{3}\right)}^{2}$
Thus, $\sqrt{5+2\sqrt{6}}=\sqrt{{\left(\sqrt{2}+\sqrt{3}\right)}^{2}}=\sqrt{2}+\sqrt{3}$
Hence, the correct answer is option (c).

Page No 63:

$\sqrt{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)}}=\sqrt{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)}×\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)}}=\sqrt{\frac{{\left(\sqrt{2}-1\right)}^{2}}{2-1}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{2+1-2\sqrt{2}}{1}}\phantom{\rule{0ex}{0ex}}=\sqrt{3-2\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{3-2×1.414}\phantom{\rule{0ex}{0ex}}=\sqrt{3-2.828}\phantom{\rule{0ex}{0ex}}=\sqrt{0.172}\phantom{\rule{0ex}{0ex}}=0.414$
Hence, the correct answer is option (c).

Page No 63:

Given: $x=3+\sqrt{8}$
$\frac{1}{x}=\frac{1}{3+\sqrt{8}}=\frac{1}{3+\sqrt{8}}×\frac{3-\sqrt{8}}{3-\sqrt{8}}=\frac{3-\sqrt{8}}{9-8}=\frac{3-\sqrt{8}}{1}=3-\sqrt{8}$
$x+\frac{1}{x}=\left(3+\sqrt{8}\right)+\left(3-\sqrt{8}\right)=6$
${\left(x+\frac{1}{x}\right)}^{2}={x}^{2}+\frac{1}{{x}^{2}}+2×x×\frac{1}{x}={x}^{2}+\frac{1}{{x}^{2}}+2\phantom{\rule{0ex}{0ex}}⇒{6}^{2}={x}^{2}+\frac{1}{{x}^{2}}+2\phantom{\rule{0ex}{0ex}}⇒36={x}^{2}+\frac{1}{{x}^{2}}+2\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\frac{1}{{x}^{2}}=36-2=34$
Hence, the correct answer is option (a).

Page No 63:

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

So, Assertion and Reason are correct (property of rational numbers). Also, Reason is the correct explanation of Assertion.

Page No 64:

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

3 3

Page No 64:

(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

It is known that e and $\mathrm{\pi }$ are irrational numbers, but Reason is not the correct explanation.

Page No 64:

(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

Page No 64:

(a) Because it is a non-terminating and repeating decimal, it is a rational number.

(b) $\mathrm{\pi }$ is an irrational number.

(c) $\frac{1}{7}=.142857142857$...
Hence, its period is 6.

(d)

Page No 64:

(a)
${\left({\left(81\right)}^{-2}\right)}^{\frac{1}{4}}\phantom{\rule{0ex}{0ex}}={\left({\left(9\right)}^{-4}\right)}^{\frac{1}{4}}={\left(9\right)}^{-4×\frac{1}{4}}={\left(9\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}$

(b)

(c)

(d)

${\left(\frac{3}{2}\right)}^{4×\frac{-3}{4}}×{\left(\frac{4}{3}\right)}^{3×\frac{-1}{3}}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{2}\right)}^{-3}×{\left(\frac{4}{3}\right)}^{-1}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{2}\right)}^{-3}×\frac{3}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\left(\frac{{3}^{-3}}{{2}^{-3}}\right)×\frac{3}{{2}^{2}}\phantom{\rule{0ex}{0ex}}=\left(\frac{{3}^{-3}×3}{{2}^{-3}×{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{3}^{-2}}{{2}^{-1}}\phantom{\rule{0ex}{0ex}}=\frac{2}{9}$

Page No 65:

Sum of a rational number and an irrational number is an irrational number.
Example: 4 + $\sqrt{5}$ represents sum of rational and an irrational number where 4 is rational and $\sqrt{5}$ is irrational.

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$\frac{665}{625}=\frac{5×19×7}{{5}^{4}}=\frac{19×7}{{5}^{3}}=\frac{19×7×{2}^{3}}{{5}^{3}×{2}^{3}}=\frac{1064}{1000}=1.064$
So, $\frac{665}{625}$ will terminate after 3 decimal places.

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${\left(1296\right)}^{0.17}×{\left(1296\right)}^{0.08}\phantom{\rule{0ex}{0ex}}={\left(1296\right)}^{0.17+0.08}\phantom{\rule{0ex}{0ex}}={\left(1296\right)}^{0.25}\phantom{\rule{0ex}{0ex}}={\left(1296\right)}^{\frac{1}{4}}\phantom{\rule{0ex}{0ex}}=\sqrt[4]{1296}\phantom{\rule{0ex}{0ex}}=6$

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$6\sqrt{36}+5\sqrt{12}\phantom{\rule{0ex}{0ex}}=6×6+5\sqrt{4×3}\phantom{\rule{0ex}{0ex}}=36+10\sqrt{3}$

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A number which is non terminating and non recurring is known as irrational number.

There are infinitely many irrational numbers between 5 and 6.

One of the example is 5.40430045000460000....

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Hence, the value of $\frac{21\sqrt{12}}{10\sqrt{27}}$ is $\frac{7}{5}$.

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Hence, the rationalised form is $\sqrt{3}-\sqrt{2}$.

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${\left(\frac{2}{5}\right)}^{2x-2}=\frac{32}{3125}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{2}{5}\right)}^{2x-2}=\frac{{2}^{5}}{{5}^{5}}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{2}{5}\right)}^{2x-2}={\left(\frac{2}{5}\right)}^{5}\phantom{\rule{0ex}{0ex}}⇒2x-2=5\phantom{\rule{0ex}{0ex}}⇒2x=5+2\phantom{\rule{0ex}{0ex}}⇒x=\frac{7}{2}$
Hence, $x=\frac{7}{2}$.

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Hence, ${\left(32\right)}^{\frac{1}{5}}+{\left(-7\right)}^{0}+{\left(64\right)}^{\frac{1}{2}}$ = 11.

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Hence, ${\left(\frac{81}{49}\right)}^{\frac{-3}{2}}=\frac{343}{729}$.

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For = 1and b = 2,
${\left({a}^{b}+{b}^{a}\right)}^{-1}\phantom{\rule{0ex}{0ex}}={\left({1}^{2}+{2}^{1}\right)}^{-1}\phantom{\rule{0ex}{0ex}}={\left(1+2\right)}^{-1}$

Thus, the value of (ab + ba)–1 when a = 1 and b = 2 is $\frac{1}{3}$.

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Let the two irrational numbers be $2+\sqrt{3}$ and $2-\sqrt{3}$.
Sum of these irrational numbers $=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4$, which is rational
Product of these irrational numbers $=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)={2}^{2}-{\left(\sqrt{3}\right)}^{2}=4-3=1$, which is rational

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Yes, the product of a rational and an irrational number is always an irrational number.
Example:
2 is a rational number and $\sqrt{3}$ is an irrational number.
Now, $2×\sqrt{3}=2\sqrt{3}$, which is an irrational number.

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The cube roots of natural numbers which are not perfect cubes are all irrational numbers.
Let $x=\sqrt[3]{2}={2}^{\frac{1}{3}}$.
Now,
${x}^{2}={\left({2}^{\frac{1}{3}}\right)}^{2}={2}^{\frac{2}{3}}={\left({2}^{2}\right)}^{\frac{1}{3}}={4}^{\frac{1}{3}}$, which is an irrational number
Also,
${x}^{3}={\left({2}^{\frac{1}{3}}\right)}^{3}={2}^{3×\frac{1}{3}}=2$, which is a rational number

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The reciprocal of $\left(2+\sqrt{3}\right)$

$=\frac{\left(2-\sqrt{3}\right)}{4-3}\phantom{\rule{0ex}{0ex}}=\frac{\left(2-\sqrt{3}\right)}{1}\phantom{\rule{0ex}{0ex}}=\left(2-\sqrt{3}\right)$

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The value of $\frac{1}{\sqrt{10}}=\frac{1}{\sqrt{10}}×\frac{\sqrt{10}}{\sqrt{10}}=\frac{\sqrt{10}}{10}=\frac{3.162}{10}=0.3162$

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We have,

${10}^{x}=64$

Taking square root from both sides, we get

$\sqrt{{10}^{x}}=\sqrt{64}\phantom{\rule{0ex}{0ex}}⇒{\left({10}^{x}\right)}^{\frac{1}{2}}=8\phantom{\rule{0ex}{0ex}}⇒{10}^{\left(\frac{x}{2}\right)}=8$

Multiplying both sides by 10, we get

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$\frac{{2}^{n}+{2}^{n-1}}{{2}^{n+1}-{2}^{n}}\phantom{\rule{0ex}{0ex}}=\frac{{2}^{n}\left(1+{2}^{-1}\right)}{{2}^{n}\left(2-1\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(1+\frac{1}{2}\right)}{1}\phantom{\rule{0ex}{0ex}}=\frac{2+1}{2}\phantom{\rule{0ex}{0ex}}=\frac{3}{2}$
${\left[{\left\{{\left(256\right)}^{-\frac{1}{2}}\right\}}^{\frac{1}{4}}\right]}^{2}\phantom{\rule{0ex}{0ex}}={\left\{{\left(256\right)}^{-\frac{1}{2}}\right\}}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}={\left(256\right)}^{-\frac{1}{4}}\phantom{\rule{0ex}{0ex}}={\left({4}^{4}\right)}^{-\frac{1}{4}}\phantom{\rule{0ex}{0ex}}={4}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}$