Q: Which of the following is not a rational number?

\sqrt { 2 } =1.4142135623730951.... \sqrt { 4 } =\sqrt { 2\times2 } =2 \sqrt { 9 } =\sqrt { 3\times3 } =3 \sqrt { 16 } =\sqrt { 4\times4 } =4 As we can see the decimal representation of \sqrt { 2 } is non-terminating non-repeating.Hence, \sqrt { 2 } is an irrational number.Therefore, option A is correct.

Question 1

State true or false.Five rational numbers between. \dfrac{2}{3}  and  \dfrac{4}{5}  are  \dfrac{41}{60},\dfrac{42}{60},\dfrac{43}{60},\dfrac{44}{60},\dfrac{45}{60}

Accepted Answer

To get the rational numbers between  \displaystyle\frac{2}{3}  and  \displaystyle\frac{4}{5} Take an LCM of these two numbers:  \displaystyle\frac{10}{15}  and  \displaystyle\frac{12}{15} Multiply numerator and denominator by 4:  \displaystyle\frac{40}{60}  and  \displaystyle\frac{48}{60} All the numbers between  \displaystyle\frac{40}{60}  and  \displaystyle\frac{48}{60}  form the answerSome of these numbers are  \displaystyle\frac{41}{60} ,  \displaystyle\frac{42}{60} ,  \displaystyle\frac{43}{60} ,  \displaystyle\frac{44}{60} ,  \displaystyle\frac{45}{60} Hence the statement is true

Question 2

State whether the given statement is true or false:Five rational numbers between  \dfrac{-3}{2}  and  \dfrac{5}{3}  are  \dfrac{-8}{6},\dfrac{-7}{6},0,\dfrac{1}{6},\dfrac{2}{6} .

Accepted Answer

Given rational numbers are  \displaystyle\frac{-3}{2}  and  \displaystyle\frac{5}{3} LCM of the denominators is  6  Hence  \displaystyle\frac{-3}{2}  and  \displaystyle\frac{5}{3}  can be written as  \displaystyle\frac{-9}{6}  and  \displaystyle\frac{10}{6} The numbers between  \displaystyle\frac{-9}{6}  and  \displaystyle\frac{10}{6}  are  \displaystyle\frac{-8}{6} ,  \displaystyle\frac{-7}{6} ,  \displaystyle{0} ,  \displaystyle\frac{1}{6} ,  \displaystyle\frac{2}{6},... Hence the statement is true.

Question 3

State True or False. \sqrt{4}  is an irrational number.

Accepted Answer

We know that, a rational number is a real number that can be represented in the form of  \dfrac pq  where both  p  and  q  are co-prime.Given number is  \sqrt 4 \Rightarrow \sqrt4=2 We can write  2  as  \dfrac21  where both  2  and  1  are co-prime.So,  \sqrt4  is a rational number.Hence, the given statement is false.

Question 4

Which of the following is not a rational number?

Accepted Answer

\sqrt { 2 } =1.4142135623730951....  \sqrt { 4 } =\sqrt { 2	imes2 } =2  \sqrt { 9 } =\sqrt { 3	imes3 } =3  \sqrt { 16 } =\sqrt { 4	imes4 } =4 As we can see the decimal representation of  \sqrt { 2 }  is non-terminating non-repeating.Hence,  \sqrt { 2 }   is an irrational number.Therefore, option  A  is correct.

Question 5

Read each of the following numbers carefully- 87, 54, 0, -13, -4.7, \sqrt{7}, 2{1}{7}, \sqrt{15}, -{8}{7}, 3\sqrt{7}, 4.807, 0.002, \sqrt{16}  and  2+\sqrt{3}. From the given numbers, write which are real numbers.

Accepted Answer

the real number includes(1) Whole numbers ( 0,1,2,3,4....etc)(2) Rational numbers  ( \frac{3}{4},0.125,0333 etc )(3) Irrational numbers ( \pi ,\sqrt{3},2+\sqrt{3}etc )Then all the number given in question are Real numbersAnswer is (D) All of above are real numbers

Question 6

Question 7

\dfrac{35}{50}  has a non-terminating decimal expansion.

Accepted Answer

Let us write the denominator of the given fraction as a product of its prime factors, 50=2	imes5	imes5        =2	imes5^2 So, we can observe that the denominator of the given fraction is of the for  2^m 	imes 5^n  so, fraction  \dfrac{35}{50}  will have terminating decimal expansion.Hence, the given statement is false.

Question 8

State the following statement is True or False \dfrac{987}{10500}  will have terminating decimal expansion.

Accepted Answer

Terminating decimal expansion, because   \dfrac{987}{10500}=\dfrac{329}{3500}=\dfrac{329}{2^2.5^3.7}=\dfrac{47}{2^2.5^3}

Question 9

Write whether every positive integer can be of the form  4q + 2 , where  q  is an integer.

Accepted Answer

No, all positive integers cannot be written in the form of   4q+2 Because,  4q+2=2\left( 2q+1 \right)  Therefore,  4q+2  is an even number.So, we can't write odd numbers in the form  4q+2  (where  q  is an integer)Also  2q+1  is an odd number, hence the maximum power of  2  that divides  4q+1  is  1 . Therefore, we can't represent the numbers divisible by  4   this form.

Question 10

Say true or false:A positive integer is of the form 3q + 1, q being a natural number, then you write its square in any form other than  3m + 1, i.e.,3m or 3m + 2  for some integer m.

Accepted Answer

Let the positive integer  n  is of the form  3q, 3q+1,  and   3q+2 If  n=3q Squaring both sides, we get,     =>{ n }^{ 2 }=9{ q }^{ 2 }      =>{ n }^{ 2 }=3\left( { 3q }^{ 2 } \right)       =>{ n }^{ 2 }=3m , where  m=3{ q }^{ 2 } Now, if  n=3q+1      =>{ n }^{ 2 }={ \left( 3q+1 \right)  }^{ 2 }      =>{ n }^{ 2 }=9{ q }^{ 2 }+6q+1      =>{ n }^{ 2 }={ 3q\left( 3q+2 \right)  }+1      =>{ n }^{ 2 }=3m+1 ,  where    m=q\left( 3q+2 \right)  Now, if  n=3q+2      =>{ n }^{ 2 }={ \left( 3q+2 \right)  }^{ 2 }      =>{ n }^{ 2 }=9{ q }^{ 2 }+12q+4      =>{ n }^{ 2 }={ 3q\left( 3q+4 \right)  }+4      =>{ n }^{ 2 }={ 3q\left( 3q+4 \right)  }+3+1      =>{ n }^{ 2 }=3m+1  where  m=\left( 3{ q }^{ 2 }+4q+1 \right)  Hence,  { n }^{ 2 }  integer is of the form  3m  and  3m+1  not  3m+2

Question 11

Use Euclid's division lemma to find the HCF of the following27727 and 53124

Accepted Answer

Here,  a=53124  and  b=27727 by using Euclid's division lemma, a=bq+r  ;  0<r<b 53124=(27727)(1)+25397 27727=(25397)(1)+2330 25397=(2330)(10)+2097 2330=(2097)(1)+233 2097=(233)(9)+0 H.C.F is  233

Question 12

Use Euclid's division lemma to find the  HCF  of the following numbers: 8068  and  12464

Accepted Answer

Here,  a=12464  and  b=8068 So, by using Euclid's division lemma, a=bq+r  ;  0<r<b \begin{aligned}{}12464 &= (8068)(1) + 4396\8068 &= (4396)(1) + 3672\4396& = (3672)(1) + 724\3672 &= (724)(5) + 52\724& = (52)(13) + 48\52 &= (48)(1) + 4\48 &= (4)(12) + 0\end{aligned} Hence,  HCF  of  8068  and  12464  is equal to  4.

Question 13

Some factor is missing in the factor tree. Fill it up.

Accepted Answer

56 = 7	imes \underline {8} = 7	imes \underline {2	imes 4} = 7	imes 2	imes \underline {2	imes 2} The missing factor is  4. Therefore ,   A  is the correct answer.

Question 14

Is  (2+\sqrt{3})  a rational number ?

Accepted Answer

Since,  \sqrt{3}  is an irrational number and we know, sum of a rational and irrational number is irrational. Hence,  (2+\sqrt{3})  is not a rational number, i.e., it is irrational.

Real Numbers