Number system class 9 important questions

Rational number between two number can be find using mean method
(i) Mean of 0.75 and 1.2 = $\frac {.75 + 1.2}{2}= .975$
So .975 is the rational lying between 0.75 and 1.2
(ii) Mean of -3/4 and -2/5 = $\frac {(-3/4) + (-2/5)}{2}= \frac {-23}{40}$
So $\frac {-23}{40}$ is the rational number

3.1,3.2,.3.3,3.4,3.5,3.6

2.101
2.102
2.103
2.104
2.105
2.106
2.107
2.108
2.109
2.110
2.111
2.112
2.113
2.114
2.115
2.116

$0.\overline{9}=0.99999..$ Let $x = 0.9999...$
$10x = 9.9999...$
$10x = 9 + x$
$9x = 9$
$x = 1$

(i) x=√ 11, So it is irrational Number
(ii) y=6, So it is rational number
(iii) z==.2, So it is rational number
(iv) u = √19/3, So it is rational number

$\frac {1}{7}=.142857142857.....= 0.\overline{142857}$
$\frac {2}{7}=.285714285714.....= 0.\overline{285714}$
Now we know that a irrational number is a non-terminating and non repeating number, So irrational number between these can be
.162857147647222977....
or
.17285716786222977....

Answer is (i)

(i)$0.\overline{36}=0.363636..$ Let $x = 0.363636.$
$100x = 36.363636.$
$100x = 36 + x$
$99x = 36$
$x = \frac {36}{99}$
(ii)$0.5\overline{6}=0.56666..$
Let $x = 0.56666..$
$10x = 5.666666....= 5 + 0.\overline{6}$
Now let $y=0.\overline{6}$
$10y=6.6666..=6+y$
or $y=\frac {6}{9}$
So $10x = 5 + \frac {6}{9}$
$x=\frac {51}{90}= \frac {17}{30}$

1.Irrational
2.Integers
3.recurring
4.Rational
5.Archimedes
6. Pythagoras
7. two
8. Natural

$\frac {\sqrt 2 -1}{\sqrt + 1} = \frac {\sqrt 2 -1}{\sqrt 2 + 1} \times \frac {\sqrt 2 -1}{\sqrt 2 - 1}=(\sqrt 2 -1)^2=3 -2 \sqrt 2$
Therefore p=3 and q=2

$x + y = \frac {2 + \sqrt 3}{ \sqrt 2 +1} + \frac {2 - \sqrt 3}{ \sqrt 2 -1} = (2+ \sqrt 3}{(\sqrt 2 -1) + (2 - \sqrt 3}{(\sqrt 2 +1)$
$=2 \sqrt 2 -2 + \sqrt 6 - \sqrt 3 + 2 \sqrt 2 +2 - \sqrt 6 -\sqrt 3=4 \sqrt 2 -2 \sqrt 3= 2( 2\sqrt 2 - \sqrt 3)$
$ x - y = \frac {2 + \sqrt 3}{ \sqrt 2 +1} - \frac {2 - \sqrt 3}{ \sqrt 2 -1} = (2+ \sqrt 3}{(\sqrt 2 -1) - (2 - \sqrt 3}{(\sqrt 2 +1)$
$=2 \sqrt 2 -2 + \sqrt 6 - \sqrt 3 - 2 \sqrt 2 -2 + \sqrt 6 +\sqrt 3=2 \sqrt 6 -4= 2( \sqrt 6 - \2)$

(p) -> (a) as $(4 -\sqrt 3) (4 + \sqrt 3)= 16-3=13$
(q) -> (a)
(r) - >(b)
(s) -> (b)

(i) $ a^3 + \frac {1}{a^3}= (\sqrt 2 -1) + \frac {1}{\sqrt 2 -1}= (\sqrt 2 - 1) + \frac {1}{\sqrt 2 -1} \times \frac {\sqrt 2 +1}{\sqrt 2 +1}= (\sqrt 2 - 1) + (\sqrt 2 + 1)=2 \sqrt 2$
(ii) $ a^3 - \frac {1}{a^3}= (\sqrt 2 -1) - \frac {1}{\sqrt 2 -1}= (\sqrt 2 - 1) - \frac {1}{\sqrt 2 -1} \times \frac {\sqrt 2 +1}{\sqrt 2 +1}= (\sqrt 2 - 1) - (\sqrt 2 + 1)=-2$