NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Excercise 1.1
Question 1. Is zero a rational number? Can you write it in the form `\frac{p}{q}`,where p and q are integers and q ≠0?
Solution: Yes, zero is a rational number it can be written in the form `\frac{p}{q}`.
0 = `\frac{0}{1}=\frac{0}{2}=\frac{0}{3}` etc. denominator q can also be taken as negative integer.
Question 2. Find six rational numbers between 3 and 4.
Solution: Let qi be the rational number between 3
and 4, where j = 1 to 6.
∴ Six rational numbers are as follows:
q1=`\frac{3+4}{2}=\frac{7}{2};3<\frac{7}{2}<4`
q2=`\frac{3+ \frac{7}{2} }{2}= \frac{ \frac{13}{2}}{2}= \frac{13}{4};3< \frac{13}{4}< \frac{7}{2}<4`
q3=`\frac{4+ \frac{7}{2} }{2}= \frac{ \frac{15}{2}}{2}= \frac{15}{4};3< \frac{13}{4}< \frac{7}{2}< \frac{15}{4} <4`
q4=`\frac{ \frac{7}{2}+ \frac{13}{4} }{2}= \frac{ \frac{14+13}{4}}{2}= \frac{ \frac{27}{4}}{2}= \frac{27}{8};3< \frac{13}{4} < \frac{27}{8}< \frac{7}{2}< \frac{15}{4}<4`
q5= `\frac{1}{2}(\frac{7}{2}+ \frac{15}{4}) = \frac{1}{2} ( \frac{14+15}{4} )=\frac{29}{8};3< \frac{13}{4} < \frac{27}{8}< \frac{7}{2} < \frac{29}{8}< \frac{15}{4}<4`
q6= `\frac{1}{2}(\frac{13}{4}+ \frac{27}{8}) = \frac{1}{2} ( \frac{26+27}{8} )=\frac{53}{16};3< \frac{13}{4} < \frac{53}{16} < \frac{27}{8} <\frac{7}{2}< \frac{29}{8}< \frac{15}{4}<4`
Thus, the six rational numbers between 3 and 4 are
`\frac{7}{2},\frac{13}{4},\frac{15}{4},\frac{27}{8},\frac{29}{8}and\frac{53}{16}`
Question 3.Find five rational numbers between `\frac{3}{5}` and `\frac{4}{5}`.
Solution:Since, we need to find five rational numbers, therefore, multiply numerator and denominator by 6.
∴ `\frac{3}{5}= \frac{3×6}{5×6}= \frac{18}{30}and \frac{4}{5}= \frac{4×6}{5×6}= \frac{24}{30}`
∴ Five rational numbers between `\frac{3}{5}and\frac{4}{5}are\frac{19}{30},\frac{20}{30},\frac{21}{30},\frac{22}{30},\frac{23}{30}`
Question 4. State whether the following statements are true or false. Give reasons for your answers.
(i) Every natural number is a whole number.
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.
Solution:(i) True
∵ The collection of all natural numbers and 0 is called whole numbers.
(ii) False
∵ Negative integers are not whole numbers.
(iii) False
∵ Rational numbers are of the form `\frac{p}{q}`, q ≠ 0 and q does not divide p completely that are not whole numbers.

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2

Question 1. State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
(ii) Every point on the number line is of the form √m , where m is a natural number.
(iii) Every real number is an irrational number.
Solution:(i) True
Because all rational numbers and all irrational numbers form the group (collection) of real numbers.
(ii) False
Because negative numbers cannot be the square root of any natural number.
(iii) False
Because rational numbers are also a part of real numbers.
Question 2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
Solution: No, if we take a positive integer, say 9, its square root is 3, which is a rational number.
Question 3. Show how √5 can be represented on the number line.
Solution: Draw a number line and take point O and A on it such that OA = 1 unit. Draw BA ⊥ OA as BA = 1 unit. Join OB = √2 units.
Now draw BB₁ ⊥ OB such that BB₁ =1 unit. Join OB₁ = √3 units.
Next, draw B₁B₂⊥ OB₁such that B₁B₂ = 1 unit.
Join OB₂ = units.
Again draw B₂B₃ ⊥OB₂ such that B₂B₃ = 1 unit.
Join OB₃ = √5 units.
Take O as centre and OB₃ as radius, draw an arc which cuts the number line at D.
Point D
represents √5 on the number line.

Question 4. Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP₁, of unit lengths Draw a line segment P₁, P₂ perpendicular to OP₁ of unit length (see figure). Now, draw a line segment P₂P₃ perpendicular to OP₂. Then draw a line segment P₃P₄ perpendicular to OP₃. Continuing in this manner, you can get the line segment P_n-1 P_n by drawing a line segment of unit length perpendicular to OP_{n – 1}. In this manner, you will have created the points P₂, P₃,…… P_n,….. and joined them to create a beautiful spiral depicting √2,√3,√4,……

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3

Question 1. Write the following in decimal form and say what kind of decimal expansion each has
`(i) \frac{36}{100} (ii) \frac{1}{11} (iii) 4\frac{1}{8} (iv) \frac{3}{13} (v) \frac{2}{11} (vi) \frac{329}{400}`
Solution:(i) We have, `\frac{36}{100}` = 0.36
Thus, the decimal expansion of `\frac{36}{100}` is terminating.
`\begin{array}{r}
\frac{536}{23) 12345} \\
\frac{115}{84} \\
\frac{69}{155} \\
\frac{138}{17}
\end{array}`
#\hspace{11.5mm}536#
#23\overline{)12345}#
#\hspace{7mm}\underline{115}#
#\hspace{11.5mm}84#
#\hspace{11.5mm}\underline{69}#
#\hspace{11.5mm}155#
#\hspace{11.5mm}\underline{138}#
#\hspace{14mm}17#
$\hspace{11.5mm}536$
$23\overline{)12345}$
$\hspace{7mm}\underline{115}$
$\hspace{11.5mm}84$
$\hspace{11.5mm}\underline{69}$
$\hspace{11.5mm}155$
$\hspace{11.5mm}\underline{138}$
$\hspace{14mm}17$
 
##\hspace{11.5mm}536##
##23\overline{)12345}##
##\hspace{7mm}\underline{115}##
##\hspace{11.5mm}84##
##\hspace{11.5mm}\underline{69}##
##\hspace{11.5mm}155##
##\hspace{11.5mm}\underline{138}##
##\hspace{14mm}17##