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

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

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

0.overline{ 1315} = 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.

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}

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

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

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

                                                    = 5sqrt{3}+3sqrt{5

Alternative Method

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

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

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

                      = 5sqrt{3 }+3sqrt{5}


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+2sqrt{6})-4}

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

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




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

Given x: = 3-2sqrt{2}

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


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


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


x^{2}= (3-2sqrt{2})^{2}=9+8-12sqrt{2}=17-12sqrt{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 [ (17-12sqrt{2})-(17+12sqrt{2}) right ] left [ (17-12sqrt{2}) +(17+12sqrt{2})right ]

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

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

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


Let  5sqrt{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



Rightarrow sqrt{6}= rational 

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

Thus, our assumption is wrong. 

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


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 nneq 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 nRightarrow 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}, qneq 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+2sqrt{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+2sqrt{3}  is a rational number.

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





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+2sqrt{3} is an irrational number .


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 bneq 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 bneq 0)

a = sqrt{2}b

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


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  whichsqrt{ n-1} + sqrt{ n + 1} is rational and equal to  frac{P}{Q},where  P and Q are positive integers (Qneq 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.