Question 1

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

Accepted Answer

Let us assume on the contrary that  \sqrt {3}  is a rational number. Then, there exist positive integers  a  and  b  such that \sqrt { 3 } =\dfrac { a }{ b }   where,  a  and  b , are co-prime i.e. their  HCF  is  1 Now, \sqrt { 3 } =\dfrac { a }{ b }  \Rightarrow \quad 3=\dfrac { { a }^{ 2 } }{ { b }^{ 2 } }    \Rightarrow \quad 3{ b }^{ 2 }={ a }^{ 2 }   \Rightarrow \quad 3  divides  { a }^{ 2 }\quad \quad \left[ \because 3 \ 	ext{divides}\ 3{ b }^{ 2 } ight]    \Rightarrow \quad 3  divides  a\quad \quad ...\left( i ight)    \Rightarrow \quad a=3c  for some integer  c \Rightarrow \quad { a }^{ 2 }=9{ c }^{ 2 }   \Rightarrow \quad 3{ b }^{ 2 }={ 9c }^{ 2 }\quad \quad \left[ \because { a }^{ 2 }=3{ b }^{ 2 } ight]    \Rightarrow \quad { b }^{ 2 }={ 3c }^{ 2 }   \Rightarrow \quad 3  divides  { b }^{ 2 }\quad \quad \left[ \because 3 \ 	ext{divides}\ 3{ c }^{ 2 } ight]    \Rightarrow \quad 3  divides  b\quad \quad ...\left( ii ight)   From  (i)  and  (ii),  we observe that  a  and  b  have at least  3  as a common factor. But, this contradicts the fact that  a  and  b  are co-prime. This means that our assumption is not correct.Hence,  \sqrt {3}  is an irrational number.

Question 2

Prove that  \sqrt {2}  is an irrational number.

Accepted Answer

Let us assume on the contrary that  \sqrt {2}  is a rational number. Then, there exist positive integers  a  and  b  such that \sqrt { 2 } =\dfrac { a }{ b }   where,  a  and  b , are co-prime i.e. their  HCF  is  1 \Rightarrow \quad { \left( \sqrt { 2 }  \right)  }^{ 2 }={ \left( \dfrac { a }{ b }  \right)  }^{ 2 }   \Rightarrow \quad 2=\dfrac { { a }^{ 2 } }{ { b }^{ 2 } }    \Rightarrow \quad 2{ b }^{ 2 }={ a }^{ 2 }   \Rightarrow \quad 2|{ a }^{ 2 }\quad \quad \left[ \because 2|2{ b }^{ 2 }\ and\ 2{ b }^{ 2 }={ a }^{ 2 } \right]    \Rightarrow \quad 2|a\quad \quad ...\left( i \right)    \Rightarrow \quad a=2c  for some integer  c \Rightarrow \quad { a }^{ 2 }=4{ c }^{ 2 }   \Rightarrow \quad 2{ b }^{ 2 }={ 4c }^{ 2 }\quad \quad \left[ \because 2{ b }^{ 2 }={ a }^{ 2 } \right]    \Rightarrow \quad { b }^{ 2 }={ 2c }^{ 2 }   \Rightarrow \quad 2|{ b }^{ 2 }\quad \quad \left[ \because 2|2{ c }^{ 2 } \right]    \Rightarrow \quad 2|b\quad \quad ...\left( ii \right)  From  (i)  and  (ii),  we obtain that  2  is a common factor of  a  and  b . But, this contradicts the fact that  a  and  b  have no common factor other than  1 . This means that our supposition is wrong.Hence,  \sqrt {2}  is an irrational number.

Question 3

Prove that  \sqrt 5  is irrational by the method of Contradiction.

Accepted Answer

Let  \sqrt5  be a rational number.then it must be in form of   \dfrac pq   where,   q
eq0      (  p  and  q  are co-prime) \sqrt5=\dfrac pq \sqrt5 	imes q=p Suaring on both sides, 5{q}^2{}={p}^{2}            --------------(1) {p}^{2}  is divisible by  5 .So,  p  is divisible by  5 . p=5c Suaring on both sides, {p}^{2}=25{c}^{2}          --------------(2)Put  {p}^{2}  in eqn.(1) 5{q}^{2}=25(c)^{2} {q}^{2}=5 c^{2} So,  q  is divisible by  5 ..Thus  p  and  q  have a common factor of  5 .So, there is a contradiction as per our assumption.We have assumed  p  and  q  are co-prime but here they a common factor of  5 .The above statement contradicts our assumption.Therefore,  \sqrt5  is an irrational number.

Question 4

Prove that  \sqrt {7}  is an irrational number.

Accepted Answer

Let us assume that  \sqrt{7}  is rational. Then, there exist co-prime positive integers  a  and  b  such that \sqrt{7} = \dfrac{a}{b} \implies a = b\sqrt{7} Squaring on both sides, we get a^2 = 7b^2 Therefore,  a^2  is divisible by  7  and hence,  a  is also divisible by 7 so, we can write  a = 7p,  for some integer  p. Substituting for  a , we get  49p^2 = 7b^2 \implies b^2 = 7p^2. This means,  b^2   is also divisible by  7  and so,  b  is also divisible by  7 .Therefore,  a  and  b  have at least one common factor, i.e.,  7 .But, this contradicts the fact that  a  and  b  are co-prime.Thus, our supposition is wrong.Hence,  \sqrt{7}   is irrational.

Question 5

In a seminar, the number of participants in Hindi, English and Mathematics are  60,\ 84  and  108  respectively. Find the minimum number of rooms required if in each room the same number of participants are to be seated and all of them being in the same subject.

Accepted Answer

The Number of room will be minimum if each room accomodates maximum number of participants. Since in each room the same number of participants are to be seated and all of them must be of the same subject. Therefore, the number of participants in each room must be the  HCF  of  60,84  and  108 . The prime factorisations of  60,84  and  108  are as under. 60={2}^{2}	imes 3	imes 5,84={2}^{2}	imes 3	imes 7  and  108={2}^{2}	imes {3}^{3} 	herefore\ HCF  of  60,84  and  108  is  {2}^{2}	imes 3=12 Therefore, in each room  12  participants can be seated. 	herefore\ Number\ of\ rooms\ required =\dfrac { Total\ number\ of\ participants  }{ 12 }                                                            =\dfrac { 60+84+108 }{ 12 } =\dfrac { 252 }{ 12 } =21

Question 6

Prove that  \sqrt 3  + \sqrt 5   is irrational.

Accepted Answer

To prove :  \sqrt{3}+\sqrt{5}   is irrational.Let us assume it to be a rational number. Rational numbers are the ones that can be expressed in  \dfrac p q  form where  p,q  are integers and  q  isn't equal to zero. \sqrt{3}+\sqrt{5}=\dfrac p q\ \sqrt{3}=\dfrac p q-\sqrt{5}\ squaring on both sides,  3= \dfrac {p^{2}}{q^{2}}-2.\sqrt{5} \left(\dfrac {p} {q}ight)+5 \Rightarrow \dfrac{(2\sqrt{5}p)}{q}=5-3+\left(\dfrac {p^2} {q^2}ight)   \Rightarrow \dfrac{(2\sqrt{5}p)}{q}= \dfrac {2q^{2}- {p^2}} {q^2} \Rightarrow \sqrt{5}= \dfrac {2q^{2}- {p^2}} {q^2}. \dfrac {q}{2p}\ \Rightarrow \sqrt{5}=\dfrac {(2q^{2}-p^{2})} {2pq}\ As  p  and  q  are integers RHS is also rational.As RHS is rational LHS is also rational i.e  \sqrt{5}  is  rational.But this contradicts the fact that  \sqrt{5}  is irrational.This contradiction arose because of our false assumption.so, \sqrt{3}+\sqrt{5}  irrational.

Question 7

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:(i)  \dfrac{13}{3125}  (ii)  \dfrac{17}{8}  (iii)  \dfrac{64}{455}  (iv)  \dfrac{15}{1600}  (v)  \dfrac{29}{343}  (vi)  \dfrac{23}{2^{3}5^{2}}  (vii)  \dfrac{129}{2^{2}5^{7}7^{5}}  (viii)  \dfrac{6}{15}  (ix)  \dfrac{35}{50}  (x)  \dfrac{77}{210}

Accepted Answer

Theorem: Let  x = \dfrac{p}{q}  be a rational number, such that the prime factorisation of q is of the form  2^{n}5^{m} , where n, m are non-negative integers. Then,  x  has a decimal expansion which terminates.(i)  \dfrac{13}{3125} Factorise the denominator, we get 3125 = 5 	imes 5  	imes 5  	imes 5  	imes 5 = 5^5 So, denominator is in form of  5^m  so,  \dfrac{13}{3125}  is terminating.(ii)  \dfrac{17}{8} Factorise the denominator, we get 8 = 2  	imes 2  	imes 2 = 2^3 So, denominator is in form of  2^n  so,  \dfrac{17}{8}  is terminating.(iii)  \dfrac{64}{455} Factorise the denominator, we get 455 = 5  	imes 7  	imes 13  So, denominator is not in form of  2^{n} 5^{m}  so,  \dfrac{64}{455}  is not terminating.(iv)  \dfrac{15}{1600} Factorise the denominator, we get 1600 = 2  	imes 2  	imes 2  	imes 2  	imes 2  	imes 2  	imes 5  	imes 5 = 2^{6} 5^{2} So, denominator is in form of  2^{n}5^{m}  so,  \dfrac{15}{1600}  is terminating.(v)  \dfrac{29}{343} Factorise the denominator, we get 343 = 7  	imes 7  	imes 7 = 7^{3} So, denominator is not in form of  2^{n}5^{m}  so,  \dfrac{29}{343}  is not terminating.(vi)  \dfrac{23}{2^{3}5^{2}} Here, the denominator is in form of  2^{n}5^{m}  so,  \dfrac{23}{2^{3}5^{2}}  is terminating.(vii)  \dfrac{129}{2^{2}5^{7}7^{5}} Here, the denominator is not in form of  2^{n}5^{m}  so,  \dfrac{129}{2^{2}5^{7}7^{5}}  is not terminating.(viii)  \dfrac{6}{15} Divide nominator and denominator both by 3 we get  \dfrac{3}{15} So, denominator is in form of   5^{m}  so,  \dfrac{6}{15}  is terminating.(ix)  \dfrac{35}{50}  Divide nominator and denominator both by 5 we get  \dfrac{7}{10} Factorise the denominator, we get 10 = 2 	imes 5 So, denominator is in form of  2^{n}5^{m}  so,  \dfrac{35}{50}  is terminating.(x)  \dfrac{77}{210} Divide nominator and denominator both by 7 we get  \dfrac{11}{30} Factorise the denominator, we get 30 = 2 	imes 3 	imes 5 So, denominator is not in the form of  2^{n}5^{m}  so  \dfrac{6}{15}  is not terminating.

Question 8

Prove that the following are irrational:(i)  \dfrac{1}{\sqrt{2}}  (ii)  7\sqrt{5}  (iii)  6+\sqrt{2}

Accepted Answer

(i)  \dfrac{1}{\sqrt{2}} Let us assume  \dfrac{1}{\sqrt{2}}  is rational.So we can write this number as \dfrac{1}{\sqrt{2}} = \dfrac{a}{b}  ---- (1)Here,  a  and  b  are two co-prime numbers and  b  is not equal to zero.Simplify the equation (1) multiply by  \sqrt{2}  both sides, we get 1 = \dfrac{a\sqrt{2}}{b} Now, divide by  b , we get b = a\sqrt{2}  or  \dfrac{b}{a}= \sqrt{2} Here,  a  and  b  are integers so,  \dfrac{b}{a}  is a rational number, so  \sqrt{2}  should be a rational number.But  \sqrt{2}  is a irrational number, so it is contradictory.Therefore,  \dfrac{1}{\sqrt{2}}  is irrational number.(ii)  7\sqrt{5} Let us assume  7\sqrt{5}  is rational.So, we can write this number as 7\sqrt{5} = \dfrac{a}{b}  ---- (1)Here,  a  and  b  are two co-prime numbers and  b  is not equal to zero.Simplify the equation (1) divide by 7 both sides, we get \sqrt{5} = \dfrac{a}{7b} Here,  a  and  b  are integers, so  \dfrac{a}{7b}  is a rational number, so  \sqrt{5}  should be a rational number.But  \sqrt{5}  is a irrational number, so it is contradictory.Therefore,  7\sqrt{5}  is irrational number.(iii)  6+\sqrt{2} Let us assume  6+\sqrt{2}  is rational.So we can write this number as 6+\sqrt{2} = \dfrac{a}{b}  ---- (1)Here,  a  and  b  are two co-prime number and  b  is not equal to zero.Simplify the equation (1) subtract 6 on both sides, we get \sqrt{2} = \dfrac{a}{b}-6 \sqrt{2} = \dfrac{a-6b}{b} Here,  a  and  b  are integers so,  \dfrac{a-6b}{b}  is a rational number, so  \sqrt{2}  should be a rational number.But  \sqrt{2}  is a irrational number, so it is contradictory.Therefore,  6+\sqrt{2}  is irrational number.

Question 9

The decimal expansion of the rational number  \displaystyle\frac{23}{2^{3}5^{2}} , will terminate after how many places of decimal?

Accepted Answer

\displaystyle\frac{23}{2^{3}5^{2}}   =  \displaystyle\frac{23}{2(2)^{2}(5)^{2}}  =  \displaystyle\frac{23}{2(2	imes5)^{2}}  =   \displaystyle\frac{23}{2(10)^{2}}  =   \displaystyle\frac{11.5}{100} = 0.115 Therefore, the decimal  expansion of the rational number  \displaystyle\frac{23}{2^{3}5^{2}} will terminate after three places of decimal.

Question 10

Prove that, if x and y are odd positive integers, then  x^2 + y^2  is even but not divisible by 4.

Accepted Answer

We know that any odd positive integer is of the form  2q + 1 , where  q  is an integer. So, let  x = 2m + 1  and  y = 2n + 1 , for some integers m and n. we have  x^2 + y^2 x^2 + y^2 = (2m + 1)^2 + (2n + 1)^2 x^2 + y^2 = 4m^2 + 1 + 4m + 4n^2 + 1 + 4n = 4m^2 + 4n^2 + 4m + 4n + 2 x^2+ y^2 = 4(m^2 + n^2) + 4(m + n) + 2 = 4\{(m^2 + n^2) + (m + n)\}+ 2 x^2 + y2 = 4q + 2 , when  q = (m^2 + n^2)+(m + n) x^2 + y^2  is even and leaves remainder 2 when divided by 4.  x^2 + y^2  is even but not divisible by 4.