Zero is a rational or an irrational number?

Is (x+y)/2 is an irrational number or a rational number?

How can we find the irrational numbers between two fractions

What is the multiplicative inverse of rational numbers

find two rationals between 0.5 and 0.55

Can u explain solving of rational numbers with appropriate property

## Is zero (0) a rational number ? justify your answer ?

Yes, zero is a rational number

zero can be expressed as    $\frac{0}{5},\frac{0}{26},\frac{0}{100}$  etc

which are in the form of, $\frac{p}{q}$  where p and q are integers and q $\neq$ 0.

## Write the simplest form of a rational number    $\frac{177}{413}$

$\frac{177}{413} = \frac{59\times 3}{59\times 7} = \frac{3}{7}$

## Is   $\frac{\sqrt{98}}{\sqrt{2}}$  a rational number or not ?

$\frac{\sqrt{98}}{\sqrt{2}} =\frac{ \sqrt{49}\times \sqrt{2}}{\sqrt{2}} =\sqrt{49}= 7$

so it is rational number.

## Identify an irrational number among the following numbers: 0.13 ,$0.13\overline{ 15}$ , $0.\overline{ 1315}$ , 0.3013001300013.....

0.13 is a terminating number. So, it is not an irrational number.

$0.13\overline{15}$ = 0.131515......, 15 is repeating continuously so it is not an irrational number.

$0.\overline{ 1315}$ = 0.13151315........is repeating continuously so it is not an irrational number.

0.3013001300013....., non terminating and non recurring decimal. Hence, it is an irrational number. So, 0.3013001300013 is an irrational number.

## Is the product of two irrational number is always an irrational number ?

No, it may be rational or irrational.

## Insert three rational numbers between $\frac{-1}{\: \: 3}$   and   $\frac{-2}{\: \: 3}$.

$\frac{-1}{\: \: 3} = \frac{-4}{12}$

$\frac{-2}{\: \: 3} = \frac{-8}{12}$

So three rational numbers are

$\frac{-5}{\: \: 3},\frac{-6}{12}\: \: and\: \: \frac{-7}{12}$

## Find two rational numbers between 4 and 5 .

$4 = \frac{4}{5} \times 5\: \: and \: \: 5 = \frac{5}{5}\times 5$

$4 = \frac{20}{5}\: \: and \: \: 5 = \frac{25}{5}$

The numbers are $\frac{21}{5}$ and $\frac{22}{5}$

## Express the rational numbers  $0 .\overline{9}$  in the form  $\frac{p}{q}$ ,  where p and q are integers and $q \neq 0$

Let  X  =  0.999............

10x  =  9.999.......

10x - x  =  (9.999....) – (0.999....)

9x  =  9

X = 1

## Calculate the irrational number 2 and 2.5

Since $\sqrt{5}= 2.236$

hence, the irrationl number between2 and 2.5 is $\sqrt{5}$

## Simplify the number   $\left ( \sqrt{2}+\sqrt{5} \right) ^{2}$

$\left ( \sqrt{2}+\sqrt{5} \right) ^{2}= \left (\sqrt{2} \right )^{2}+\left ( \sqrt{5} \right )^{2}$$+2\times \sqrt{2}\times \sqrt{5}$

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

$= 7+2\sqrt{10}\: \: \left ( irrational\: number \right )$

## Find the rational numbers between 0.121221222122221...and  0.141441444144441... in the p form, where p and q integers and $q \neq 0$

Two rational numbers between0.121221222122221...and 0.141441444144441...are 0.13 and 0.14

$\frac{13}{10} \: \: and\: \: \frac{14}{100}$

$\frac{13}{100}\: \: and\: \: \frac{7}{50}$

## Find four rational numbers between   $\frac{1}{5}$   and   $\frac{1}{6}$

Since LCM of 5 and 6 is 30

$\frac{1}{6} = \frac{1}{6} \times \frac{5}{5} = \frac{5}{30} \times \frac{5}{5} = \frac{25}{150}$

$\frac{1}{5} = \frac{1}{5} \times \frac{6}{6} = \frac{6}{30} \times \frac{5}{5} = \frac{30}{150}$

Hence , four rational numbers between   $\frac{1}{6}$   and   $\frac{1}{5}$

are $\frac{26}{150} , \frac{27}{150} , \frac{28}{150} , \frac{29}{150}$

## Find six rational numbers between 3 and 4.

let  a = 3 , and b = 4

Here, we find six rational numbers i,e n = 6

$d = \frac{b - a}{n + 1} = \frac{4 - 3}{6 + 1} = \frac{1}{7}$

1st rational number =   $a + d = 3 + \frac{1}{7} = \frac{22}{7}$

2nd rational number =  $a + 2d = 3 + \frac{2}{7} = \frac{23}{7}$

3 rd rational number =  $a + 3d = 3 + \frac{3}{7} = \frac{24}{7}$

4th rational number  =  $a + 4d = 3 + \frac{4}{7} = \frac{25}{7}$

5th rational number  =  $a + 5d = 3 + \frac{5}{7} = \frac{26}{7}$

6th rational number  =   $a + 6d = 3 + \frac{6}{7} = \frac{27}{7}$

So six rational numbers are $\frac{22}{7} , \frac{23}{7} , \frac{24}{7} , \frac{25}{7} , \frac{26}{7} , \frac{27}{7}$

## Insert three rational numbers between  and   $\frac{3}{5}$  and $\frac{5}{7}$

LCM of 5 and 7 is  35

$\frac{3}{5} = \frac{3}{5} \times \frac{7}{7} = \frac{21}{35}$

and     $\frac{5}{7} = \frac{5}{7} \times \frac{5}{5} = \frac{25}{35}$

so      $\frac{21}{35} < \frac{22}{35} < \frac{23}{35} < \frac{24}{35} < \frac{25}{25}$

The required three rational numbers are

$\frac{22}{35} , \frac{23}{35}\: \: and\: \: \frac{24}{35}$

## $\frac{\sqrt{147}}{\sqrt{75}}$  is not a rational number as $\sqrt{147}$  and $\sqrt{75}$ are not rational. State whether it is true or false. Justify your answer.

false

justification  :     $\frac{\sqrt{147}}{\sqrt{75}} = \sqrt{\frac{147}{75}} = \sqrt{\frac{49}{25}} = \frac{7}{5}$

Which is a rational number

## simplify   $\frac{6-4\sqrt{3}}{6+4\sqrt{3}}$   by rationalizing the denominator

$\frac{6-4\sqrt{3}}{6+4\sqrt{3}} \times \frac{6-4\sqrt{3}}{6-4\sqrt{3}}$

$= \frac{(6-4\sqrt{3})^{2})}{36-48}$

$= \frac{36+48-48\sqrt{3}}{-12}$

$= \frac{84-48\sqrt{3}}{-12} =-(7-4\sqrt{3})=4\sqrt{3}-7$

## Find three rational numbers between  $\frac{5}{7}$  and  $\frac{9}{11}$

Since LCM of 7 and 11 is 77

$\frac{5}{7} = \frac{5}{7}$$\times \frac{11}{11} = \frac{55}{77}$

$\frac{9}{11} = \frac{9}{11}\times \frac{7}{7} = \frac{63}{77}$

Hence three rational numbers between  $\frac{5}{7}$  and  $\frac{9}{11}$  are  $\frac{56}{77} , \frac{57}{77} , \frac{58}{77}$

## Rationalize the denominator  $\frac{1}{2\sqrt{7}+3\sqrt{3}}$

$\frac{1}{2\sqrt{7}+3\sqrt{3}} =\frac{1}{(2\sqrt{7}+3\sqrt{3})}\times \frac{(2\sqrt{7}-3\sqrt{3})}{2\sqrt{7}-3\sqrt{3}}$

$=\frac{2\sqrt{7}-3\sqrt{3}}{(2\sqrt{7})^{2}-(3\sqrt{3})^{2}}$

$=\frac{2\sqrt{7}-3\sqrt{3}}{4\times 7-9\times 3}$$=\frac{2\sqrt{7}-3\sqrt{3}}{1}$$= \frac{2\sqrt{7}-3\sqrt{3}}{28-27} = 2\sqrt{7}-3\sqrt{3}$

## Rationalize the denominator of   $\frac{30}{5\sqrt{3}-3\sqrt{5}}$

$\frac{30}{5 \sqrt{3}-3\sqrt{5}} \times \frac{5\sqrt{3 }+3\sqrt{5}}{5\sqrt{3 }+3\sqrt{5}}$ $=\frac{30(5(\sqrt{3 }+3\sqrt{5})}{(5\sqrt{3 })^{2}-(3\sqrt{5})^{2}}$

$= \frac{30(5\sqrt{3}+3\sqrt{5})}{30}$

$= 5\sqrt{3}+3\sqrt{5$

Alternative Method

$\frac{30}{5\sqrt{3}-3\sqrt{5}} = \frac{30}{(5\sqrt{3}-3\sqrt{5})}\times$ $\frac{(5\sqrt{3}+3\sqrt{5})}{(5\sqrt{3}+3\sqrt{5})}$

$= \frac{30(5\sqrt{3}+3\sqrt{5})}{(5\sqrt{}3)^{2}-(3\sqrt{5})^{2}}$

$= \frac{30(5\sqrt{3}+3\sqrt{5})}{75-45} =\frac{30(5\sqrt{3}+3\sqrt{5})}{30}$

$= 5\sqrt{3 }+3\sqrt{5}$

## Give two rational numbers whose  : (1)  Difference is a rational number (2)  Sum is a rational number (3)  Product is a rational number (4)  Division is a rational  number               Justify also.

Any example and verification of example

$Let\: \: \: m = \frac{4}{5},\: \: \: n = \frac{9}{2}$

$difference\: \: =\frac{9}{2}-\frac{4}{5}=\frac{37}{10}\: \: \: (Rational number )$

$sum\: \: =\frac{4}{5}+\frac{9}{2}=\frac{53}{10}\: \: \: (Rational number )$

product          =       4/5 x 9/2 = 36/10 (Rational number)

Division         =       9/2  $\div$ 4/5 = 45/8  ( Rational number)

## Rationalize the denominator of  $\frac{\sqrt{3}+\sqrt{2}}{5+\sqrt{2}}$

$\frac{\sqrt{3} +\sqrt{2}}{5+\sqrt{2} }= \frac{\sqrt{3} + \sqrt{2}}{5+ \sqrt{2}} \times \frac{5-\sqrt{2}}{5 -\sqrt{2}}$

$= \frac{5\sqrt{3} +5\sqrt{2} - \sqrt{6}-2}{25-2}$

$=\frac{5\sqrt{3}+5\sqrt{2} -\sqrt{6}-2}{23}$

## Given two rational numbers whose (1) difference is a rational number (2) sum is a rational number (3) product is a rational number (4) division is a rational number

Any example and verification of example

let m = 4/5, n= 9/2

$\: difference\: = \: \frac{9}{2}-\frac{4}{5}=\frac{37}{10}\: \: (\: rational \: \: number\: )$

$\: sum\: = \: \frac{4}{5}+\frac{9}{2}=\frac{53}{10}\: \: (\: rational \: \: number\: )$

$\: product\: = \: \frac{4}{5}\times \frac{9}{2}=\frac{36}{10}\: \: (\: rational \: \: number\: )$

$\: division\: = \: \frac{9}{2}\div \frac{4}{5}=\frac{45}{8}\: \: (\: rational \: \: number\: )$

## Find any two irrational numbers between 0.1 and 0.12.

Required to irrational number are :

i)  0.10100100010000......

ii) 0.1020020002000.......

## Find the values of a and b when   $\frac{5+\sqrt{6}}{5-\sqrt{6}} = a+b \sqrt{6}$

$\frac{5+\sqrt{6}}{5-\sqrt{6}} =\frac{5+\sqrt{6}}{5-\sqrt{6}} \times \frac{5+\sqrt{6}}{5+\sqrt{6}}$

$= \frac{(5+\sqrt{6})^{2}}{(5)^{2}-(\sqrt{6}^{2})}$

$= \frac{25+6+10\sqrt{6}}{25-6}$

$= \frac{31+10\sqrt{6}}{19}$

$\therefore a+b \sqrt{6}=\frac{31}{19} +\frac{10}{19}\sqrt{6}$

Compairing the rational and irrational parts  of both sides we get

$a =\frac{31}{19} \: \: b\: ,=\: \frac{10}{19}$

## Rationalize the denominator of $\frac{1}{(\sqrt{2}+\sqrt{3})-\sqrt{4}}$

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

$=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{1+2\sqrt{6}}\times \frac{1-2\sqrt{6}}{1-2\sqrt{6}}$

$=\frac{\sqrt{2}+ \sqrt{3}+\sqrt{4}-2\sqrt{12}-2\sqrt{18}-4\sqrt{6}}{1^{2}-(2\sqrt{6})^{2}}$

$=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-4\sqrt{3}-6\sqrt{2}-4\sqrt{6}}{1-24}$

$=\frac{-5\sqrt{2}-3\sqrt{3}+2-4\sqrt{6}}{-23}$

$=\frac{+5\sqrt{2}+3\sqrt{3}+4\sqrt{6}-2}{23}$

## if  $x =3-2\sqrt{2}$  then find the value of  $x^{4}-\frac{1}{x^{4}}$

Given $x\: = 3-2\sqrt{2}$

$\frac{1}{x}= \frac{1}{(3-2\sqrt{2})}\times \frac{(3+ 2\sqrt{2})}{(3+2\sqrt{2})}\: \: (\: rationalizing\: )$

$\frac{1}{x}= \frac{(3+2\sqrt{2})}{9-8}= 3+2\sqrt{2}$

$\frac{1}{x^{2}}= (3+2\sqrt{2})^{2}=9+8+12\sqrt{2}=17+12\sqrt{2}$

$x^{2}= (3-2\sqrt{2})^{2}=9+8-12\sqrt{2}=17-12\sqrt{2}$

Now , $x^{4}-\frac{1}{x^{4}}=\left ( x^{2}-\frac{1}{x^{2}} \right )\left ( x^{2}+\frac{1}{x^{2}} \right )$

$=\left [ (17-12\sqrt{2})-(17+12\sqrt{2}) \right ] \left [ (17-12\sqrt{2}) +(17+12\sqrt{2})\right ]$

$= 17-12\sqrt{2}-17-12\sqrt{2}\left ( 17-12\sqrt{2}+17+12\sqrt{2} \right )$

$=(-24\sqrt{2})\times 34=-816\sqrt{2}$

## (i)Find the six rational  numbers bbetween 3 and 4  (ii) Which mathematical concept is used in this problem (iii) Which value is depticted in this question

(i) We known that between two rational numbers  x and y such that x< y there is a rational number $\frac{x+y}{2}$.

ie,       $3< \frac{7}{2}< 4$

Now a rational number between 3 and   $\frac{7}{2}< 4$   is :

$\frac{1}{2}\left ( 3+\frac{7}{2} \right )=\frac{1}{2}\times \left ( \frac{6+7}{2} \right )=\frac{13}{4}$

A rational nummber berween  $\frac{7}{2}\: \: and \: \: 4 \: \: is$

$\frac{1}{2}\left ( \frac{7}{2}+4 \right )=\frac{1}{2}\times \left ( \frac{7+8}{2} \right )=\frac{15}{4}$

$3< \frac{13}{4}< \frac{7}{2}< \frac{15}{4}< 4$

Further a rational number between 3 and  $\frac{13}{4}\: \: is :$

$\frac{1}{2}\left ( 3+\frac{13}{4} \right )=\frac{1}{2}\left ( \frac{12+13}{4} \right )=\frac{25}{8}$

A rational number between  $\frac{15}{4}\: \: and\: \: 4\: \: is :$

$\frac{1}{2}\left ( \frac{15}{4}+4 \right )=\frac{1}{2}\times \frac{15+16}{4}=\frac{31}{8}$

A rational number between  $\frac{31}{8}\: \: and\: \: 4\: \: is$

$\frac{1}{2}\left ( \frac{31}{8}+4 \right ) = \frac{1}{2}\times \left ( \frac{31+32}{8} \right )=\frac{63}{16}$

$\therefore 3< \frac{25}{8}< \frac{13}{4}< \frac{7}{2}< \frac{15}{4}< \frac{31}{8}< \frac{63}{16}< 4$

Hence , six rational numbers between 3 and 4 are

$\frac{25}{8}, \frac{13}{4},\frac{7}{2}, \frac{15}{4}, \frac{31}{8}\: \: and\: \: \frac{63}{16}$

(ii) Number syatem

(iii) Rationality is always welcomed

## Find the smallest natural number by which 1,200 should be multiplied so that the square root of the product is a rational number.

$1,200= 4\times 3\times (2\times 5)^{2}$

$=2^{4}\times 3\times 5^{2}$

The smallest natural number is 3.

## What is the condition  for the decimal expansion of a rational number to terminate? Explain with the help of an example.

The decimal expansion of a rational  number terminates, if the denominator of rational no $\frac{p}{q}$, When p and q  are co-primes and q can be expressed as 2m 5n . where m and n are non- negative integers.

eg-

$\frac{3}{10}=\frac{3}{2^{1}\times 5^{1}}=0.3$

## Find the smallest positive rational number by which  $\frac{1}{7}$ should be multiplied so that its decimal   expansion terminates after 2 places of decimal

Since  $\frac{1}{7}\times \frac{7}{100} = \frac{1}{100} = 0.01$

Thus smallest rational number is $\frac{7}{100}$

## What type of decimal expansion does a rational  number has? How can you distinguish it from decimal expansion of irrational number?

A rational number is either terminating or non- terminating repeating.

An irrational number is non-terminating and non- repeating.

## Show that $5\sqrt{6}$ is an irrational number.

Let  $5\sqrt{6}$  be a rational number, which can be put in the form $\frac{a}{b}$ , where   $b \neq 0$ , a and b   are co-prime

$5\sqrt{6}=\frac{a}{b}$

$\sqrt{6}=\frac{a}{5b}$

$\Rightarrow \sqrt{6}$= rational

But, we know that $\sqrt{6}$ is an irrational number.

Thus, our assumption is wrong.

Hence, 5$\sqrt{6}$ is an irrational number.

## Write the denominator of the rational number $\frac{257}{500}$  in the form 2m x 5n , where  m and n are non- negative integers. Hence write its decimal expansion without actual division.

Denominator  = 500

= 22 x 53

Decimal expansion,  $\frac{257}{500}=\frac{257\times 2}{2\times 2^{2}\times 5^{3}} = \frac{514}{10^{3}}$

$= 0.514$

## Express the number $0.3\overline{178}$ in the form of rational number $\frac{a}{b}$.

Let   $x =0.3 \overline{178}$

$\Rightarrow x = .3178178178....................$

$\Rightarrow 10,000x = 3178.178178....................$

$\Rightarrow 10x = 3.178178.........$

Substracting $9990x = 3175$

$\Rightarrow x = \frac{3175}{9990} = \frac{635}{1998}$

## Prove that $\sqrt{2}$ is an irrational number

Let $\sqrt{2}$ be a rational number

$\therefore \sqrt{2}=\frac{p}{q}$

Where p and q are co-prime integers and  $q \neq 0$

$\Rightarrow 2 =\frac{p^{2}}{q^{2}}$

$\Rightarrow p^{2} = 2q^{2}$

$\Rightarrow P^{2}$ is divisible by 2.

$\therefore$  P is divisible by 2.

Let p = 2r for some positive integer r

$\Rightarrow p^{2} = 4r^{2}$

$\therefore 2q^{2} = 4r^{2}$

$\Rightarrow q^{2} = 2r^{2}$

$\Rightarrow q^{2}$ is divisible by 2.

$\therefore$ q is divisible by 2.

From (i) and (ii), p and q are divisible by 2, which contradicts the fact that p and q are co-primes.

Hence, our assumption is false.

So, $\sqrt{2}$ is irrational

## Prove that $\sqrt{3}$  is an irrational number.

Let $\sqrt{3}$  be a rational number

$\sqrt{3}= \frac{a}{b}$

(a and b are integers and co-primes )

On squaring both the sides,

$3 = \frac{a^{2}}{b^{2}}$

$\Rightarrow 3b^{2}= a^{2}$

$\Rightarrow a^{2}$ is divisible by 3

$\therefore a$  is divisible by 3

We can write  a = 3c for some integer

$\Rightarrow a^{2}=9c^{2}$

$\Rightarrow 3b^{2}=9c^{2}$

$\Rightarrow b^{2}=3c^{2}$

$b^{2}$ is divisible by 3

b is divisible by 3.

From equations (i) ad (ii),  we get 3 as a factor of "a"  and  "b"

Which is contradicting the fact that a and b are co-primes. Hence our assumption that  $\sqrt{3}$ is a rational number, is false. So  $\sqrt{3}$ is an irrational number.

## if p is a prime number, then prove that $\sqrt{p}$ is irrational.

Let p be a prime number and if possible, let $\sqrt{p}$ be rational.

Let  $\sqrt{p}= \frac{m}{n}$  where m and n are  integers having no

common factor other than 1 and $n\neq 0$

Then,  $\sqrt{p}= \frac{m}{n}$

Squaring on both sides, we get

$\frac{(\sqrt{p})^{2}}{1}= \left ( \frac{m}{n} \right )^{2}$

$\Rightarrow \frac{p}{1} = \frac{m^{2}}{n^{2}}$

$\Rightarrow pn^{2} = m^{2}$

$\therefore$ P divides  m2 and p  divides m . [ $\because$ p divides pn2 ]

[$\because$ P is prime and p divides m2 $\Rightarrow$ p divides m]

Let              m = pq  for some integer q

on putting  m = pq  [in eq.(i) we get ]

pn2 = p2q2

n2 = pq2

p divides n2

and p divides n.

[ $\because$ p is prime and p divides n$\Rightarrow$ p divides n ]

Thus p is a common factor of m and n but this contraficts  the fact that m and n have no common factor other than 1.

The contradiction arises by assuming that $\sqrt{p}$ is a rational .

Hence, $\sqrt{p}$ is irrational.

## Prove that   $3+\sqrt{5}$ is an irrational number.

Let 3+$\sqrt{5}$ is  a rational number

$\therefore 3+\sqrt{5} = \frac{p}{q}, q\neq 0$

$3+\sqrt{5} = \frac{p}{q}$

$\Rightarrow \sqrt{5} = \frac{p}{q} -3$

$\Rightarrow \sqrt{5} = \frac{p-3q}{q}$

$\sqrt{5}$ is irrational  and   $\frac{p-3q}{q}$  is a rational

But the rational number cannot be equal to an irrational number.

$\therefore 3+\sqrt{5}$ is an irrational number.

## Prove that  $\sqrt{3}$  is an irrational number. Hence show that 7+2$\sqrt{3}$  is also an irrational number

If possible let $\sqrt{3}$  be a rational number.

(i)   $\therefore \frac{a}{b} = \sqrt{3}$, where a  and b are integers and co -primes

squaring both sides, we have

$\frac{ a^{2}}{b^{2}} = 3$

$\Rightarrow a^{2} = 3b^{2}$

$\Rightarrow a^{2}$ is divisible by 3

$\therefore$ a is divisible by 3    .............................(1)

We can write a = 3c for some c (integer)

(3c)2 = 3b2

$\Rightarrow 9c^{2} = 3b^{2}$

$\Rightarrow b^{2} = 3c^{2}$

$\Rightarrow b^{2}$ is divisible by  3.

$\Rightarrow b$ is divisible by 3   ...........................(2)

From eq (i) and (ii) we have,

3 is a factor a and b which is contradicting the fact that 'a' and 'b' are co-prime.

Thus our assumption that  $\sqrt{3}$  is rational numbers is wrong.

Hence, $\sqrt{3}$ is an irrational number.

(ii) Let us assume to the contrary that $7+2\sqrt{3}$  is a rational number.

$7+2\sqrt{3}=\frac{p}{q},\: q\: \neq 0\: and \: p\: ,\: q\epsilon \: Integer$

$7+2\sqrt{3}=\frac{p}{q}$

$2\sqrt{3}=\frac{p}{q}-7$

$2\sqrt{3}=\frac{p-7q}{q}$

$\sqrt{3}=\frac{p-7q}{2q}$

p- 7q and  2q both are integers hence $\sqrt{3}$  is a rational number.

But this contradicts the fact that is $\sqrt{3}$ is  irrational number. Hence is $7+2\sqrt{3}$ is an irrational number .

## Find the three rational numbers lying between 3 and 4.

LCM of 3 and 4 = 12

$3 = \frac{36}{12}$

$4 = \frac{48}{12}$

So rational nos. can be

$\frac{37}{12} , \, \, \frac{38}{12} ,\, \, \, \frac{39}{12}$

## Find the 6 rational numbers between $\frac{-3}{2}$ and   ​$\frac{5}{3}$

$\frac{-3}{2} = \frac{-3 \times 3}{2\times 3 } = \frac{-9}{6}$

$\frac{5}{3} = \frac{5 \times 2}{3\times 2 } = \frac{10}{6}$

The rational numbers are $\frac{-8}{6}\: ,\: \frac{-7}{6}\: ,\: \frac{-6}{6}\: ,\: \frac{-5}{6}\: ,\: \frac{-4}{6}\: ,\: \frac{-3}{6}\: ,\:$$\frac{-2}{6}\: ,\frac{-1}{6}\: ,\: \frac{1}{6}\: ,\: \frac{2}{6}\: ,\: \frac{3}{6}\: ,\: \frac{4}{6}\: ,\: \frac{5}{6}\: ,\: \frac{6}{6}$ , $\frac{7}{5}\: ,\frac{8}{6}\: ,\frac{9}{6}$

## Find 3 rational numbers between 1/4 and 1/2

$\frac{1}{4}=\frac{1\times 4}{4\times 4}=\frac{4}{16}$

$\frac{1}{2}=\frac{1\times 8}{2\times 8}=\frac{8}{16}$

3 rational numbers between $\frac{1}{4}$ and $\frac{1}{2}$ are $\frac{5}{16}\: ,\frac{6}{16}\: ,\frac{7}{16}$

1

13

19

35

## Prove that $\sqrt{5}$ is an irrational number and hence show that $2-\sqrt{5}$ is also an irrational number.

Let $\sqrt{5}$ be a rational number

$\therefore \sqrt{5} = \frac{a}{b}$

(a, b are co -prime integers and $b\neq 0$)

$\Rightarrow a = b \sqrt{5}$

$\Rightarrow a ^{2}= 5b^{2}$

$\Rightarrow$ 5 is a factor of $a^{2}$

$\Rightarrow$ 5 is a factor of a

let  a = 5c,

$\Rightarrow a^{2} = 25c^{2}$

$\Rightarrow 5b^{2} = 25c^{2}$

$\Rightarrow b^{2} = 5c^{2}$

$\Rightarrow$ 5 is a factor of $b^{2}$

$\Rightarrow$ 5 is a factor of b

5 is a common factor of a, b

But this contradicts the fact that a, b are co-primes

$\therefore \sqrt{5}$ is irrational

Let $2- \sqrt{5}$ be a rational

$\therefore 2- \sqrt{5} = a$

$\Rightarrow 2- a = \sqrt{5}$

2 - a is rational, so is $\sqrt{5}$

But $\sqrt{5}$ is not rational contradiction

$\therefore 2-\sqrt{5}$ is irrational.

## Prove that $\sqrt{2}$ is an irrational number. Hence show that $\frac{3}{\sqrt{2}}$ is alos an irrational number.

Let $\sqrt{2}$ be a rational number

$\sqrt{2} = \frac{a}{b}$

(a, b are co-prime integers and $b\neq 0$)

$a = \sqrt{2}b$

$a^{2} = 2b^{2}$

Squaring,

$\Rightarrow$ 2 divides a2

$\Rightarrow$ 2 divides a

So we can write a = 2c for some integer c,, substitute for $a, 2b^{2} = 4c^{2} , b^{2} = 2c^{2}$

This means 2 divides b2, so 2 divides b.

a and b have '2' as a common factor

But this contradicts that a, b have no commom factor other than 1.

Our assumption is wrong.

Hence, $\sqrt{2}$ is irrational.

Let $\frac{3}{ \sqrt{2}}$ be rational

$\frac{3}{ \sqrt{2}} = \frac{a}{b}$ where a and b are integers, $b \neq 0$

$\Rightarrow 3b = \sqrt{2a}$

$\Rightarrow \sqrt{2a} = \frac{3b}{a}$

$\frac{3b}{a}$ is rational but $\sqrt{2}$ is not rational

Our assumption is wrong

$\frac{3}{ \sqrt{2}}$ is rational.

## Show that there is no positive integer n, for which $\sqrt{n-1}+\sqrt{n+1}$ is rational.

Let us assume that there is a positive integer n for  which$\sqrt{ n-1} + \sqrt{ n + 1}$ is rational and equal to  $\frac{P}{Q}$,where  P and Q are positive integers ($Q\neq 0$)

$\sqrt{ n-1} + \sqrt{ n + 1} = \frac{P}{Q}$................(i)

$\Rightarrow \frac{P}{Q} = \frac{1}{\sqrt{n-1} + \sqrt{n+1}}$

$= \frac{\sqrt{n-1} -\sqrt{n+1}}{(\sqrt{n-1} + \sqrt{n+1}) (\sqrt{n-1} - \sqrt{n+1})}$

$= \frac{\sqrt{n-1} -\sqrt{n+1}}{(n-1) - (n+1) } = \frac{\sqrt{n-1} - \sqrt{n+1}}{-2}$

$\Rightarrow \sqrt{n +1} - \sqrt{n-1} = \frac{2Q}{P}$................(ii)

Apply (i) + (ii) we get

$2 \sqrt{n+1} = \frac{P}{Q} + \frac{2Q}{P} = \frac{P^{2} + 2Q^{2}}{PQ}$

$\Rightarrow \sqrt{n+1} = \frac{P^{2}+2Q^{2}}{2PQ}$.............(iii)

Apply (i) and (ii) we get

$\sqrt{n-1} = \frac{P^{2} + 2Q^{2} }{2PQ}$.................(iv)

From (iii) and (iv), we can say $\sqrt{n +1}$ and $\sqrt{n -1}$ both are rational because P and Q both are  rational. But it is possible only when (n+1) and (n-1)  both  are perfect squares. But they differ by 2 and two perfect square never differ by 2. So both (n+1) and (n-1) cannnot be perfect squares, hence there is no positive integer n for which $\sqrt{n -1} + \sqrt{n+1}$ is rational.